1 mf-852 financial econometrics lecture 4 probability distributions and intro. to hypothesis tests...

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MF-852 Financial Econometrics

Lecture 4 Probability Distributions and Intro. to

Hypothesis Tests

Roy J. EpsteinFall 2003

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Distribution of a Random Variable A random variable takes on

different values according to its probability distribution.

Certain distributions are especially important because they describe a wide variety of random variables.

Binomial, Normal, student’s t

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Binomial Distribution Random variable has two

outcomes, 1 (“success”) and 0 (“failure”) Coin flip: heads = 1, tails = 0 P(success) = p P(failure) = q = (1 – p)

Binomial distribution yields probability of x successes in n outcomes.

Excel will do the calculations.

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Tails and Body of a Distribution

Binomial Distributionp = 0.4, n = 8

0.0%

5.0%

10.0%

15.0%

20.0%

25.0%

30.0%

0 1 2 3 4 5 6 7 8

Successes

Pro

bab

ility

upper tail

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Binomial Example (RR p. 20) Medical treatment has p = .25. n = 40 patients What is probability of at least 15

successes (cures) I.e, P(x 15)?

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Normal Distribution A normally distributed random

variable: Is symmetrically distributed around

its mean Can take on any value from – to + Has a finite variance Has the famous “bell” shape

“Standard normal:” mean 0, variance 1.

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Standard Normal Distribution

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0.350

0.400

0.450

-3

-2.9

-2.7

-2.6

-2.4

-2.3

-2.1 -2

-1.8

-1.7

-1.5

-1.4

-1.2

-1.1

-0.9

-0.8

-0.6

-0.5

-0.3

-0.2 0

0.1

5

0.3

0.4

5

0.6

0.7

5

0.9

1.0

5

1.2

1.3

5

1.5

1.6

5

1.8

1.9

5

2.1

2.2

5

2.4

2.5

5

2.7

2.8

5 3

z

f(z)

tail area

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N(0, .5) Distribution

0.000

0.200

0.400

0.600

0.800

1.000

1.200

1.400

1.600

1.800

-3

-2.9

-2.7

-2.6

-2.4

-2.3

-2.1 -2

-1.8

-1.7

-1.5

-1.4

-1.2

-1.1

-0.9

-0.8

-0.6

-0.5

-0.3

-0.2 0

0.15 0.3

0.45 0.6

0.75 0.9

1.05 1.2

1.35 1.5

1.65 1.8

1.95 2.1

2.25 2.4

2.55 2.7

2.85 3

z

f(z)

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N(0,1) Probabilities Suppose z has a standard normal

distribution. What is: P(z 1.645)? P(z –1.96)? Excel will tell us!

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N(0,1) and Standardized Variables Suppose x is N(12,10).

What is P(x 24.8) ?

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Key Properties of Normal Distribution Sum of 2 normally distributed

random variables is also normally distributed.

The distribution of the average of independent and identically distributed NON-NORMAL random variables approaches normality. Known as the Central Limit Theorem Explains why normality is so

pervasive in data

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Sample Mean Take a sample of n independent

observations from a distribution with an unknown . Data are n random variables x1, …

xn.

We estimate the unknown population mean with the sample mean “xbar”:

n

1in

1xx

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Properties of Sample Mean

Sample mean is unbiased!

11)(

11)(

111

n

nnxE

nxE

nxE

nn

i

n

i

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Properties of Sample Mean

Sample mean has variance. But the variance is reduced with more data.

n

nnn

xVarn

xVarn

xVarnn

i

n

i

2

2

1

2

12

12

11)(

11)(

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Null Hypothesis “Null hypothesis” (H0) asserts a

particular value (0) for the unknown parameter of the distribution.

Written as H0 : = 0 E.g., H0 : = 5

H0 usually concerns a value of particular interest (e.g., given by a theory)

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Null Hypothesis xbar is unlikely to equal 0

exactly. Samples have sampling error, by

definition. Is xbar still consistent with H0

being a true statement? This involves a hypothesis test.

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Hypothesis Testing Hypothesis testing finds a range

for called the confidence interval.

The confidence interval is the set of acceptable hypotheses for , given the available data.

H0 is accepted if the confidence interval includes 0.

Otherwise H0 is rejected.

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Confidence Interval confidence interval = xbar

allowable sampling error How wide should the interval be

around xbar? Customary to use a 95%

confidence interval. The interval will include the true

95% of the time Each tail probability is 2.5%.

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Construction of Confidence Interval If x1, … xn are normally distributed

then xbar is normally distributed. Then:

The 95% confidence interval is

%95)96.1)/(

96.1( 0

n

xP

%95)96.196.1( 0 n

xn

xP

nx

96.1

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Confidence Interval Example You are a restaurant manager. Burgers

are supposed to weigh 5 ounces on average. The night shift makes burgers with a standard deviation of 0.75 ounces.

You eat 12 burgers from the night shift and xbar is 5.4 ounces. What is a 95% confidence interval for the weight of the night shift burgers?

You eat 8 more burgers that have an average weight of 5.25 ounces. What is a 95% confidence interval for this sample?

What is a 95% confidence interval based on all 20 burgers?

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Sample Variance Usually the population variance, as

well as the mean, is unknown. Estimate 2 with the sample

variance:

We divide by n-1, not n. What is the sample variance of xbar?

n

i xxn

s1

22 )(1

1

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Sample Variance Usually the population variance, as

well as the mean, is unknown. Estimate 2 with the sample

variance:

We divide by n-1, not n. What is the sample variance of xbar?

n

i xxn

s1

22 )(1

1

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t-distribution Confidence intervals use the t-

distribution instead of the normal when the variance is estimated from the sample.

T-distribution has fatter tails than the normal.

Confidence intervals are wider because we have less information.

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t distribution (3 dof)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

-3

-2.9

-2.7

-2.6

-2.4

-2.3

-2.1 -2

-1.8

-1.7

-1.5

-1.4

-1.2

-1.1

-0.9

-0.8

-0.6

-0.5

-0.3

-0.2 0

0.15 0.3

0.45 0.6

0.75 0.9

1.05 1.2

1.35 1.5

1.65 1.8

1.95 2.1

2.25 2.4

2.55 2.7

2.85 3

t

f(t)

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Confidence Interval with t-distribution You hired Leslie, a new salesperson.

Leslie made the following sales each month in the first half:

January — $25,000 April — $20,000 February — $27,000 May — $22,000 March — $29,000 June —

$35,000 What is a 95% confidence interval for

Leslie’s monthly sales? (assume monthly sales are normally distributed)

Suppose you knew that the standard deviation of sales was $1,500. How would your conclusion change?

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Significance Levels Assuming H0, what is the

probability that the sample value would be as extreme as the value we actually observed? Alternative to confidence interval

Equal to

variatesnormalfor ))/(

( 0

n

xzP

atesfor t vari))/(

( 0

ns

xtP

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Type 1 and Type 2 Error Accept or reject H0 based on the

confidence interval. Type 1 error: reject H0 when it is

true. What is probability of this?

Type 2 error: accept H0 when it is false. How important is this?