review midtermii summer09

7/28/2019 Review MidtermII Summer09

1/51

Introduction to Probability and Statistics

Probability & Statistics for Engineers & Scientists, 8th Ed.

2007

Review II

Instructor: Kuo-Jung Lee

TA: Brian Shea

The pdf file for this class is available on the class web page.

http://www.stat.umn.edu/~kjlee/STAT3021_Summer2009.html

1


2/51

Marginal Distribution

The marginal distributions of X alone and of Y alone are

Discrete case:

g(x) =

yf(x, y) and h(y) =

x

f(x, y),

Continuous case:

g(x) =

f(x, y)dy and h(y) =

f(x, y)dx,

2


3/51

Conditional Distribution

Let X and Y be two random variables, discrete or continuous.

The conditional distribution of the random variable Y given

that X = x is

f(y|x) = f(x, y)g(x)

, g(x) > 0,

Similarly the conditional distribution of the random variable X

given that Y = y is

f(x|y) = f(x, y)h(y)

, h(y) > 0.

3


4/51

Statistical Independent

Let X and Y be two random variables with joint probability dis-

tribution f(x, y) and marginal distributions g(x) and h(y), respec-tively. The random variables X and Y are said to be statistically

independent if and only if

f(x, y) = g(x)h(y)

for all (x, y) within their range.

4


5/51

Example 1

Consider the following joint probability density function of the

random variables X and Y

f(x, y) =

12ye

x, 0 < x, 0 < y < 2;0, elsewhere.

1. Find the marginal density functions of X and Y.

2. Are X and Y are independent?

3. Find P(X > 2).

5


6/51

Solution:

1.

g(x) =

yf(x, y)dy = ex, x > 0;

h(y) =

xf(x, y)dy =

y

2, 0 < y < 2.

2. Since f(x, y) = g(x)h(y), they are independent.

3.

P(X > 2) = e2.


7/51

Definition: Expectation

Let X be a random variable with probability distribution f(x).

The mean or expected value of X is:

if X is discrete = E(X) =

x

xf(x)

if X is continuous

= E(X) =

xf(x)dx

6


8/51

Theorem

Let X be a random variable with probability function f(x). The

expected value of the random variable g(X) is

Discrete: if X is discreteg(X) = E[g(X)] =

x

g(x) f(x)

Continuous: if X is continuous

g(X) = E[g(X)] =

g(x) f(x)dx

7


9/51

Definition

Let X and Y be random variables with joint probability distribu-

tion f(x, y). The mean or expected value of g(X, Y) is:

if X and Y are discreteg(X,Y) = E[g(X, Y)] =

x

y

g(x, y)f(x, y)

if X and Y are continuousg(X,Y) = E[g(X, Y)] =

g(x, y)f(x, y)dxdy

8


10/51

Definition: Variance

Let X be a random variable with probability function f(x) and

mean . The variance of the random variable X is

Discrete: if X is discrete2 = Var(X) = E[(X )2] =

x(x )2 f(x)


2

= Var(X) = E[(X )2

] =(x )

2

f(x)dx

The positive square root of the variance, , is called the stan-

dard deviation of X.9


11/51

Theorem

The variance of a random variable X is

2 = Var(X) = E(X2) 2

10


12/51

Theorem

Let X be a random variable with probability function f(x). The

variance of the random variable g(X) is

Discrete: if X is discrete2g(X) = E[g(X) g(X)]2 =

x

[g(x) g(X)]2 f(x)


2g(X) = E[g(X) g(X))]2 =

[g(x) g(X))]2 f(x)dx

11


13/51

Definition: Covariance

Let X and Y be random variables with joint probability distribu-

tion f(x, y). The covariance of the random variables X and Y

is

Discrete: if X and Y are discrete is discreteCov(X, Y) = E(XX)(yY) =

x

y

(xx)(yy)f(x, y)

Continuous: if X and Y are continuousCov(X, Y) = E(XX)(yY) =

(xx)(yy)f(x, y)dxdy

12


14/51

Theorem

The covariance of two random variables X and Y with means

X and Y, respectively, is given by

XY = Cov(X, Y) = E(XY) XY.

13


15/51

Theorem

If a and b are constants, then

E(aX + b) = aE(X) + b E(aX + bY) = aE(X) + bE(Y).

Theorem

The expected value of the sum or difference of two or morefunctions of two random variable X and Y is the sum or difference

of the expected values of the functions. That is,

E[g(X) h(Y)] = E[g(X)] E[h(Y)].Theorem

Let X and Y be two independent random variables. Then

E(XY) = E(X)E(Y).

14


16/51

Theorem

If a and b are constants, then

2aX+b = Var(aX + b) = a2Var(X).

Theorem

If X and Y are random variables with joint probability distribution

f(x, y) and a and b are constants, then

2aX+bY = Var(aX+bY) = a2Var(X)+2abCov(X, Y)+b2Var(Y)

Cov(aX + b,cY + d) = acCov(X, Y)

15


17/51

Example 2

Consider the following joint probability density function of the

random variables X and Y

f(x, y) = 1

2yex

, 0 < x, 0 < y < 2;0, elsewhere.

1. Find means and variance X and Y.

2. Find covariance of X and Y.

16


18/51

Solution:

1.

E(X) =

0xg(x)dx = 1;

Var(X) = 1.

E(Y) =2

0yh(y)dy =

1

6;

Var(Y) = E(Y2) (16

)2 =7

72.

2. Since f(x, y) = g(x)h(y), they are independent. Cov(X, Y) =

0.

17


19/51

Example 3

Let X and Y have joint probability density function (p.d.f.)

f(x, y) = 3x2y

32 , if 0 < x < 2 and 1 < y < 3;

0, otherwise.

1. Find the marginal probability density functions of X and Y,

respectively, and determine if X and Y are independent ?

2. Calculate Cov(0.652X,

17Y).

18


20/51

Solution:

1.

g(x) = 3

1

3x2y

32

dy =3x2

8

, 0 < x < 2;

h(y) =2

0

3x2y

32dy =

y

4, 1 < y < 3.

Since f(x, y) = g(x)h(y), they are independent.

2. Since X and Y are independent from (a), Cov(X, Y) = 0.

Cov(0.652X,

17Y) = 0.652

17 Cov(X, Y) = 0.

19


21/51

Example 4

Let X and Y be jointly distributed with (X, Y) = 1/2, X = 2,

andY

= 3. Find Var(2X

4Y

+ 3).

Solution:

Var(2X 4Y + 3) = Var(2X 4Y)= 22Var(X) + 2 2 (4)Cov(X, Y) + 42Var(Y)= 132.

20


22/51

Markovs Inequality

Let X be a nonnegative random variable; then for any t > 0,

P(X t) E(X)t

.

Theorem (Chebyshevs Inequality)

The probability that any random variable X will assume a value

within k standard deviations of the mean is at least 1 1k2

. That

is,

P( k < X < + k) 1 1k2

21


23/51

Discrete Probability Distributions

Name Notation P.D.F. f(x) = P(X = x)Uniform 1k , x = x1, . . . , xk.

Bernoulli Ber(p) px(1 p)1x, x = 0, 1.Binomial Bin(n, p) px(1 p)nx, x = 0, 1, . . . , n

Multinomial nx1, x2, . . . , xk

px11 px22 p

xk

kGeometric Geo(p) pqx1, x = 1, 2, 3, . . . .

Negative Binomial NB(k, p)

x 1k 1

pkqxk, x = k, k + 1, . . . .

Hypergeometric Hyp(N , n , k)

k

x

N kn x

Nn

Poisson Poi(t) et(t)x

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

22


24/51

Discrete Probability Distributions

Name Mean Variance M.G.F.

Bernoulli p p(1

p) pet + qBinomial np np(1 p) (pet + q)n

Geometric 1p1pp2

pet

1qet

Negative Binomial kpk(1p)

p2

pet

1

qet

k

Poisson t t e(es1)

23


25/51

Example 5: Binomial Distribution

The probability that a student is accepted to a prestigious college

is 0.3. If 5 students from the same school apply, what is the

probability that at most 2 are accepted?

Solution:

To solve this problem, we compute 3 individual probabilities,

using the binomial formula. The sum of all these probabilities

is the answer we seek. Let X be the number of students are

accepted, X

Bin(5, 0.4). Thus,

P(X 2) =2

x=0

b(x; 5, 2) = 0.8369.

24


26/51

Example 6: Multinomial Distribution

Suppose a card is drawn randomly from an ordinary deck ofplaying cards, and then put back in the deck. This exercise is

repeated five times. What is the probability of drawing 1 spade,

1 heart, 1 diamond, and 2 clubs?

Solution:

The experiment consists of 5 trials, so n = 5.

The 5 trials produce 1 spade, 1 heart, 1 diamond, and 2clubs; so n1 = 1, n2 = 1, n3 = 1, and n4 = 2.

On any particular trial, the probability of drawing a spade,heart, diamond, or club is 0.25, 0.25, 0.25, and 0.25, respec-

tively. Thus, p1 = 0.25, p2 = 0.25, p3 = 0.25, and p4 = 0.25.

25


27/51

5

1, 1, 1, 3

(0.25)1(0.25)1(0.25)1(0.25)3 = 0.05859.

Thus, if we draw five cards with replacement from an ordinary

deck of playing cards, the probability of drawing 1 spade, 1 heart,

1 diamond, and 2 clubs is 0.05859.


28/51

Example 7: Hypergeometric Distribution

Suppose we randomly select 5 cards without replacement from

an ordinary deck of playing cards. What is the probability of

getting exactly 2 red cards (i.e., hearts or diamonds)?

Solution:

This is a hypergeometric experiment in which we know the fol-

lowing:

N = 52; since there are 52 cards in a deck.

k = 26; since there are 26 red cards in a deck.

n = 5; since we randomly select 5 cards from the deck.26


29/51

x = 2; since 2 of the cards we select are red.

We plug these values into the hypergeometric formula as follows:

h(X = x; N , n , k) =

kx

N kn

x

Nn

h(X = 2; 52, 5, 26) =

262

52 26

5

2

525

= 0.32513.


30/51

Example 8: Negative Binomial & Geometric Distributions

Bob is a high school basketball player. He is a 70% free throw

shooter. That means his probability of making a free throw is

0.70. During the season, what is the probability that Bob makes

his third free throw on his fifth shot?Solution:

This is an example of a negative binomial experiment. The

probability of success p is 0.70, the number of trials x is 5, and

the number of successes k is 3.

nb(5;3, 0.7) =

42

(0.7)3(0.3)2 = 0.18522.

27


31/51

Example 9: Poisson Distribution

The average number of homes sold by the Acme Realty companyis 2 homes per day. What is the probability that exactly 3 homes

will be sold tomorrow?

Solution:

This is a Poisson experiment in which we know the following:

= = 2; since 2 homes are sold per day, on average.

t = 1; since unit time is one day.

x = 3; since we want to find the likelihood that 3 homes willbe sold tomorrow.

p(x = 3; t = 2) = 0.180

28


32/51

Continuous Probability Distributions

Name Notation P.D.F. f(x)

Uniform U[a, b] 1ba, a x b.

Normal N(, 2)1

2 exp{(x

)2

22 }, < x < .Exponential Exp() 1e

x/, x > 0.Gamma (, ) 1

()x1ex/, x > 0.

Chi-Squared 2 =

2, 2

1

2/2(/2)x/21ex/2, x > 0.

Lognormal LogN(, 2) 1x

2e 122 [ln(x)]2, x > 0

29


33/51

Continuous Probability Distributions

Name Mean Variance M.G.F.

Uniform a+b2 (ba)2

12

Normal 2 exp{t + 2t22 }Exponentail 2

Gamma 2

Chi-Squared 2

30


34/51

Normal Approximation to the Binomial-I

Let X be a binomial random variable with parameters n and p.

Then X has approximately a normal distribution with = np and

2 = npq = np(1 p) and

P(X x) = xk=0

b(k; n, p) (1)

area under normal curve to the left of x + 0.5 (2) P

Z x + 0.5 np

npq

(3)

where Z N(0, 1), and the approximation will be good if np andn(1 p) are greater than or equal to 5.

31


35/51

Normal Approximation to the Binomial-II

Let X be a binomial random variable with parameters n and p.

Then X has approximately a normal distribution with = np and

2 = npq = np(1 p) and

P(x1 X x2) =x2

k=x1b(k; n, p)

area under normal curve tothe right ofx1 0.5 and left ofx2 + 0.5.

Px1 0.5 npnpq Z

x2 + 0.5 npnpq

where Z N(0, 1), and the approximation will be good if np andn(1 p) are greater than or equal to 5.

32


36/51

Memorylessness for Geometric & Exponential Distribution

A nonnegative random variable X is called memoryless if for all

s, t 0,P(X t) = P(X t + s|X s)

33


37/51

Example 10

The position X of the first defect on a digital tape (in cm)

has the exponential distribution with mean = 50. Find the

probability that X < 200 given X > 150.Solution:

P(X < 200|X > 150) = 1 P(X > 200|X > 150)= 1 P(X > 50)

= 1 e1

.

34


38/51

Example 11

The lifetime of a TV tube (in years) is an exponential random

variable with mean 10. If Jim bought his TV set 10 years ago,what is the probability that its tube will last another 10 years?

Solution:

Let X be the lifetime of a TV tube. X Exp( = 10).P(X > 20

|X > 10) = P(X > 10) = e1.

35


39/51

Example 12

Let the probability density function of a random variable X be

f(x) =1

2

2e(x2)2

222 , if < x < .

Show that

P(|X 2| < 4) 34

.

Solution:

Since

X N(2, 22),by Chebyshevs inequality, we have

P(|X 2| < 4) = P(|X 2| < 2 2) 1 122

34

.

36


40/51

If we are sampling from a population with unknown distribu-

tion, either finite or infinite, the sampling distribution of X will

be approximately normal with mean and variance 2/n pro-

vided that the sample size is large (n > 30).

Central Limit Theorem

If X is the mean of a random sample of size n taken from a

population with mean and finite variance 2, then the limiting

form of the distribution of

Z =

X /nas n , is the standard normal distribution N(0, 1).

37


41/51

Example 13

A pair of fair 4-sided dice is rolled 192 times. Let T be the

number that a total of 5 occurs.

1. Find the probability function of T?

2. What are the mean (expected value) and variance of T?

3. What is the probability that a total of 5 occurs at most 49

times?

38


42/51

Solution:

1. T Bin

192, 14

. That is

f(x) =

nx

1

4

x 34

192x, x = 0, . . . , 192.

2. E(T) = 48, Var(T) = 36

3. Normal Approximation to Binomial:

P(T 49) P

T 486

review midtermii summer09

Documents