1 chapter 18 the lognormal distribution. 2 the normal distribution normal distribution (or density):
TRANSCRIPT
3
The normal distribution
Normal density is symmetric: If a random variable x is normally distributed
with mean and standard deviation , z is a random variable distributed standard
normal: The value of the cumulative normal [P(z<a)]
distribution function N(a) or NormSDist(a) in Excel equals to the probability P of a number z drawn from the normal distribution to be less than a.
( ; , ) ( ; , )x x
x N~ ( , ) 2
z N~ ( , )0 1
N a e dxxa
( )
1
2
1
22
4
The normal distribution
The Normal Distribution allows us to make statements and inferences regarding future stock price levels.
We are able to estimate probabilities of the terminal stock price being between two values, above a value, or below a value.
Note: the probability of reaching an EXACT value is zero.
6
The normal distribution (cont.)
The probability of a number drawn from the standard normal distribution of being between a and –a is:
Prob (z < –a) = N(–a)
Prob (z < a) = N(a)
thereforeProb (–a < z < a) =
N(a) – N(–a) = N(a) – [1 – N(a)] = 2·N(a) – 1
Example 18.1: Prob (–0.3 < z < 0.3) = 2·0.6179 – 1 =
0.2358
7
The normal distribution (cont.)
Converting a normal random variable to standard normal: If , then if
And vice versa: If , then if
Example 18.2: Suppose and , then
and
z N~ ( , )0 1x N~ ( , ) 2zx
x N~ ( , ) 2z N~ ( , )0 1 x z
x N~ ( , )3 5
z N~ ( , )0 1
xN
3
50 1~ ( , ) 3 5 3 25 z N~ ( , )
8
The normal distribution (cont.)
Example:The number 7 is drawn from a Normal
distribution of mean 4 and variance 9.
What is the equivalent draw from a standard normal distribution?
What is the probability of drawing a number larger than 7 ? Lower than 7 ? Exactly the number 7 ?
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Cumulative Normal Table
d N(d) d N(d) d N(d) d N(d) -3 0.00135 -1.45 0.07353 0.05 0.51994 1.55 0.93943
-2.95 0.00159 -1.4 0.08076 0.1 0.53983 1.6 0.94520 -2.9 0.00187 -1.35 0.08851 0.15 0.55962 1.65 0.95053 -2.85 0.00219 -1.3 0.09680 0.2 0.57926 1.7 0.95543 -2.8 0.00256 -1.25 0.10565 0.25 0.59871 1.75 0.95994 -2.75 0.00298 -1.2 0.11507 0.3 0.61791 1.8 0.96407 -2.7 0.00347 -1.15 0.12507 0.35 0.63683 1.85 0.96784 -2.65 0.00402 -1.1 0.13567 0.4 0.65542 1.9 0.97128 -2.6 0.00466 -1.05 0.14686 0.45 0.67364 1.95 0.97441 -2.55 0.00539 -1 0.15866 0.5 0.69146 2 0.97725 -2.5 0.00621 -0.95 0.17106 0.55 0.70884 2.05 0.97982 -2.45 0.00714 -0.9 0.18406 0.6 0.72575 2.1 0.98214 -2.4 0.00820 -0.85 0.19766 0.65 0.74215 2.15 0.98422 -2.35 0.00939 -0.8 0.21186 0.7 0.75804 2.2 0.98610 -2.3 0.01072 -0.75 0.22663 0.75 0.77337 2.25 0.98778 -2.25 0.01222 -0.7 0.24196 0.8 0.78814 2.3 0.98928 -2.2 0.01390 -0.65 0.25785 0.85 0.80234 2.35 0.99061 -2.15 0.01578 -0.6 0.27425 0.9 0.81594 2.4 0.99180 -2.1 0.01786 -0.55 0.29116 0.95 0.82894 2.45 0.99286 -2.05 0.02018 -0.5 0.30854 1 0.84134 2.5 0.99379
-2 0.02275 -0.45 0.32636 1.05 0.85314 2.55 0.99461 -1.95 0.02559 -0.4 0.34458 1.1 0.86433 2.6 0.99534 -1.9 0.02872 -0.35 0.36317 1.15 0.87493 2.65 0.99598 -1.85 0.03216 -0.3 0.38209 1.2 0.88493 2.7 0.99653 -1.8 0.03593 -0.25 0.40129 1.25 0.89435 2.75 0.99702 -1.75 0.04006 -0.2 0.42074 1.3 0.90320 2.8 0.99744 -1.7 0.04457 -0.15 0.44038 1.35 0.91149 2.85 0.99781 -1.65 0.04947 -0.1 0.46017 1.4 0.91924 2.9 0.99813 -1.6 0.05480 -0.05 0.48006 1.45 0.92647 2.95 0.99841 -1.55 0.06057 0 0.50000 1.5 0.93319 3 0.99865 -1.5 0.06681
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The normal distribution (cont.)
Answer:
Equivalent standard draw is: z* = (7- 4)/9 = 1.
The probability of drawing a number larger than 7 is Prob(x>7) or Prob(z>z*). This is equal to 1-N(z*) or
1-N(1) = 1 - 0.841345 = 15.87%.
The probability of drawing a number lower than 7 is Prob(x<7) or Prob(z<z*). This is equal to N(z*) or N(1) = 0.841345 = 84.13%.
Obtaining exactly the number 7 has a zero probability.
11
The normal distribution (cont.)
The sum of normal random variables is also normal:
where xi, i = 1,…,n, are n random variables, with mean E(xi) = i, variance Var(xi) =i
2, covariance Cov(xi,xj) = ij = ijij
Ex: ax1+bx2 ~ N(a1+b2, a21+b2
2+2ab12)
Question: how is this variance obtained ?
i i i i i j ijj
n
i
n
i
n
i
nx N~ ,
1111
12
The normal distribution (cont.)
Example:
The variance of the sum of the two variables is:
Var(ax1+bx2) = Cov(ax1+bx2 , ax1+bx2)
= a2Cov(x1,x1) + 2abCov(x1,x2) + b2Cov(x2,x2)
= a21 + 2ab12 + b2
2
= a21 + b2
2 + 2ab12
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The normal distribution (cont.)
Example: take the case of two identically and independently distributed returns x1 and x2
(assume we are talking about continuously compounded returns, so you can add them).
Using the last two slides, what would be the resulting means and variances?
14
The normal distribution (cont.)
If x1 and x2 are iid and are summed, we have:
1=2= 1=2= =0 (since they are independent) a=1 and b=1
Hence we get:
x1+x2 ~ N(2, 2)
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The lognormal distribution
A random variable x is lognormally distributed if ln(x) is normally distributed If x is normal, and ln(y) = x (or y = ex), then y is
lognormal If continuously compounded stock returns are normal
then the stock price is lognormally distributed
Product of lognormal variables is lognormal If x1 and x2 are normal, then y1=ex
1 and y2=ex2 are
lognormal. The product of y1 and y2: y1 x y2 = ex
1 x ex2 = ex
1+x
2
Since x1+x2 is normal, ex1+x
2 is lognormal
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The lognormal distribution (cont.)
The lognormal density function:
where S0 is initial stock price, and ln(S/S0)~N(m,v2), S is the future stock price, m is the mean of the continuously compounded return and v is standard deviation of the continuously compounded return
If x ~ N(m,v2), then
20ln( / )1
20
1( ; , , )
2
S S m
vg S m v S eSv
E e ex m v( )
1
22
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A lognormal model of stock prices If the stock price St is lognormal, St / S0 = ex, where x,
the continuously compounded return from 0 to t is normal
If R(t, s) is the continuously compounded return from t to s, and, t0 < t1 < t2, then R(t0, t2) = R(t0, t1) + R(t1, t2)
From 0 to T, E[R(0,T)] = nh , and Var[R(0,T)] = nh2
(we showed it in the two-period cases earlier)
If returns are iid, the mean and variance of the continuously compounded returns are proportional to time
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A lognormal model of stock prices (cont.)
If we assume that
ln(St / S0) ~ N [( – – 0.52)t, 2t]
then ln(St / S0) = ( – – 0.52)t + tz
and therefore St = S0e( – – 0.52)t + tz
Exercise: compute the value of E(St) by using the fact that if x ~ N(m,v2), then
E(ex) = em+(1/2)v2 . (Note that S0 is constant)
20
A lognormal model of stock prices (cont.)
Let x = ln(St / S0) ~ N [( – – 0.52)t, 2t]
Thus m = ( – – 0.52)t
and v2 = 2t
Then E(ex) = E(St / S0) = em+(1/2)v2
= e( – – 0.52)t +(1/2)2t = e( – )t
and therefore E(St ) = S0e( – )t
21
A lognormal model of stock prices (cont.)
If the stock price is lognormally distributed, we can use the fact that the distribution is known to compute a number of probabilities and expectations:
Probability that the option will expire in-the-money.
Expected stock price, given that the option expires in the money.
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A lognormal model of stock prices (cont.)
If current stock price is S0, the probability that the option will expire in the money, i.e.,
where the expression for d2 contains , the true expected return on the stock instead of the risk-free rate r.(if you used r, you would obtain the risk-neutral probability of expiring in the money)
Prob( ) ( )S K N dt 2
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A lognormal model of stock prices (cont.)
Using the lognormal distribution, confidence intervals can be derived.
One might be interested in computing the prices St
L and StU such that:
Prob (St<StL) = p/2 and Prob (St
U < St ) = p/2
The probability of being in the tails of the distribution is split in two, half for each tail.
We will then be (1-p)% confident that the final stock price will be between SL and SU.
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Lognormal probability calculations Prices St
L and StU such that Prob (St<St
L) = p/2 and Prob (St
U < St ) = p/2 are:
Example: use the cumulative normal distribution table seen earlier to derive the 95% confidence interval if S0=100, t=2, =0.10, =0, =0.30
2 11( ) ( / 2)
20
t tN pLtS S e
2 11( ) ( / 2)
20
t tN pUtS S e
25
Lognormal probability calculations
Answer:
There is a 95% probability that in two years the stock will be between $48.60 and $256.40
21(0.10 0 0.3 )(2) (0.3) 2 (1.96)
20 48.6L
tS S e
21(0.10 0 0.3 )(2) (0.3) 2 (1.96)
20 256.4U
tS S e
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Lognormal probability calculations (cont.)
Given the option expires in the money, what is the expected stock price? The conditional expected price
where the expressions for d1 and d2 contain , the true expected return on the stock in place of r, the risk-free rate
E S S K SeN d
N dt t
t( | )( )
( )( ) 1
2