ise option traders statistics with applications to options march 2004copyright (c) 2001-2004 by...

March 2004 Copyright (c) 2001-2004 by Marshall, Tucker & Associates, LLC All rights reserved. 1

ISE Option Traders

Statistics with Applications to Options

Alan L. Tucker, Ph.D.631-331-8024 (tel)631-331-8044 (fax)

[email protected]

www.mtaglobal.com



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Alan L. Tucker, Ph.D.

Alan Tucker was the Founding Editor of the Journal of Financial Engineering and is currently on the editorial boards of the Journal of Derivatives and Global Finance Journal. He is the author of three textbooks and numerous articles appearing in such journals as the Journal of Finance, Review of Economics and Statistics, Journal of Financial and Quantitative Analysis, the Virginia Tax Review, and others. Dr. Tucker is a tenured professor of finance at the Lubin School of Business, Pace University, and an adjunct professor of finance at the Stern School of Business, New York University. As a consultant, he has worked for the US Treasury, the US Department of Justice, JP Morgan, Morgan Stanley, UBS, Deutsche Bank, TIAA-CREF, Lazard, Merrill Lynch, LG Securities, and others.


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Purpose:

The purpose of this presentation is refresh registrants about a number of key statistical concepts.

Our approach is to impart said concepts within the context of a specific application germane to option traders, namely, the computation of a

trader’s Value-at-Risk. Other applications are also addressed.


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What is Value-at-Risk?

Value-at-risk (abbreviated either VAR or VaR) is a measure of risk that boils risk down to a single easy-to-understand number.

Simply put, an option trader’s VAR is the maximum number of dollars that the trader might lose over a specified period of time (called the risk horizon) at a specified level of confidence.

For example:

Suppose that a trader’s one-day VAR is $25,000 at a 95% level of confidence. This says that the trader expects to lose no more than $25,000 over a one-day period 95 out of each 100 days. Of course, this also implies that the trader can expect to lose more than $25,000 in a single day five days out of each 100 days.


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How is VAR measured?

A trader’s VAR can be measured for different risk horizons and different levels of confidence.

It can be measured for a single trader, a single group of traders (a desk), a single division (group of desks), or for the trading firm as a whole.

VAR is used as a risk monitoring tool but also as a risk management tool.

For example, if the trader’s VAR turns out to be $35,000 and management believes that a VAR above $25,000 is excessive, it can order a reduction in the level of risk bearing to no more than a VAR of $25,000.


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Management can set VAR limits for each trader, for each desk, for each division and for the firm as a whole.

Thus, it is possible to monitor each trader’s VAR, aggregate these to a desk VAR, and so forth. Presumably, if an option trader’s Greeks (delta, gamma and vega) are within their limits, said trader’s VAR will be acceptable.

The aggregation of VARs is not linear. For example if a desk consists of two traders and each trader has a VAR of $30,000, the desk VAR will likely be less than $60,000. Indeed, it is possible that the desk VAR is less than $30,000 and could even be zero!

Additionally, VAR is not linear with respect to risk horizons. For example, if a trader’s one-day VAR is $10,000, his or her two-day VAR would likely be less than $20,000.


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Our Goals:

To get a good grasp on the basic statistical concepts associated with measuring and interpreting an option trader’s VAR;

to understand some of the different ways that VAR can be measured;

to understand the aggregation properties and the horizon properties of VAR;

to appreciate other applications of basic statistical concepts that are of interest to option traders.


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Basic Statistical Concepts


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The Basics: Statistical Concepts and Risk Concepts

Risk measures, including VAR, are based on statistical concepts and tools. We therefore need to develop the more important of these concepts and tools. Later, we will see how they are applied to measure VAR. The concepts we will need are:

random variableprobabilityprobability distributioncumulative probability distributionmeanvariancestandard deviationcovariancecorrelation


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Random Variable:

In the language of statistics, a random variable is anything that will take on a numeric value at some point in the future where the precise value that it will take on is not presently known. At the same time, we might know the probabilities associated with the different values the random variable can take on.

We have to distinguish a random variable from a number. Once the random variable takes on a numeric value it is no longer a random variable. It is now a number.

We begin with the simplest type of random variable.


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An experiment with a die:

If I throw a die, there are six possible numeric values that could result. Each of these values has an associated probability.

As the die is rolling, it is a random variable. Once it stops, it is a number. The number that results is called the “outcome.”

What are the values and the associated probabilities that this random variable might take on?


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A probability distribution: The complete set of values the random variable might take on together with their associated probabilities. This is sometimes called a probability density function (pdf).

probabilities

values1 2 3 4 5 6

1/6


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A cumulative probability distribution: The cumulative probability that the value of a random variable will lie at or below each point.

cumulative probabilities

values1 2 3 4 5 6

1/6

2/6

3/6

4/6

5/6

6/6


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Mean: The mean is a measure of the center of a distribution in the sense that 50% of the outcomes (weighted by their probabilities) will be at or above this central value and 50% will be at or below this value.

Where is the center of this distribution?

probabilities

values1 2 3 4 5 6

1/6

The center, or mean, of the distribution is sometimes called the expected value.It is also called the first moment of the distribution.


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Calculating a mean:

The calculation of a mean depends on whether we know the true probabilities associated with the random variable.

For example, in the case of the die, if I tell you the die is “fair,” I am telling you that the probability associated with each individual outcome is 1/6. But it is also possible that I don’t know if the die is fair (that is, certain outcomes might have a higher probability and others a lower probability).

The calculation of the mean of a random variable depends on whether we do or do not know the probabilities.


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If we do know the true probabilities:

Denote the random variable by X and the different values the random variable might take on by Xi . If we do know the true probabilities associated with the outcomes the random variable might take on, then we can calculate the “true” mean (sometimes called a population mean). The mean is denoted by the Greek letter and calculated as follows:

N

= Xi × Prob[Xi] i=1

For example, in the case of the die, the mean would be computed as:

= (1 × 1/6) + (2 × 1/6) + (3 × 1/6) + (4 × 1/6) + (5 × 1/6) + (6 × 1/6)

= 3.5


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If we do not know the true probabilities:

If we do not know the true probabilities, we cannot know the true mean. But, we can formulate an estimate of it by conducting an experiment. Such an estimate is called a sample mean.

The estimation process can also generate estimates of the probabilities of each possible outcome. For example, we could roll the die 600 times. Denote the kth outcome of this test by Ok, and the number of times the distribution is sampled by Z. Then the sample mean is given by:

Z

Ok k=1

= Z


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We tossed the die 600 times. We count the number of times each of the six possible outcomes occurred and we plot them. This is called a histogram.

number of times

1 2 3 4 5 6

103 10690 102 95 104


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From the histogram we can estimate the probabilities of each of the six possible outcomes. For example, the estimated probability of getting a 1 is (103/600). The estimated probability of getting a 2 is (106/600). This implies certain assumptions. What are they?

number of times

1 2 3 4 5 6

103 10690 102 95 104


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From this we can see that we can get the sample mean by adding up all the outcomes and dividing by 600. Or, we can multiply each of the six possible outcomes by their associated estimated probabilities. In either case we get 3.48667.

number of times

1 2 3 4 5 6

103 10690 102 95 104


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Variance: Another important measure associated with a probability distribution is called the variance. The variance, and a related measure called the standard deviation, are both measures of dispersion. By measures of dispersion, we mean measures of how far away from the mean a single outcome drawn at random from the distribution (e.g., throw the die once) might be.

Calculating a Variance:

The calculation of a variance depends on whether we know the true probabilities associated with the outcomes of the random variable. If we do, we can calculate a true variance (sometimes called a population variance) If we don’t, the best we can do is calculate a sample variance.

Note: Variance is also called the second moment of the distribution.


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If we know the true probabilities:

The variance of a random variable is usually denoted 2. To calculate a true variance from the known probabilities, we employ the following formula:

N

2 = (Xi - )2 × Prob[Xi]

i=1

Where X denotes the random variable, Xi denotes the ith possible outcome, denotes the mean of the random variable (assumed to have already been calculated), and N is the number of different values that the random variable can take on.


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N

2 = (Xi - )2 × Prob[Xi]

i=1

Let’s use the variance formula above to calculate the variance associated with a fair die. Again, there are only six possible outcomes (1,2,3,4,5,6) and each has a probability of 1/6.

2 = [(1 - 3.5)2 × 1/6] + [(2 - 3.5)2 × 1/6] + [(3 - 3.5)2 × 1/6]

+ [(4 - 3.5)2 × 1/6] + [(5 - 3.5)2 × 1/6] + [(6 - 3.5)2 × 1/6]

= 2.91667


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If we do not know the true probabilities:

If we do not know the true probabilities, we cannot know the true variance. But, we can formulate an estimate of it by conducting an experiment. Such an estimate is called a sample variance. We will denote the sample variance by 2.

Z

(Ok - )2

k=1

2 = Z

Where Z is the number of times the sampling is repeated, and Ok is the outcome from the kth sampling.


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Example:

Consider again the die that we did not know was fair. We tossed the die 600 times. From the 600 outcomes, we calculated a sample mean of 3.48667. We now use this sample mean to help get the sample variance.

The sample variance formula produces the following result:

[103 × (1 - 3.48667)2] +[106 × (2 - 3.48667)2] + ….+ [104 × (6 - 3.48667)2] 2 =

600

= 2.98982


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Standard deviation:

The variance is a measure of dispersion, but it suffers from the fact that it is computed from squared values. To eliminate this squaring effect, we compute a new measure of dispersion called the standard deviation. This measure is simply the positive square root of the variance.

= + 2

In the case of the fair die: In the case of the sampling experiment:

= 2.91667 = 2.98982

= 1.70783 = 1.72911


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• Alternative Methods of Estimating Volatility

– Implied Volatility

– Historic Volatility (RiskMetrics)

– GARCH(1,1)


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Building an Implied Volatility Matrix

CSCO Calls, Closing Prices, 28 February 2001Strike Mar (16) Apr(41) July(132)October(223)

17.50 2 1/8 3 1/8 6 6 1/4

20.00 1 1/16 1 5/8 3 7/16 5

22.50 3/16 9/16 2 1/4 2 3/4

25.00 1/16 5/16 1 9/16 2 1/8

27.50 NA 1/8 15/16 1 1/2

30.00 NA NA 9/16 1

Numbers in parentheses represent days until option expiration. CSCO closing price on 28 February 2001 was $19.25. The interest rate term structure was essentially flat at 4.5% with continuous compounding. A calendar year is used.


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Implied Volatility MatrixStrike March April July October

17.50 64.46% 85.14% 114.76% 91.49%

20.00 84.98 74.03 78.84 86.12 (term structure)

22.50 65.63 60.83 71.28 61.88

25.00 73.25 67.23 69.80 62.43

27.50 NA 66.54 65.10 60.14

30.00 NA NA 62.44 57.41

(skew)

Example: 74.03% obtained by entering S = 19.25, X = 20.00, T - t = .1123 years (41 days), C = 1.625, and solving for implied volatility.


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• Term structure effects are principally occasioned by:

– Mean Reversion

– Scheduled Informational Events

– Market Segmentations


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• Skews and Smiles are most often occasioned by:

– Violations of the Assumption that Prices are Log-normally

Distributed

– Leverage

– Market Segmentation

– Option Maturity


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Forward implied volatility is computed from spot implied volatility much like a forward interest rate is computed from spot rates. Keep in mind however that option volatility refers to the annualized standard deviation of the continuously compounded rate of return of the underlying asset, whereas most forward rate computations involving interest rates involve a different compounding frequency (such as a semi-annual periodicity in the US Treasury bond market).

Question: What is the forward implied volatility of the nearest at-the-money CSCO call between the third Friday of July and the third Friday of October?


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Answer: The two relevant spot implied volatility measures are those of the CSCO July 20 and CSCO October 20 calls. These are 78.84% and 86.12% respectively. Recall there are 132 days until the July expiry and 223 days until the October expiry, and thus we are looking for an estimate of the 91-day forward vol that begins 132 days hence. Call this FV. The answer is obtained as follows:

exp(.8612)(223/365) = exp(.7884)(132/365) x exp(FV)(91/365)(.8612)(223/365) = (.7884)(132/365) + (FV)(91/365)FV = .9668 = 96.68%.

Note that, just like with interest rate term structures, if the relevant segment of the spot implied volatility term structure is upward sloping, then the forward implied volatility term structure that it begets is even more steeply upwardly sloped. Also note that it is possible to compute other forward implied vols and therefore entire forward implied volatility term structures. These may be useful, for example, for pricing forward-starting options such as executive stock options that have a vesting period.


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Measuring Historic Volatility. There are many ways to measure volatility using historic data. We will focus on methods originally introduced by Engle (Econometrica, 1982) and used today in applications such as JP Morgan’s RiskMetrics.

Define V(n) as the variance of a stock, stock index, or other market variable on day n, as estimated at the end of day n - 1. Let SD(n) be its square root; SD(n) is commonly called “vol”.

Let the market variable at the end of day i be S(i). Let the variable U(i) denote the continuously compounded return during day i (between the end of day i - 1 and the end of day i):

U(i) = ln[S(i)/S(i - 1)].


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A model that estimates V(n) while giving more weight to more recent data is:

(1) V(n) = E{a(i)[U(n - i)^2]},

where E is a summation operator where i = 1 to m, m is the total number of observations (sample size) of the daily U(i), and the variable a(i) represents the amount of weight given to the observation i days ago. The a’s are positive and a(i) < a(j) when i > j (because we want to assign less weight to older observations). The weights must sum to one, that is, E[a(i)] = 1.


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An extension of equation (1) is obtained by assuming that there is a long-run average volatility and that this should be given some weight:

(2) V(n) = b(LV) + E{a(i)[U(n - i)^2]},

where LV is long-run volatility and b is the weight assigned to LV. (Now the sum of b and the a’s must be one.) Equation (2) is known as an ARCH(m) model, where the acronym ARCH stands for “AutoRegressive Conditional Heteroscedasticity”. In practice, b(LV) is replaced by a single variable, say w, when parameters are estimated.


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The exponentially weighted moving average (EWMA) model is a particular case of equation (1) where the weights, a(i) decrease exponentially as we move back through time. Specifically, a(i + 1) = k[a(i)] where k is a constant between zero and one. This weighting scheme occasions the following simple formula for updating volatility estimates:

(3) V(n) = k[V(n - 1)] + (1 - k)[U(n - 1)^2].

Here the estimate of the variance for day n, V(n), which is made at the end of day n - 1, is calculated from V(n - 1) (the estimate that was made one day ago of the variance for day n - 1) and U(n - 1) (the most recent observation on changes in the market variable).


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For example, suppose that k is 0.94, which is precisely the value that JP Morgan uses to update daily volatility estimates in its RiskMetric database. This value, being so close to 1, produces estimates of daily volatility that respond relatively slowly to new information provided by the U(i)^2. Also suppose that the volatility estimate for day n - 1 is 1% per day, and that the proportional change in the market variable during day n - 1 is 2%. So V(n - 1) = (0.01)^2 = .0001 and U(n - 1)^2 = (0.02)^2 = .0004. Equation (3) gives V(n) = 0.94 x .0001 + 0.06 x .0004 = .000118. The estimate of volatility is therefore (0.000118)^(0.50) = 0.01086278, or about 1.086% per day, or about 17.24% per annum using a trading day year (252 days).


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A GARCH(1,1) model is a Generalized version of the EWMA model - itself a particular ARCH model - just described (see Bollerslev, J. of Econometrics, 1986). Here V(n) is calculated from a long-run average variance, LV, as well as from V(n - 1) and [U(n - 1)^2]. The model is:

(4) V(n) = b(LV) + a[U(n - 1)^2] + c[V(n - 1)].

Here c is a weight assigned to V(n - 1) and now a, b and c must sum to one. The EWMA model is a nested version of the GARCH model of equation (4) where b = 0, a = (1 - k) and c = k. [Under GARCH(1,1), V(n) is based on the most recent observation of U^2 and V. A more general GARCH(p,q) model calculates V(n) from the most recent p observations on U^2 and the most recent q observations of V. For asymmetric GARCH models and other variants, see Nelson (Econometrica, 1990) and Engle and Ng (J. of Finance, 1993).]


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To accommodate parameter estimation, the term b(VL) is usually replaced by a single parameter w. Once w, a and c have been estimated, b is given by 1 - a - c. The long-term variance LV is then calculated as w/b. For example, suppose that a GARCH(1,1) model is estimated from daily data (using maximum likelihood estimation or variance targeting techniques, c.f. Engle and Mezrich, RISK, 1996) as:

V(n) = .000002 + 0.13 x U(n - 1)^2 + 0.86 x V(n - 1).

This corresponds to w = .000002, a = 0.13, c = 0.86, b = 0.01 and LV = 0.0002. In other words, the long-run average variance per day is 0.0002, corresponding to a vol of 1.4% per day. Now suppose that the estimate of the vol on day n - 1 is 1.6% per day so that V(n -1) = 0.000256 and that the proportional change in the market variable on day n - 1 is 1% so that [U(n - 1)]^2 is 0.0001. Then:

V(n) = 0.000002 + 0.13 x 0.0001 + 0.86 x 0.000256 = 0.00023516.

The new estimate of the volatility is therefore (0.00023516)^(0.50) = 0.0153 or 1.53% per

day.


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Recall that the GARCH(1,1) model is similar to the EWMA model except that, in addition to assigning weights that decline exponentially to past U^2, it also assigns some weight to the long-run average volatility. Because of this added feature, GARCH models can accommodate mean reversion in the volatility. Indeed, the parameter c in the model is a type of “decay rate” similar to the parameter k in the EWMA model. And for GARCH(1,1) per se, the variance V(n) exhibits mean reversion with a reversion level of LV and a reversion rate of 1 - (a + c). (In the EWMA model, (a + c) = 1 so the reversion rate is zero.)


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Building an Historic Volatility Term Structure. Because GARCH models accommodate mean reversion, they can be used to build forecasts of entire volatility term structures. These are call “historic volatility term structures” because they are built using historic data samples and are therefore not to be confused with implied volatility term structures. Still, one might already envision trading strategies based on comparisons of the two term structures.

It can be shown that the GARCH(1,1) model described in equation (4) occasions the following estimate of future variance:

(5) V(n + f) = LV + {[(a + c)^f] x (V(n) - LV)}

where V(n + f) is an estimate of the variance to occur on day n + f in the future. Notice that when a + c < 1, the final term in equation (5) becomes progressively smaller as f increases. (If a + c > 1 then the variance would not be reverting but would be “fleeing”.)


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To build a historic volatility term structure suitable for options, consider an option that lasting between day n and day n + N. One can use equation (5) to compute the expected variance during the life of the option as:

(6) (1/N) x S[V(n + f)],

where S is a summation operator for f = 0,…,N - 1.


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Suppose that for a particular stock, a + c = 0.9602 and LV = 0.00004422 (a long-run daily vol of 0.66498%, or a trading-day yearly vol of 10.556%). Also suppose that the current variance per day, V(n), is 0.00006. This corresponds to a daily vol of 0.77460%, which is greater than LV, and an annual vol of 12.30% based on a 252 trading-day year. The first table below shows the historic volatility term structure (% per annum for a calendar day year) based on these data and equation (6), while the second table shows the impact on the term structure of a 1% change in the instantaneous volatility. Of course, once one has a volatility term structure like that in the first table below, one can always compute a forward vol and indeed an entire forward vol term structure a la getting a forward implied vol from an implied volatility term structure.

Notice in first table how the vol predicted in 500 days is closing in on the LV value of 10.556%. Also notice that the term structure is downward sloping, so the forward term structure would be even more steeply downwardly sloped.


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Option Life (Days) 10 30 50 100 500

Option Volatility 12.03 11.61 11.35 11.01 10.65

(% per annum, 252-day trading year)

Option Life (Days) 10 30 50 100 500

Option Volatility Now 12.03 11.61 11.35 11.01 10.65

After 1% Change 12.89 12.25 11.83 11.29 10.71

Increase in Volatility 0.86 0.64 0.48 0.28 0.06

Note: 12.30% current daily vol moves to 13.30%. So the daily vol becomes 0.84% and the daily variance becomes 0.00007016. Applying equation (6) to this new situation produces the second row in the second table.


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Multiple Random Variables:

In situations involving statistics, we are often called upon to examine the statistical behavior of multiple random variables. Specifically, we might be interested in knowing how two random variables vary with respect to one another.

For example, if we were to randomly select children at a local elementary school and we were to measure their ages in months and their heights in inches, we have in essence obtained joint observations on two different random variables.

What would we expect to find?

Do the two variables tend to move together?

How so?

Is the relationship perfect?


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What would it mean for the relationship to be perfect?

Height

Age


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What would it mean for the relationship to be perfect?

Height

Age

Notice that the random variable calledage has a mean and a standard deviation.

Notice that the random variable calledheight has a mean and a standard deviation.

Is this what we are likely to find?


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Age

Height

How could we describe this in a statistical sense?


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Denote the two random variables X and Y

X

Y

X

Y

X

Y

Positive Covariance Negative Covariance Zero Covariance


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Measuring Covariance:

True Covariance:

N M

CovX,Y = (Xi - µX) × (Yj - µY) × Prob[Xi and Yj] i=1 j=1

Sample Covariance:

Z

(Ox,k - uX) × (OY,k - uY) k=1CovX,Y = Z


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The problems with covariance:

As a measure of the degree to which two random variables move together, covariance has two weaknesses.

The first of these concerns the choice of units of measure:

Age: Years Height: FeetMonths InchesWeeks MetersDays CentimetersSeconds Millimeters

Second, there is no theoretical upper or lower limit to a covariance. That is, covariances can be extremely large positive numbers or extremely large negative numbers.

Both of these make it inconvenient to work with covariances.


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Correlation Coefficients:

To address the weaknesses of covariance, statisticians developed the concept of a correlation coefficient. A correlation coefficient is sometimes called a “standardized covariance.” Correlation is usually symbolized by the Greek letter rho ():

CovX,Y

X,Y =

X × Y

True correlations are obtained from true covariances and true standard deviations

Sample correlations are obtained from sample covariances and sample standard deviations.


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Correlation coefficients are bounded between +1 (called perfect positive correlation) and –1 (perfect negative correlation).

CovX,Y

X,Y

= X × Y

When the CovX,Y is positive the correlation is positive.

When the CovX,Y is negative, the correlation is negative.

When the CovX,Y is zero, the correlation is zero.


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• Simple Linear Regression

Two variables X and Y be said to be linearly related if their relationship can be expressed by the following simple linear model:

yi = α + βxi + ei

Where yi is the value of the Y (“dependent”) variable for a typical unit of association from the population for Y, xi is the value of the X (“independent” or “explanatory”) variable for that same unit of association, α and β are parameters called the regression constant and slope coefficients, respectively, and e i is a random variable that is i.n.d. with mean 0 and variance equal to that of Y.

Under some classical assumptions regarding X and Y, we can estimate α and β, estimate the strength of the linear relationship between X and Y, and conduct certain significance tests.


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Using the least-squares criterion for best fit (whereby we estimate the intercept and slope coefficients by minimizing the sum of the squared errors, where said errors represent the distances between the X,Y coordinates and the fitted line), we have the following formulas for a (our estimate of α) and b (our estimate of β):

a = y – bx (1)

b = [nΣxiyi - ΣxiΣyi]/[nΣxi2 – (Σxi)2] (2)

Where y is the mean value of yi (i = 1 to n, our sample size and the limits of the summation operators), and x is the mean value of x i.


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For example, suppose that we have the following data relating production (X) and manufacturing expenses (Y) for 10 firms:

X Y

Production (thousands of units) Expenses (thousands of dollars)

40 150

42 140

48 160

55 170

65 150

79 162

88 185

100 165

120 190

140 185


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Then our estimates of a and b from equations 1 and 2 above are: a = 134.72, b= 0.3978. And thus we have our estimated equation:

yc = 134.72 + 0.3978x,

Where yc is the calculated value of Y for a given X.

The “coefficient of determination” (usually denoted R2) provides an objective measure of the goodness of the fit of the estimated equation. To understand how it is computed, first consider how the “total deviation” for a particular value of y i from its calculated or “predicted” value yc is decomposed partly into “explained deviation” and partly into “unexplained deviation”:

(yi – y) = (yc – y) + (yi – yc)

total deviation explained deviation unexplained deviation


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For example, in the case of our ninth observation (y9 = 190), we have 24.3 = 16.8 + 7.5.

Performing similar calculations for all ten observations in our sample, and squaring every measure (so as to ensure that positive and negative total deviations do not offset), we get:

Σ(yi – y)2 = Σ(yc – y)2 + Σ(yi – yc)2

Total Sum Explained Sum Unexplained Sum

of Squares of Squares of Squares

For our sample, TSS = 2554.10, ESS = 1666.33, and USS = 887.77. In turn, the coefficient of determination is simply the ratio of the ESS over TSS:

R2 = ESS/TSS = 1666.33/2544.10 = 0.6538 = 65.38%.

Here we would say that, subject to the assumptions invoked, about 65% of the total variability in production expenses (Y) is due to/explained by the amount of units produced (X).


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The correlation coefficient between Y and X is simply the square root of the coefficient of determination, or here 0.8086 or about 81%. Here we say that production costs and output are about 81% correlated.

Now we turn our attention to tests of significance. Such tests require a procedure known as “analysis of variance” or ANOVA. You may recall the follow ANOVA table for simple linear regression:

Source of

Variation SS df F-statistic

RegressionESS 1 ESS/[USS/(n – 2)]

Error USS n – 2

For our sample, the F-statistic (with 1 and 8 degrees of freedom) is 1666.33/[(887.77/8)] = 14.99, which is greater than 14.69, thus indicating that we can reject the null hypothesis of no linear relationship between Y and X (R2 = 0) at the 0.01 level.


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We can similarly test for the significance of the estimated coefficients. For instance, the test statistic for determining whether or not b is significantly different from zero is simply the square root of the F-statistic, or 3.87. This figure is greater than 2.306 – the critical value of t for a two-sided test with 8 degrees of freedom at the 5% level. Hence we conclude that b is positive and statistically significant.

The t-statistic for b is also given by the estimated value of b (0.3978) divided by the standard deviation of b. We can solve for the latter now: 3.87 = 0.3978/s, so s = 0.1028. This in turn permits us to compute a confidence interval for b. For example, the 95% confidence interval for b is 0.3978 +/- 2.306(0.1028) = [0.1608,0.6348].

We could go on and on here.


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Types of distributions:

We started out our discussion of statistical distributions using one of the simplest types of distributions.

This distribution had the properties that it was discrete and uniform.

By discrete distribution we mean that there are only a finite number of outcomes.

By uniform, we mean that every possible outcome has the same probability.

Some real life distributions do have these characteristics. But, most don’t.


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As an example, define a new random variable as the number of dots facing up when a pair of dice are thrown together. There are a total of twelve things that can happen and they do not all have the same probability:

Possible Outcome Probability

2 1/36 [1,1] 3 2/36 [1,2 2,1] 4 3/36 [1,3 2,2 3,1] 5 4/36 [1,4 2,3 3,2 4,1] 6 5/36 [1,5 2,4 3,3 4,2 5,1] 7 6/36 [1,6 2,5 3,4 4,3 5,2 6,1] 8 5/36 [2,6 3,5 4,4 5,3 6,2] 9 4/36 [3,6 4,5 5,4 6,3] 10 3/36 [4,6 5,5 6,4] 11 2/36 [5,6 6,5] 12 1/36 [6,6]

What do the probability distribution and the cumulative probability distribution look like?


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This is discrete, but it is not uniform!

value 2 3 4 5 6 7 8 9 10 11 12

probability

6/36

5/36

4/36

3/36

2/36

1/36


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value 2 3 4 5 6 7 8 9 10 11 12

cumulativeprobability

36/36

30/36

24/36

18/36

12/36

6/36


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Continuous distributions:

The distribution that we just described for the pair of dice is an example of a type of discrete distribution called a binomial distribution. It is discrete, but the outcomes do not have equal probability.

We now consider distributions that are not discrete.

Distributions that are not discrete are said to be continuous. A random variable is said to be continuous if it can take on an infinite number of different values over some range of values.


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For example, the height of a child selected at random could have any value ranging from the height of the smallest child on earth to the height of the tallest child on earth. Suppose that the shortest child is 15 inches and the tallest child is 70 inches. How many different heights are there between 15 inches and 70 inches if we measure with absolute precision?

Answer: an infinite number of heights.


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There are many types of continuous distributions but the most common type is one called a normal distribution.

The normal distribution has several important characteristics:

It is continuous

It is symmetric

It can take on any value from negative infinity to positive infinity

It is fully described by its mean and its standard deviation (or variance)

It is extremely well defined mathematically


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value

probability(pdf)

X ~ N(µ, )


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value

probability(pdf)

µ


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value

µ

X

X

YY

X ~ N(µX, X)Y ~ N(µY, Y)

probability(pdf)


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value

probability(pdf)

Confidence Interval: 90%

µ

µ + 1.645µ – 1.645


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value

probability

Confidence Intervals: The role of z scores

µ

µ + zµ – z

Confidence IntervalProbability

z = 1.645 90% confidence intervalz = 1.960 95% confidence intervalz = 2.326 98% confidence intervalz = 2.576 99% confidence interval


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value

probability

Example: A random variable that is normally distributed has a mean of 50 and a standard deviation of 10. What is the 90% confidence interval?

µ + 1.645 66.45

µ – 1.645 33.55

X ~ N(50, 10)

µ


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0

0.2

0.4

0.6

0.8

1

1.2Cumulativeprobability

value


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Skewness (a measure of symmetry, called the third moment)

Skewed left

Symmetric (no skew)

Skewed right

Kurtosis (the tails, called the fourth moment)

Platykurtic: Data distributed heavily in midregion (thin tails)

Mesokurtic: Data is consistent with a normal distribution

Leptokurtic: Data is heavily distributed in the tails (fat tails)


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Risk Profiles: Depicting an Exposure to a Single Market Risk


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Commercial activity often exposes a corporation to one or more market risks.

Example:

Suppose that a farmer has 1000 acres of land on which he can plant wheat. Supposethat the cost of growing a bushel of wheat is $4.00. Each acre of land will generate30 bushels of wheat. Thus, his maximum production is 30,000 bushels.

Assume that there is no quantity uncertainty.


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The farmer would plant the wheat in April and harvest it in October (six months later).

The spot price of wheat at the time of decision making (planting) is $4.90 a bushel.The forward price of wheat (for October delivery) is $5.00 a bushel.

Which price counts?

If the spot price at harvest is equal to the current forward price for October,how much profit does he make?

What risks does he run?

How can we depict this risk?


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Profit = Quantity × Profit per Bushel

= Quantity × (Market Price - Cost of Production)

= 30,000 × (Price - $4.00)

What if the price is $5.00 in October? Profit =




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Profit = 30,000 × (Market Price - $4.00)

Price in October

$0 $1 $2 $3 $4 $5 $6 $7 $8 $9

Profit

$60,000

$120,000

0

-$60,0000

$180,000

Risk Profile


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At the time of planting, the price of wheat in October is a random variable:

It has a mean, it has a standard deviation, and it has some type of probabilitydistribution.

Let’s assume that the distribution is normal.

The mean is $5.

The standard deviation is $0.75.

Can we depict this with a normal curve and show the 90% confidence interval?


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valueµ

$6.234$5 + 1.645($0.75)

$3.766$5 – 1.645($0.75)

Price ~ N(5, 0.75)

$2 $3 $4 $5 $6 $7 $8


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probability

µ $6.234$5 + 1.645($0.75)

$3.766$5 – 1.645($0.75)

$2 $3 $4 $5 $6 $7 $8

Notice that the translation from price per unit to profit is a linear transformation!

Profit

$30,000$60,000

-$60,0000

$90,000

-$30,0000

value


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How can we convert the mean price and the standard deviation of price to a meanprofit and a standard deviation of profit?

Mean profit = Quantity × (mean price - cost of production)

= 30,000 × ($5.00 - $4.00)

= $30,000

Standard deviation = Quantity × standard deviation of per unit profit

= 30,000 × $0.75

= $22,500


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Price per bushel

probability

-$60000 $0 $30000 $90,000

What is this farmer’s six-month VAR @ 95%?

$2 $3 $4 $5 $6 $7 $8

Profit

$30000 - 1.645($22,500) 30000 + 1.645($22,500)


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Would every wheat farmer have the same six-month VAR?

Suppose another farmer had 2000 acres on which he is planting wheat,what does his risk profile look like relative to the first farmer?

What is the second farmer’s six-month VAR?


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$0 $1 $2 $3 $4 $5 $6 $7 $8 $9

Profit

$60,000

0

-$60,0000

$120,000

Risk Profile

Farmer 1

Farmer 2


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-$60000 $0 $30000 $90000 Profit Farmer 1

$30000 - 1.645($22,500) $30000 + 1.645($22,500) $60000 - 1.645($45,000) $60000 + 1.645($45,000)

-$120000 $0 $60000 $180000 Profit Farmer 2

probability


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The Roots of VAR: Portfolio Theory


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– Portfolio Theory is concerned with

• Risk = standard deviation of return

– for single assets

– for portfolios of assets

• Reward = mean return

– for single assets

– for portfolios of assets

– Portfolio theory was originally developed in the context of stock portfolios and we will initially adopt this for convenience.


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• Measuring total return on a stock

– components of return• dividend component• gain component

– dollar returns vs percentage returns

D + ( PE - PB) R = PB

– annualizing the percentage return• cumulative total return• annual total return (we will call it the rate of return)


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• Return as a random variable

– why is the rate of return a random variable?

– measuring expected (mean) return

– measuring the standard deviation of return


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Combining stocks to form portfolios:

– Harry Markowitz (1952)– return on a portfolio– mean return of a portfolio– variance of a portfolio– standard deviation of a portfolio


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• Let there be N stocks.

• Denote the return on stock i by Ri

• Denote the mean return on stock i by i

• Denote the standard deviation of return on stock i by i

• Denote the covariance of returns of any two stocks i and j by covi,j

• Denote the correlation coefficient for the returns on any two stocks by i,j

• where i = 1,2,3,...,N and j = 1,2,3,...,N in all cases


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Let Rp denote the return on a stock portfolio

Let p denote the mean return on a stock portfolio

Let p2 denote the variance of return on a stock portfolio

Markowitz showed:

N

Rp = wiRi

i=1

N

p = wii

i=1

N N

p2 = wiwj covi,j

i=1 j=1


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Suppose that you plan to invest your money in a portfolio of three stocks as below. By coincidence, each of the three stocks has an expected return of 15%. What is the expected return on your portfolio?

Stock Company Weight Mean Return 1 IBM 30% (.3) 15% 2 MSFT 20% (.2) 15% 3 INTC 50% (.5) 15%

N

p = wii

i=1

= (.3 15%) + (.2 15%) + (.5 15%)

= 15.0%


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Suppose that after one year, the actual returns on the stocks in your portfolio are as below.

Stock Company Weight Actual return

1 IBM .3 14%

2 MSFT .2 40%

3 INTC .5 10%

What was the actual return on your portfolio?


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Stock Company Weight Actual return

1 IBM .3 14%

2 MSFT .2 40%

3 INTC .5 10%

What was the actual return on your portfolio?

N

Rp = wiRi

i=1

= (.3 14%) + (.2 40%) + (.5 10%)

= 17.2%


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Notice that the actual return on the portfolio deviated from the expected return on the portfolio. That is, there was deviation from the expected. It is this potential for deviation from expected return that we understand as risk.

Portfolio risk is measured as the variance (or its square root, the standard deviation) of return.

N N

p2 = wiwjcovi,j

i=1 j=1


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N N

p2 = wiwj covi,j

i=1 j=1

Recall that a covariance is related to a correlation coefficient as follows:

covi,j i,j = i × j

Therefore, covi,j = ij i,j


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Therefore: N N

p2 = wiwj ij i,j

i=1 j=1


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N N

p2 = wiwj ij i,j

i=1 j=1

When i = j then ij = i2 , wiwj = wi

2 , and i,j = 1.

N N N

p2 = wi

2 i2 + wiwj ij i,j i =1 i j


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N N N

p2 = wi

2 i2 + wiwj ij i,j i=1 i j

Stock Company Weight Std. Dev. Correlation IBM MSFT INTC

1 IBM .3 19% 1.0 0.5 0.4 2 MSFT .2 19% 0.5 1.0 0.3 3 INTC .5 19% 0.4 0.3 1.0


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N N N

p2 = wi



1 IBM .3 19% 1.0 0.5 0.4 2 MSFT .2 19% 0.5 1.0 0.3 3 INTC .5 19% 0.4 0.3 1.0

p2 = (.32 .192) + (.22 .192) + (.52 .192)

+ (.3 .2 .19 .19 .5) + (.3 .5 .19 .19 .4) + (.2 .3 .19 .19 .5) + (.2 .5 .19 .19 .3) + (.5 .3 .19 .19 .4) + (.5 .2 .19 .19 .3)


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1 IBM .3 19% 1.0 0.5 0.4 2 MSFT .2 19% 0.5 1.0 0.3 3 INTC .5 19% 0.4 0.3 1.0

p2 = (.32 .192) + (.22 .192) + (.52 .192)

+ (.3 .2 .19 .19 .5) + (.3 .5 .19 .19 .4) + (.2 .3 .19 .19 .5) + (.2 .5 .19 .19 .3) + (.5 .3 .19 .19 .4) + (.5 .2 .19 .19 .3)

= 0.013718 + 0.008664

= 0.022382


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p2 = 0.022382

Therefore:

p = 0.1496

= 14.96%


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Summarizing:

p = 15.00%

p = 14.96%

Conclusions:

What is the effect of diversification on expected return?

What is the effect of diversification on risk?


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Statistics with Applications to Options N N N

p2 = wi


Diversification and risk

– unsystematic risk (diversifiable risk, also called specific risk)• that risk which is company specific (e.g., a fire at a plant)• caused by changes in those variables that impact a single company or

only a small group of companies (e.g., an explosion at a production facility)

– systematic risk (non-diversifiable risk, also called market risk)• that risk which is shared by all or most firms• caused by changes in those variables that impact most companies

simultaneously (e.g., interest rates)


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Standard deviation of return

Number of Stocks in Portfolio 10 20 30 40

systematic risk

unsystematic risk

If a portfolio is well diversified, then the portfolio’s standard deviation is all systematic risk. If a portfolio is not well diversified, then the portfolio’s standard deviation is partly unsystematic risk and partly systematic risk.


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The Markowitz measures of portfolio risk and return can be used to develop a confidence interval and a VAR-like measure.

Suppose we ask:

“What is the 90% confidence interval for the rate of return on a stock portfolio if the portfolio’s annual rate of return has a mean of 15.00% and a standard deviation of 14.96%?”

To answer this question, we must make some assumption about the nature of the underlying distribution. We will assume that it is normal.


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Annual rate of return

probability

15.00%

14.96%

90% confidence interval = 15.00% ± (1.645 × 14.96%)

– 9.61% 39.61%

5% 5% 90%


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From Portfolio Theory to Parametric VAR

(Variance-Covariance VAR)


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The first approach to measuring VAR was based on Markowitz’s measures of a portfolio’s mean return and a portfolio’s standard deviation of return.

Because this measure was based on two statistical parameters (mean and standard deviation) and the assumption that the probability distribution is normal, the method is known as parametric VAR or Variance-Covariance VAR.

The calculation of parametric VAR requires two important adjustments to Markowitz original measures of a portfolio’s parameters.

1. VAR is measured in number of monetary units (e.g., USD, JPY, EUR, etc.). Markowitz measures are in terms of rates of return.

2. VAR is measured over any desired risk horizon (1 day, 2 days, 3 days, etc.). Markowitz assumed the investment horizon was one year in

length.


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The first adjustment requires that we re-state the mean and the standard deviation in terms of monetary units, instead of percentage returns. We will use dollars as the monetary units.

This adjustment is straightforward enough. Since percentage returns are nothing more than the return per $1, we simply substitute the number of dollars invested in each asset for the weight on that asset.


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The expected profit becomes:

N

µ = Ii µi

i=1

where µ denotes the expected (mean) dollar profit over the risk horizon

Ii denotes the number of dollars invested in asset i.

µi denotes the expected per dollar return on asset i over the risk horizon. This is the same as the percentage return.


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The variance of profit becomes:

N N

2 = IiIjCovi,j

i=1 j=1

where 2 denotes the variance of dollar profit measured over the risk horizon

Ii and Ij denote the number of dollars invested in assets i and j respectively.

Covi,j denotes the covariance of the returns (measured over periods that

correspond to the length of the risk horizon) per dollar invested in asset i and per dollar invested in asset j. This is the same as percentage returns.


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The second adjustment is also straightforward. Markowitz assumed, for simplicity, that percentage returns would be measured over a one-year period. This is why they are called “rates of return.” But there is no reason that percentage returns need to be measured over a period of one year. They could just as easily be measured over a period of 1 hour, 1 day, 1 week, 1 month, and so forth.

When using VAR analysis, risk managers may use several risk horizons. That is, they may state a one-day VAR, a one-week VAR, and a one-month VAR.


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It turns out that, if you know the statistical parameters to get a one-day VAR it is a simple matter to get a VAR for any horizon one likes. For this reason, most publicly available databases that can be tapped for the necessary covariance information use covariances for $1 investments measured over one-day. This information is presented in matrix form and is called a variance-covariance matrix.

The first firm to provide this data in a form that would feed into a parametric VAR model was JP Morgan. The JP Morgan system together with its variance-covariance matrix database is called RiskMetrics.


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Riskmetrics, and other systems like it, provide covariances that are based on some number of days of data.

For example, suppose that the covariances are calculated using daily data for the past 45 market days. Each day one new day of data is added and the earliest day of data is deleted. Thus, these are moving covariances.

1 2 3 4 5 • • • 44 45 46 47


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Example of a Variance-Covariance Matrix (values measured in dollars per $1):

1-Day Variance-CovarianceAsset 1 2 3 4

1 0.000010 0.000008 –0.000002 0.000003

2 0.000008 0.000018 –0.000005 0.000004

3 –0.000002 –0.000005 0.000025 –0.000001

4 0.000003 0.000004 –0.000001 0.000015

Note: This variance-covariance matrix is a description of the market environment.


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1-Day Variance-CovarianceAsset 1 2 3 4

1 0.000010 0.000008 –0.000002 0.000003

2 0.000008 0.000018 –0.000005 0.000004

3 –0.000002 –0.000005 0.000025 –0.000001

4 0.000003 0.000004 –0.000001 0.000015

Suppose that a cash market trader has positions in these four assets only. He has a $10 mm long position in Asset 1, a $15 mm short position in Asset 2, a $5 mm long position in Asset 3, and a $10 mm short position in Asset 4. The expected returns on these positions are as follows:

Asset 1-Day expected Return per $1 of investment (assumes long position) 1 $0.00027 2 –$0.00021 3 $0.00035 4 –$0.00030

firm’s estimates of theexpected profits from itspositions (per $1 invested).


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Calculate the Trader’s One Day VAR at 95%:

Step 1: Calculate the expected 1-day dollar return

Asset Size of Position × Expected Return Per $1 = Product

1 $10,000,000 $0.00027 $2,700 2 – $15,000,000 –$0.00021 $3,150 3 $5,000,000 $0.00035 $1,750 4 –$10,000,000 –$0.00030 $3,000

$10,600


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Step 2: Calculate the Variance and Standard Deviation of 1-day profit:

Dollars Asset i × Dollars Asset j × Covariance per dollar =

1,1 10,000,000 10,000,000 0.000010 10000000001,2 10,000,000 -15,000,000 0.000008 -12000000001,3 10,000,000 5,000,000 -0.000002 -1000000001,4 10,000,000 -10,000,000 0.000003 -3000000002,1 -15,000,000 10,000,000 0.000008 -12000000002,2 -15,000,000 -15,000,000 0.000018 40500000002,3 -15,000,000 5,000,000 -0.000005 3750000002,4 -15,000,000 -10,000,000 0.000004 6000000003.1 5,000,000 10,000,000 -0.000002 -1000000003,2 5,000,000 -15,000,000 -0.000005 3750000003,3 5,000,000 5,000,000 0.000025 6250000003,4 5,000,000 -10,000,000 -0.000001 500000004,1 -10,000,000 10,000,000 0.000003 -3000000004,2 -10,000,000 -15,000,000 0.000004 6000000004,3 -10,000,000 5,000,000 -0.000001 500000004,4 -10,000,000 -10,000,000 0.000015 1500000000

2 = 6,025,000,000

= $77,621


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Step 3: Calculate the 1-day VAR at the desired confidence level:

Note:95% VAR z-score corresponds to 90% confidence interval z-score.

99% VAR z-score corresponds to 98% confidence interval z-score.

95% 1 day VAR z-score is 1.645, therefore 1-day VAR is calculated as follows:

VAR(1 day, 95%) = µ – (1.645 × )

= $10,600 – (1.645 × $77,621)

= –$117,087

Thus, the maximum 1-day loss at a 95% level of confidence, is $117,087.


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Note: Because a “loss” is understood to be a negative number, it would not be appropriate to say:

The VAR is –$117,087, since VAR is generally understood to be a loss.

Thus, we usually drop the negative sign when quoting VAR.

An alternative way to avoid this sign confusion is to measure the VAR as follows:

VAR = (1.645 × ) – µ

This is the usual convention and we will use it from here on.


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Note: Some providers of VAR variance-covariance matrices assume that the VAR will be measured at a specific level (usually 95%) and they pre-multiply each covariance in the matrix by 1.6452.

When this is done, the variance-covariance matrix is a VAR-scaled variance-covariance matrix. As a result, the standard deviation obtained is a VAR-scaled standard deviation and has to be interpreted as such.

VAR = (VAR-scaled) – µ

It is important to read the technical document that accompanies variance-covariance matrix data to know whether or not it has been scaled. This is available from the data supplier.


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Adjusting VAR for longer risk horizons:

The procedure we just described provided a one-day VAR at a 95% level of confidence. We know that we can get a one-day VAR at a higher or lower level of confidence by simply adjusting the z-score.

But, what if the risk manager has been asked for say a 2-day VAR or a one-week (7-day) VAR or a 1-month VAR.† How do we make these adjustments?

This turns out to be very simple. It will be approximately true that the T-day mean profit µ(T) will be related to the one-day mean profit µ and the T-day standard deviation (T) will be related to the one-day standard deviation as follows:

µ(T) = T × µ (T) = T ×

† It is important to know whether the variance-covariance matrix was generated on the basis of a trading day year (251 days) or a calendar day year (365 days). For example, a week is 5 days in the former case and 7 days in the latter case.


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µ(T) = T × µ (T) = T ×

Consider again the case we looked at earlier in which µ = $10,600 and = $77,671

Risk Horizon Mean Standard Deviation VAR @ 95%

1 day $10,600 $77,621 $117,087 2 days 21,200 109,773 159,376 7 days 74,200 205,366 263,627 30 days 318,000 425,148 381,368

VAR = 1.645(T) – (T)


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VAR

Risk Horizon in days 1 2 7 30


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A note on daily expected profit:

Because daily expected profit is generally quite small relative to daily standard deviation and daily expected profit is very hard to estimate, many users of VAR simply assume that daily expected profit is zero.


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Alternative Approaches to Measuring VAR


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Alternative Approaches to Measuring VAR:

The parametric approach to measuring VAR assumes that the risk manager has available the relevant variance-covariance matrix and assumes that the distribution is normal (or at least approximately so).

If either of these conditions is not satisfied, the parametric method would not seem to fit the bill.

Another inherent problem with parametric VAR is that it assumes a linear correspondence between the source of market risk and the size of the trader’s exposure. This is not always the case. For example, if the trader’s portfolio contains options or embedded options, this condition is violated.

There are several other methods for obtaining a VAR. Each has its strengths and each its weaknesses. We will look at two of these: Historic VAR and Simulated VAR.


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Historic VAR


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Historic VAR:

Historic VAR is a method of calculating VAR which assumes that the future will be very much like the past.

In such a scenario, we can ask “What was the trader’s daily P&L each day for the last 100 days?”

This information would be available to any trader that marks all of his or her positions to market at the end of each day to generate a daily P&L.


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Once we have the daily P&Ls, we simply list them (rank them) from the highest to lowest:

Rank Daily P&L 1 +$182,300

2 +$174,100 3 +$164,900 4 +$164,500

5 +$151,000 6 +$142,500 • •

• • • • 94 –$107,400

95 –$112,400 96 –$115,300 97 –$120,200 98 –$127,000 99 –$134,100 100 –$142,500

VAR @ 95%


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Historic P&L data can also be organized into a histogram. For example, we might say what percentage of the time did the daily P&L lie between +$185,000 and $190,000 (answer 0)? What percentage of the time did it lie between +$180,000 and $185,000, and so forth? This can be plotted.

percentage

daily P&L ranges

We can now ask “How doesthis compare with a normaldistribution?”


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Weaknesses of Historic VAR:

Historic VAR ignores important issues:

Has the scale of operations changed?

Has the trading strategy changed?

Has the use of leverage changed?

Have the instruments we trade changed?

Is the past a good indicator of the future?

shocks to the system

state of the economy


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Simulated VAR(Monte Carlo VAR)


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Simulated Value-at-Risk:

Simulated VAR (also called Monte Carlo VAR) is most useful when the trader whose VAR we are attempting to estimate holds positions that have non-linear payoffs (also called non-linear risk profiles).

This will be the case whenever the trader holds positions in options, securities with embedded options, or securities other than options that have non-linear payoffs--such as bonds (e.g., convexity).


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To demonstrate why parametric VAR fails to work properly in cases involving non-linear payoffs, let’s consider a trader with a long position in a straddle.

Straddles:

A long straddle involves a long position in both a call option and a put option on the same underlying asset with the same expiration date and the same strike price.

A short straddle involves a short position in both a call option and a put option on the same underlying asset with the same expiration date and the same strike price.


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Because a straddle is a combination of a call and a put, we need to take a moment to look at the value diagrams of calls and puts.

A call option grants its owner the right but not the obligation to buy the underlying asset from the option writer for the strike price written into the option contract. This right is good for a limited time, called the time to expiry.

At expiration, the value of a call option on a single unit of the underlying is given by:

Value = max[S-X, 0]

where S = spot price of the underlying asset

X = strike price of the option

Prior to expiry, the value of the option is given by:

Value = max[S-X, 0] + time value


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A put option grants its owner the right but not the obligation to sell the underlying asset to the option writer for the strike price written into the option contract. As with a call, this right is good for a limited time called the time to expiry.

At expiration, the value of a put option is given by: Value = max[X-S, 0]

where S = spot price of the underlying asset

X = strike price of the option

Prior to expiry, the value of the option is given by: Value = max[X-S, 0] + time value.


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The valuation of an option at any time prior to expiry is complex and requires the employment of an appropriate model.

Models are an attempt to mathematically capture the importance of the various factors1 that determine an instrument’s value. In this case, the instrument is the option. Often, there is less than universal agreement as to which of several models is the “best” model for valuing a particular option.

The most widely known of option pricing models is the Black-Scholes-Merton model.

1 The factors that determine the value of an instrument are sometimes called value drivers.


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Building a model requires that we make assumptions. For example, the Black-Scholes-Merton model assumes:

The price of the underlying follows a random walk through time

The future price of the underlying asset is lognormally distributed

The volatility of the price of the underlying is constant

The risk-free rate of interest is constant

Trading is continuous (in both time and in the size of transactions)

There is no transaction cost associated with trading

Black/Scholes assumes no payout

Merton allows for a payout


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While the details will differ a bit, the value of an option, irrespective of which model is employed, will be a function of five key variables, which constitute the value drivers:

the current price of the underlying (denoted S)

the strike price of the option (denoted X)

the annual volatility (denoted v)

the time to option expiration as a fraction of a year (denoted )the rate of interest (denoted r)

This can be expressed in function form:

Value = f(S, X, v, , r)

If the underlying is a payout asset, then the payout must also be built into the model. We assume that the underlying asset does not pay out anything.


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The Mechanics of Monte Carlo Simulation


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The Mechanics of Simulation:

In simulation analysis, the first thing we must do is determine the relevant variables. Next, we must determine the type of distribution each variable has, the parameters of the variables (e.g., , ), and the degree of correlation between the random variables ().

Example:

Suppose that X and Y are two normally distributed random variables. X has a mean of 5 and a standard deviation of 7. Y has a mean of 30 and a standard deviation of 4. X and Y have a linear correlation of 0.6.

Simulate 100 joint outcomes on X and Y.


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Random Number Generator:

To simulate the outcomes on X and Y, we will employ a random number generator. Excel has such a random number generator function built into it. It is called RAND().

RAND() generates observations on a continuous uniform distribution bounded between 0 and 1. Nevertheless, the function can be used to generate observations on a standard normal distribution as follows:

Let Ri be the ith value generated from RAND(). Now define E as follows

12

E = Ri – 6

i = 1

E will have a standard normal distribution. That is E ~ N(0, 1)


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Suppose now that we generate two observations by this process. Let’s call the first E1 and the second E2.

Each of these random variables is distributed as a standard normal:

E1 ~ N(0,1) E2 ~ N(0,1)

While these two random variables are both standard normal, they are independent of one another. That is, their correlation is 0.

We now want to add the appropriate degree of correlation.


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Adding the Correlation:

Define two new random variables as follows:

A = E1 and B = ×E1 + E2×(1 - 2)½

A and B, like E1 and E2, are still standard normal random variables, but unlike E1 and E2, which are independent of one another, A and B are correlated with one another and the degree of correlation between them is given by .

In our example, the correlation is 0.6 so the relationship above would look as follows:

A = E1 and B = 0.6×E1 + E2×(1 - 0.62)½


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Adjusting the Means and Standard Deviations:

A and B have standard normal distributions with the appropriate degree of correlation. We now want to adjust their means and their standard deviations so that they have the parameters that X and Y are supposed to have. Define X and Y as follows:

X = A×A + A Y = B×B + B

= A×7 + 5 = B×4 + 30

At this point we have simulated the outcomes that we want:

X ~ N(5, 7) Y ~ N(30, 4) X,Y = 0.6


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Consider the following:

Suppose that we have an option written on a stock. The option’s strike price is $100. The stock’s current price is $102. The option covers 1 share. Suppose that the option expires in 182 days, its annual volatility is 30%, and the risk-free rate of interest is 5%.

Using the Black-Scholes-Merton Model

What is the value of the option if it is a call? Answer: $10.8333

What is the value of the option if it is a put? Answer: $6.3710


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What value might the option have tomorrow if the option is a call?

The value of the option tomorrow will depend on the same value drivers as does the value of the option today. The only difference is that we may not know today what the inputs for those value drivers will be tomorrow.

The value drivers that might change between today and tomorrow are:

The amount of time remaining until the option expires

The price of the underlying asset

The volatility of the price of the underlying asset

The rate of interest

With the exception of the time to expiry, each of these “value-drivers” can change in an unexpected way. Thus, they are each a source of “risk.” Simulation modeling of VAR attempts to simulate input values for these value drivers in order to simulate a value for the option.


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How does each value driver affect the value of an option?

We know that tomorrow the option will have only 181 days to expiry. Let’s suppose that the volatility and risk-free rate of interest are constant, so that only the price of the underlying is uncertain.

What might happen to the value of the call?

What might happen to the value of the put?


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Value

0

5

10

15

20

25

86 89 92 95 98 101

104

107

110

113

Price of underlying

Days = 181

r = 5%

vol = 30%

X = 100

Long Call


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0

2

4

6

8

10

12

14

16Value

Price of underlying

Days = 181

r = 5%

vol = 30%

X = 100

Long Put


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A Closer Look at the Call:

Holding volatility and the interest rate constant, what happens with each passing day?


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Value

0

5

10

15

20

25

86 89 92 95 98 101

104

107

110

113

Price of underlying

r = 5%

vol = 30%

X = 100

181 days

120 days

60 days0 days


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Holding time to expiry constant, and the interest rate constant, what happens if volatility increases or decreases?


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0

5

10

15

20

25

86 89 92 95 98 101

104

107

110

113

Price of underlying

r = 5%

Days = 181

X = 100

Value vol = 40%

vol = 30%

vol = 20%


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Holding time to expiry and volatility constant what happens if interest rates increase or decrease?


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Value

0

5

10

15

20

25

Price of underlying

vol = 30

Days = 181

X = 100


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A Closer Look at the Put:

Puts will exhibit similar behaviors:

As time grows shorter, the puts value will decline.

If volatility increases, the puts value will increase.

If interest rates rise, the value of a put goes down.


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The Straddle:

We are now ready to look at the straddle. The value of the straddle is the sum of the values of the call and the put.

Holding volatility (30%), time to expiry (182 days), and the interest rate (5%) constant, what does the value diagram of the straddle look like?

Suppose that the trader’s straddle covers 500,000 units.


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Now consider what happens to the value of our position between today and tomorrow.

Today, the value of the call is $10.8333

Today, the value of the put is $6.3710

Thus, today the value of the straddle is $10.8333 + $6.3710 = $17.2043

And, the value of the position today is = $17.2043 × 500,000 = $8,602,150

The change in the value of the position = Value Tomorrow - Value Today


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We can graphically illustrate the change in the value of the position tomorrow but only with respect to a change in at most two of the value drivers at a time.

Assume that volatility and the interest rate do not change.

We can then depict the change in value of the position with respect to a change in the underlying’s price, remembering that one day will have passed.

This represents a “risk profile” with respect to one source of market risk (the price of the underlying). That is, we are looking at the risk profile with respect to one of the risk factors.


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Profit/Loss

-1000000

-500000

0

500000

1000000

1500000

2000000

2500000

3000000

Price of underlying

102100989694 90 92 104 106 108 88

vol = 30%

r = 5%

Days = 181

X = 100


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Profit/Loss

-1000000

-500000

0

500000

1000000

1500000

2000000

2500000

3000000

Price of underlying

102100989694 90 92 104 106 108 88

vol = 30%

r = 5%

Days = 181

X = 100

What happens if we overlay this risk profile with a normal distribution?


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Profit/Loss

-1000000

-500000

0

500000

1000000

1500000

2000000

2500000

3000000

Price of underlying

102100989694 90 92 104 106 108 88


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What happens if the underlying’s price changes, the volatility changes, and the interest rate changes?

The multidimensionality of the problem and the non-linearities result in a situation that is too complex to map into a parametric VAR.

This is where simulation analysis provides a solution.


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Simulating changing market conditions:

What we want to do is simulate vectors of possible outcomes for all the value drivers that will influence the market value of the straddle. These outcomes include: The price of the underlying, the volatility of the underlying, and the interest rate.

This requires that we make decisions about:

1. The types of distributions describing these three value drivers

2. The mean of each value driver

3. The standard deviation of each value driver

4. The values of other possible parameters

5. The correlations among the different value drivers.


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Example:

Suppose that a DB trader is long a straddle on GTV stock. The straddle covers 500,000 shares. GTV is currently priced at $102, GTV has an annual vol of 30%, the rate of interest is 5%. The straddle has a strike of $100 and 182 days to go.

Assuming that this is the only position this trader holds, determine the trader’s one-day VAR at 95%.


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Under Black-Scholes-Merton, the future price of the underlying (GTV stock) is log-normally distributed, so that the natural logarithm of the future price is normally distributed as follows:

lnST ~ N[lnS0 + (r – (σ2/2))T,σ√T]

Assuming a 365-day year, then the natural logarithm of the one-day future price of GTV is distributed normally with mean 4.625 and standard deviation 0.0157:

lnS1 ~ N[ln(102) + (.05 – (.302/2))(1/365),.30√(1/365)] ~ N(4.625,0.0157)

Recall that the interest rate and volatility are constant in the Black-Scholes-Merton framework. Thus there are no correlated variables in this particular illustration.


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Now, using the technique described above for sampling a standard normal variate and then mapping it into one with mean 4.625 and standard deviation 0.0157, suppose that we obtain an estimate of lnS1 of 4.6407. Then the estimate of the price of the stock in one day’s time is e4.6407 = 103.6169. Under our new parameter set (S = 103.6169, r = .05, T = 181 days, σ = 0.30, and X = 100), the values of the call and put are $11.84 and $5.77, respectively. Thus the estimate of the value of the straddle contract in one day is ($11.84 + $5.77) x 500,000 = $8,805,000. The one-day P&L is a gain of $202,850.

This entire process would be repeated over and over, perhaps 10000 times. The profits and losses would be ranked (from greatest profit to greatest loss). The 95% one-day VAR of the straddle would correspond to the outcome ranked 9501 out of the 10000.


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Suppose that we wanted an estimate of the long straddle position’s 95% VAR, but for 5 days rather than 1 day. From the previous equation, the natural logarithm of the underlying share price in five days would be normally distributed with a mean of 4.625 and a standard deviation of 0.0351. So now we would generate a random standard normal variate, and transform it to one with mean 4.625 and standard deviation 0.0351. This in turn would give us one estimate of the share price in five days, and thus one measure of the 5-day P&L on the long straddle position. Again, this process would be repeated 10000 times, with the 95% VAR being the 9501st ranked P&L.

If we wished, we could use a forward interest rate and a forward volatility (instead of the 5% and 30% “spot” measures, respectively), in order to obtain our estimate of a future stock value. Using forward values for r and σ may be prudent if the VAR horizon is long and/or the interest rate and volatility term structures are steeply sloped.


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While we have demonstrated the calculation of simulated VAR for a position consisting of only one position (a straddle on one stock), there is no reason why we could not have simultaneously generated potential values for all of the positions in the trader’s book (where a book is defined as one wherein all options have the same underlying stock, e.g., DELL).

Is it also possible to accommodate the calculation of simulated VAR when we blend books, that is, when we have a portfolio of option positions and there is more than one underlying stock (e.g., DELL and CSCO)? The answer is “yes”. Here we must accommodate the correlations among all the stock prices. We would do so for two stocks as described earlier, that is, when accommodating two correlated and normally distributed variables. (Recall that we must work with the natural logarithms of the two stock prices. That is, we must treat the natural logs as normally distributed, and accommodate the correlation among the natural logs of the prices.)


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What if we wanted to accommodate more correlated variables, for example, by permitting multiple underlying stocks? Or, if we assume that the interest rate and volatility are log-normally distributed, how can we accommodate these variables and their possible correlations with each other and with stock values? This can be accomplished as follows:

Consider the situation where we require n correlated samples from normal distributions where the coefficient of correlation between sample i and sample j is ρi,j. We first sample n independent variables xi (1 ≤ i ≤ n), from univariate standardized normal distributions. The required samples are εi (1 ≤ i ≤ n), where

εi = Σαikxk

where Σ is a summation operator for k = 1 to i. For εi to have the correct variance and correct correlation with εj (1 ≤ j < i), we must have

Σα2ik = 1 (where the summation factor is again for k = 1 to i),

and, for all j < i,


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Σαikαjk = ρij (where here the summation factor is for k = 1 to j).

The first sample, ε1, is set equal to x1. These equations for the α’s can be solved so that ε2 is calculated from x1 and x2, ε3 is calculated from x1, x2, and x3, and so on. The procedure is known as the Cholesky decomposition.

Note: Monte Carlo simulation tends to be numerically more efficient than other simulation techniques when there are three or more stochastic variables, because the time required to conduct a Monte Carlo simulation only increases about linearly with the number of variables. It can also accommodate products like path-dependent options.


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Potential Problems with Simulated VAR:

Don’t be fooled into believing that simulated VAR is non-parametric. It is.

Simulated VAR required knowledge of both the distributions and the parameters of the distributions of the drivers (input values). Thus, simulated VAR is parametric in nature. If parameters are misestimated or distributions are incorrectly specified, simulation can generate faulty estimates of VAR.


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Model Risk:

Simulated VAR generates a change in value by deducting the potential value of the option tomorrow from the value of the option today. The value today is often observable in the market today, but the potential value tomorrow has to be inferred from the simulated outcomes of the value drivers and the model that takes the value driver inputs and produces an option value output.

This begs the question “What if we are using the wrong model?” The very real possibility that we are using the wrong model is one form of a broader class of problems called model risk.


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Blended VAR:

Some risk managers form a “blended” or “composite” VAR from VAR estimates obtained using several different approaches.

For example, suppose that the following methods lead to the following 1-day VARs at 95%:

Parametric VAR: $117,087

Historic VAR: $115,300

Simulated VAR: $121,200

We could formulate a weighted average of these three VAR measures (either equally weighted or by giving more weight to one method or another).


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What VAR is Not


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We have been trying to answer the questions: What is VAR and how is it measured?

We have concluded that VAR is a measure of the maximum loss a trader might experience over a given risk horizon at a specified level of confidence.

It is equally important to understand what VAR is not.

VAR is not a statement of the absolute maximum loss the trader might suffer. VAR does not attempt to address this question.

For example if the 1-day VAR at 95% is $1.2mm, we may conclude that the trader will lose no more than $1.2 mm in a single day on 95 out of each 100 days. But, it does not attempt to say how much might be lost on the other 5 days!

Thus, VAR is not a panacea for risk managers. It is merely one tool.

What VAR is Not


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Of course, one could use a 99% VAR or a 99.9% VAR in an effort to measure the potential for loss under extreme conditions.

There are two problems with this. The first is that VAR estimates become more and more unreliable the higher the level of confidence that is sought. The reason for this has to do with departures of the underlying distributions from the assumed normality. It is known, for example, that price distributions often exhibit leptokurtosis (fat tails).

The one-day stock market declines in October 1929 and again in October 1987 are examples of this.


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Second, there is considerable risk that the relationships among the value drivers (i.e., risk factors) might suddenly change. For example, under conditions of stress in the market place (e.g. periods of flight to safety) the correlations among prices sometimes rise sharply. When this happens, the potential for loss can greatly exceed that implied by VAR.


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Stress Testing:

One way to measure the potential to experience a loss in excess of that implied by VAR is to engage in stress testing.

In stress testing, we simulate VAR but we assume conditions of market stress. By market stress, we mean we go back and look for periods when the markets were under unusual stress (such as periods when there was a flight to quality). We then measure the correlations among prices during those periods and the standard deviations of prices during those periods. (It is important that we employ values that are internally consistent.)

From these extreme-condition parameters, we can test the values produced by our models and estimate VARs.

ise option traders statistics with applications to options march 2004copyright (c) 2001-2004 by...

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