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Review Discrete Distributions Binomial distribution, Negative binomial distribution, Hypergeometric distribution, Poisson distribution

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Review

Discrete Distributions

• Binomial distribution,

• Negative binomial distribution,

• Hypergeometric distribution,

• Poisson distribution.

Expected Value

• If X is a discrete rv and p(x) is the value of its probability distribution at x, the expected value of X is defined as

x

X xpxXE )()(

Example

• Toss a coin 4 times. X = number of heads. What’s E(X) ?

• The pmf of X is

x: 0 1 2 3 4

p(x): 1/16 4/16 6/16 4/16 1/16

• So,

.216

14

16

43

16

62

16

41

16

10)( XE

Example

• Let X be a Bernoulli rv with pmf

Then E(X) = 0p(0) + 1p(1) = p. So the expected value of X is just the probability that X takes on the value 1.

1

0 1)(

xp

xpxp

Example• X = number of children born up to and

including the first boy. The pmf of X is

• Then

,...3,2,1 ,)1()( 1 xppxp x

.1

])1( [

)1( )()(

1

1

1

pp

dp

dp

ppxxpxXE

x

x

x

x

Expected Value of a Function of a RV

• If a rv X has a pmf p(x), then the expected value of any function h(X) is computed by

• Special case: h(x) = a·x + b. E(a X + b) = a·E(X) + b. Why?

x

Xh xpxhXhE )()()]([( )(

Variance

• The expected value measures the center of a probability distribution.

• Variance measures the variability of a pmf.

Variance

• Let X have pmf p(x) and expected value . Then the variance of X, denoted by

• The standard deviation (SD) of X is

).()-(x ])[( V(X)

is ,just or ,or )(22

22

xpXE

XV x

).(XVx

Example

• If X has pmf :

x 1 2 6 8

p(x) .4 .1 .3 .2

Then = 1×.4 + 2×.1 + 6×.3 + 8×.2 = 4 . 2 = (1 - 4)2×.4 + (2 - 4)2 × .1 + (6 - 4)2 ×.3

+ (8 - 4)2 ×.2 = 8.4.

and = 2.90.

A Shortcut Formula

• Proof:

.)]([)()( 222 XEXEXV

.)]([)()(2)(

) X2-(])[( V(X)2222

222

XEXEXEXE

XEXE

Rules of Variance

• In particular,

.||

,)(

222

xbaX

XbaX

a

abaXV

.

|| ,22

222

XbX

xaXXaX aa

Moments

• The kth moment about the origin of a rv X, denoted by µk

’ , is the expected value of Xk, , symbolically,

µk’ = E(Xk) = x xk · p(x).

• The kth moment about the mean of a rv X, denoted by µk, is the expected value of (X - µ)k, , symbolically,

µk = E[(X - µ)k] = x (x - µ)k · p(x).

Special Cases

• The expectation, or the mean, is the 1st moment about the origin.

µ = µ1’ = E(X) = x x · p(x).

• The variance is the 2nd moment about the mean

2 = µ2 = E[(X - µ)2] = x (x - µ)2 · p(x).

The Binomial Distribution

Binomial Distribution

• For X ~ Bin(n,p), the cdf will be denoted by

.,...,2,1,0 )1(),;( nxppx

npnxb xnx

x

y

pnybpnxBxXP0

),;(),;()(

Mean & Variance

If X ~ Bin(n, p), then

• E(X) = np,

• V(X) = npq (where q = 1-p.)

.npqx

Example(Cont)

• n = 5, p = 11/32 . Then

• E(X) = n · p = 5 · 11/32 = 1.72.

• V(X) = n · p · q = 5 · 11/32 · 21/32 = 1.13. = (1.13)1/2 = 1.06.

Hypergeometric and Negative Binomial

Distribution

Introduction

• The hypergeometric and negative binomial distribution are both closely related to the binomial distribution.

Introduction

• The negative binomial distribution arises from fixing the number of S’s and letting the number of trials to be random.

• The hypergeometric distribution is the exact probability model for sampling without replacement from a finite dichotomous (S,F) population.

Negative Binomial Dist’n

• The experiment consists of a sequence of independent trials.

• Each trial results in either S or F.

• The probability of success, p, is constant from trial to trial.

• Trials are performed until a total of s successes have been observed, where s is a prespecified positive integer.

Negative Binomial RV

• X = the number of F’s that precede the rth success, is called a negative binomial rv.

• Possible values of X are 0, 1, 2, …

pmf

• Denote by nb(x; r, p) the pmf of X. Then

• Why?

• Total # of trials = x; The last trial must be a success. Among the first (x-1) trials, there are (s - 1) successes & x-s failures.

,...2,1 ,)1(1

1),;(

xpps

xpsxnb sxs

Review of Chapter 3• Hypergeometric distribution,

• Poisson distribution.

2,... 1, 0, x,!

);(

x

exp

x

).,min(),0max(for

),,;()(

MnxMNn

n

Nxn

SN

x

S

NSnxhxXP

Example

• What’s the probability that < 3 requests are received during a particular hour?

• P( X < 3) = P(0) + P(1) + P(2)

= e-5 + 5· e-5 + 52 · e-5/2

= 0.125.

Example

• What’s the probability that exactly 10 requests are received during a particular 2-hour period?

• Rate = 2 × 5 = 10.

• P(X = 10) = e-10 1010/10! = 0.125.

Example

• How many calls do they expect to get during a 45-min period?

• E(X) = (3/4) · 5 = 3.75.

Continuous RVs&

Probability Distributions

Continuous RV

• An rv X is continuous if its set of possible values is an entire interval of numbers.

Example:

• X = the pH of a random soil sample

• X = the weight of a randomly selected

person.

pdf

• Let X be a continuous rv. Then a probability density function (pdf) of X is a function f(x) such that for any two numbers a and b with a b,

• For f(x) to be a pdf, f(x) must satisfy:

f(x) 0 for all x, and

b

adxxfbXaP .)()(

.1)(

dxxf

Example• Waiting time at a bus station. A bus

arrives every 10 minutes. So the waiting time is from 0 to 10. One possible pdf for waiting time X is

• The probability of waiting between 3 to 5 minutes is:

otherwise. 0

100 ,10/1)(

xxf

.2.03

5 1.0 1.0)53(

5

3 xdxXP

Uniform Distribution

• A continuous rv X is said to have a uniform distribution on the interval [A, B] if the pdf of X is

• Graphs of uniform distributions.

otherwise 0

)/(1) , ;(

BxAABBAxf

Probability at a Point

• When X is a discrete rv, each possible value is assigned positive probability. This is no longer true for continuous rv.

• If X is a continuous rv, then for any number c, P(X = c) = 0. Consequently, P(a X b) = P(a < X b) = P(a X < b)

= P(a < X < b).

Example

• Let X = the “time headway” for two randomly chosen consecutive cars on a freeway during a period of heavy flow. Suppose the pdf of X is given by:

f(x) = 0.15 e-0.15( x - 0.5), x 0.5.

f(x) = 0 for x < .5 and f(x) decreases exponentially fast as x increase from .5.

Example• First, it clear that f(x) 0. Now we verify

• The probability that headway time is at most 5 seconds is

.115.

115.

15.15.)(

)5)(.15(.075.

5.

15.075.

5.

)5.(15.

ee

dxeedxedxxf xx

.491.)(5.

5)

15.

1(15.

15.15.)()5(

075.75.075.15.075.

5

5.

15.075.5 5

5.

)5.(15.

eee

x

xee

dxeedxedxxfXP

x

xx

CDFs & Expected Values

cdf

• The cumulative distribution function (cdf) F(x) for a continuous rv X is defined for every number x by

• For each x, F(x) is the area under the density curve to the left of x. It is the probability of observing X a value smaller than or equal to x.

xdyyfxXPxF .)()()(

Example

• Let X have a uniform distribution on the interval [A, B]. Then

• So, for x < A, F(x) = 0 and for x B, F(x) = 1. For A x B,

otherwise 0

)/(1) , ;(

BxAABBAxf

.1

)()(

x x

A AB

Axdy

ABdyyfxF

Example

• The entire cdf is:

• The graph of the cdf looks like:

. 1

0

)(

Bx

BxAAB

AxAx

xF

Propositions

• Compute probabilities using F(x): P(a x b) = F(b) - F(a).

• Obtaining pdf from cdf:

• If X is a continuous rv with cdf F(x) differentiable at every point x, then the pdf f(x) =F ’(x).

Example

• For uniform distribution on [A, B], the cdf is

• So, for example, if A < a < b < B, then P(a < X < b) = F(b)-F(a) = (b-a)/(B-A).

• The pdf

• f(x) = F ’(x) = 1/(B-A) for A < x < B.

. 1

0

)(

Bx

BxAAB

AxAx

xF

Expected Values

• The expected value (or, mean) of a continuous rv X with pdf f(x) is

• If X is a continuous rv with pdf f(x) and h(X) is any function of X, then

.)()( dxxfxXEX

.)()())(()( dxxfxhXhEXh

Example

• The pdf of the waiting time (in minutes) at a checkout is given by

f(x) = x/8 for 0 x < 4.

• What’s the probability of waiting less than 3 min?

• What’s the expectation of the waiting time?

Example

• What’s the probability of waiting less than 3 min?

• What’s the expectation of the waiting time?

.5625.16

9

0

3

16

8)3(

23

0

x

xxdxx

XP

.667.224

64

0

4

24

8)(

34

0

x

xxdxx

xXE

Variance & S.D.

• The variance of a continuous rv X with pdf f(x) and mean is

• The standard deviation (S.D.) of X is

• V(X) = E(X2) - [E(X)]2.

].)[()()()( 222 xEdxxfxXVX

).(XVx

Linear Transformation

• If h(X) = a X + b and V(X) = 2, then

V(h(X))=V(a X + b) = a 2 2

and

aX+b = |a| .

Example(Cont)

• The pdf of the waiting time at a checkout:

f(x) = x/8 for 0 x < 4.

• Find the variance of the waiting time. = E(X) = 2.667.

.889.667.28)]([)()(

.832

256

0

4

32

8)(

222

44

0

22

XEXEXV

x

xxdxx

xXE

Normal Distribution

Introduction

• The normal distribution is the most important distribution in all of probability and statistics.

• Many numerical populations have distributions that can be approximated very well by a normal curve.

Example

• Scores of standardized tests,

• Measurements of intelligence & aptitude,

• Returns of a stock (or a portfolio),

• Measurement errors …

Definition

• A continuous rv X is said to have a normal distribution with parameters and 2 if the pdf of X is

. 2

1), ;(

2

2

2

)(2

xexfx

Remarks

• Notation: X ~ N(, 2).

• It’s clear that f(x; , 2) 0 and it can be shown that

• E(X) = , and V(X) = 2.

.1),;( 2 dxxf

Standard Normal Dist’n

• With = 0 and = 1, the normal distribution is called a standard normal distribution.

• The pdf of a standard normal rv Z is

• The cdf of Z is denoted by (z).

. 2

1)1 ,0 ;( 2/2

zezf z

Normal Probability Table

• Table A.3 on page 704 of the text tabulates the standard normal probabilities (cdf). This is one of the most useful statistical tables.

• Example: Using the table to compute:* P(Z < 1.20), * P(Z > 1.68), (= 1 - P( Z 1.68)) * P(-1.96 < Z < 1). (= P( Z < 1) - P( Z -1.96))

Inverse Reading of Table A.3

• Z denotes the (100)th percentile of the standard normal distribution.

• The area under the standard normal curve to the right of Z (tail probability) is .

• Find: Z.30, Z.90.

Standardization

• If Z ~ N(0, 1), then X = + Z ~ N(, 2).

• Inversely, if X ~ N(, 2), then

Z = (X - )/ ~ N(0, 1).

• The transformation

Is called standardization.

• P(X x) =P[Z (x - )/] = [(x - )/].

X

ZX

Standardization

• (100p)th percentile for N(, 2)

= + · (100p)th percentile for N(0, 1).

• So if X ~ N(, 2), then

X = + · Z .

Rule of Thumb

If X is (approximately) normal, then

• about 68% of the x's are within 1 SD of the mean;

• about 95% of the x's are within 2 SDs of the mean;

• about 99.7% of the x's are within 3 SDs of the mean;

Example(Fish)

The lengths of fish in a certain fish population follows a normal distribution with = 54 mm and = 4.5 mm.

• What percentage of the fish are between 50 and 60 mm long?

* Let Z = (X - )/. Then

z1=(50 - 54)/4.5= -.89, z2=(60 - 54)/4.5=1.33.

Use Table A.3: P(50 X 60)=P(-.89Z1.33) =.9082- .1867 = .7215.

Example(Fish)

• What percentage of the fish are more than 48 mm long?

* z1 = (48 - 54)/4.5 = - 1.33.

P( X > 48) = 1- .0918 = .9082.

• What percentage of the fish are between 58 and 60 mm long?

* z1 = (58 - 54)/4.5 = 0.89, z2 = 1.33. P(58 < X < 60) = .9082 - 0.8133 = .0949.

Example(Fish)

• What is the 70th percentile of the fish length ? What is the 90th percentile?

* From Table A.3, Z.70 = 0.52. So, X.70 = 54 + 4.5 ·0.52 = 56.3

* Similarly, Z.90 = 1.29. and

X.90 = 54 + 4.5 ·1.29 = 59.80.

Example (Height)

Among American women aged 18 - 24, 10% are less than 61.2 inches tall; 80% are between 61.2 and 67.4 inches and 10% are more than 67.4 inches. Assume the height can be well approximated by a normal distribution.

• Find the mean and the SD .

Example(Height)

* Z.10 = -1.29 and Z.90 = 1.29 , so

Solving for and , we have = (67.4 - 61.2)/(1.29 + 1.29) = 2.4, and

= 64.3.

29.1 4.67

29.12.61

Normal Approximation

• The normal distribution is often used to approximate the distribution of discrete populations.

• In particular, under certain conditions, the normal distribution can be used as an approximation to the binomial distribution.

Normal Approximation to Binomial Distribution

• For a binomial rv X , we have

• When both np and nq are relatively large, the normal distribution with the same mean and SD is a very good approximation to Bin(n, p).

. and , npqnp XX

Normal Approximation to Binomial Distribution

• Let X ~ Bin(n, p), Then if np 5 and nq 5, X has approximately a normal distribution with

• This is the area under the normal curve to the left of x+.5. “+.5” is the correction for discreteness. This is called continuity correction.

. and , npqnp XX

).5.

(),;()(npq

npxpnxBxXP

Example • X ~ Bin(30, 0.3). Want: P(6 X 10).

• Mean=30 × .3 = 9, SD = (30× .3× .7)1/2=2.51.

• P(6 X 10) = P(X 10) - P(X 5) ((10 + .5 - 9)/2.51) - ((5 + .5 - 9)/2.51)

= (.598) - (-1.394)= .7257 - .0832= .6425.

• Direct calculation yields

P(6 X 10) = P(6) + … + P(10) = .6437.

• The results are very close.