Jointly Distributed Random Variableslzhang/teaching/3070spring2009... · Two components of a...

Jointly Distributed Random Variables

Example (variant of Problem 12)Two components of a minicomputer have the following joint pdffor their useful lifetimes X and Y :

f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise

a. What is the probability that the lifetimes of both componentsexcceed 3?

b. What are the marginal pdf’s of X and Y ?

c. What is the probability that the lifetime X of the firstcomponent excceeds 3?

d. What is the probability that the lifetime of at least onecomponent excceeds 3?



f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise







f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise







f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise







f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise







f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise






Recall the following example (variant of Problem 12):Two components of a minicomputer have the following joint pdffor their useful lifetimes X and Y :

f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise

The marginal pdf’s of X and Y are

fX

(x) = xe−x and fY

(y) = e−y ,

respectively. We see that

fX

(x) · fY

(y) = f (x , y).



f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise


fX

(x) = xe−x and fY

(y) = e−y ,


fX

(x) · fY

(y) = f (x , y).



f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise


fX

(x) = xe−x and fY

(y) = e−y ,

respectively.

We see that

fX

(x) · fY

(y) = f (x , y).



f (x , y) =

{xe−(x+y) x ≥ 0 and y ≥ 0

0 otherwise


fX

(x) = xe−x and fY

(y) = e−y ,


fX

(x) · fY

(y) = f (x , y).


DefinitionTwo random variables X and Y are said to be independent if forevery pair of x and y values,

p(x , y) = pX

(x) · pY

(y) when X and Y are discrete

or

f (x , y) = fX

(x) · fY

(y) when X and Y are continuous

If the above relation is not satisfied for all (x , y), then X and Yare said to be dependent.


DefinitionTwo random variables X and Y are said to be independent if forevery pair of x and y values,

p(x , y) = pX

(x) · pY

(y) when X and Y are discrete

or

f (x , y) = fX

(x) · fY

(y) when X and Y are continuous

If the above relation is not satisfied for all (x , y), then X and Yare said to be dependent.


Examples:Problem 12. The joint pdf for X and Y is

f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise


fX

(x) = −e−x and fY

(y) =1

(1 + y)2,


fX

(x) · fY

(y) 6= f (x , y).



f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise


fX


(y) =1

(1 + y)2,


fX

(x) · fY

(y) 6= f (x , y).



f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise


fX


(y) =1

(1 + y)2,

respectively.

We see that

fX

(x) · fY

(y) 6= f (x , y).



f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise


fX


(y) =1

(1 + y)2,


fX

(x) · fY

(y) 6= f (x , y).


Examples:Our first example: tossing a fair die. If we let X = the outcome

of the first toss and Y = the outcome of the second

toss, then we will have

p(x , y) = pX

(x) · pY

(y)

Obviously, we know the two toss should be independent.

Our second example: dinning choices.

y

p(x , y) 12 15 20 p(x , ·)12 .05 .05 .10 .20

x 15 .05 .10 .35 .5020 0 .20 .10 .30

p(·, y) .10 .35 .55p

X(12) · p

Y(12) 6= p(12, 12).





p(x , y) = pX

(x) · pY

(y)



y

p(x , y) 12 15 20 p(x , ·)12 .05 .05 .10 .20

x 15 .05 .10 .35 .5020 0 .20 .10 .30

p(·, y) .10 .35 .55p

X(12) · p

Y(12) 6= p(12, 12).





p(x , y) = pX

(x) · pY

(y)



y

p(x , y) 12 15 20 p(x , ·)12 .05 .05 .10 .20

x 15 .05 .10 .35 .5020 0 .20 .10 .30

p(·, y) .10 .35 .55p

X(12) · p

Y(12) 6= p(12, 12).


DefinitionIf X1,X2, . . . ,Xn are all discrete random variables, the joint pmf ofthe variables is the function

p(x1, x2, . . . , xn) = P(X1 = x1,X2 = x2, . . . ,Xn = xn)

If the random variables are continuous, the joint pdf ofX1,X2, . . . ,Xn is the function f (x1, x2, . . . , xn) such that for any nintervals [a1, b1], . . . , [an, bn],

P(a1 ≤ X1 ≤ b1, . . . an ≤ Xn ≤ bn) =∫ b1

a1

· · ·∫ bn

an

f (x1, . . . , xn)dxn . . . dx1


DefinitionIf X1,X2, . . . ,Xn are all discrete random variables, the joint pmf ofthe variables is the function

p(x1, x2, . . . , xn) = P(X1 = x1,X2 = x2, . . . ,Xn = xn)

If the random variables are continuous, the joint pdf ofX1,X2, . . . ,Xn is the function f (x1, x2, . . . , xn) such that for any nintervals [a1, b1], . . . , [an, bn],

P(a1 ≤ X1 ≤ b1, . . . an ≤ Xn ≤ bn) =∫ b1

a1

· · ·∫ bn

an

f (x1, . . . , xn)dxn . . . dx1


Example:Consider tossing a particular die six times. The probabilities foroutcomes of each toss are given as following:

x 1 2 3 4 5 6

p(x) .15 .20 .25 .20 .15 .05

If we are interested in obtaining exactly three “1’s”, then thisexperiment can be modeled by the binomial distribution.However, if the question is “what is the probability for obtainingexactly three 1’s, two 5’s and one 6”, the binomial

distribution can not do the job.Let Xi = number of i’s from the experiment (six

tosses). Then

P(X1 = 3,X2 = 0,X3 = 0,X4 = 0,X5 = 2,X6 = 1) =

6!

3!2!1!(.15)3(.15)2(.05)1



x 1 2 3 4 5 6

p(x) .15 .20 .25 .20 .15 .05

If we are interested in obtaining exactly three “1’s”, then thisexperiment can be modeled by the binomial distribution.

However, if the question is “what is the probability for obtainingexactly three 1’s, two 5’s and one 6”, the binomial


tosses). Then

P(X1 = 3,X2 = 0,X3 = 0,X4 = 0,X5 = 2,X6 = 1) =

6!

3!2!1!(.15)3(.15)2(.05)1



x 1 2 3 4 5 6

p(x) .15 .20 .25 .20 .15 .05


distribution can not do the job.

Let Xi = number of i’s from the experiment (six

tosses). Then

P(X1 = 3,X2 = 0,X3 = 0,X4 = 0,X5 = 2,X6 = 1) =

6!

3!2!1!(.15)3(.15)2(.05)1



x 1 2 3 4 5 6

p(x) .15 .20 .25 .20 .15 .05



tosses). Then

P(X1 = 3,X2 = 0,X3 = 0,X4 = 0,X5 = 2,X6 = 1) =

6!

3!2!1!(.15)3(.15)2(.05)1


Multinomial Distribution:1. The experiment consists of a sequence of n trials, where n isfixed in advance of the experiment;2. Each trial can result in one of the r possible outcomes;3. The trials are independent;3. The trials are identical, which means the probabilities foroutcomes of each trial are the same. We use p1, p2, . . . , pr todenote them. (pi > 0 and

∑ri=1 pi = 1)

DefinitionAn experiment for which Conditions 1 — 4 are satisfied is called amultinomial experiment.Let Xi = the number of trials resulting in outcome i ,then the joint pmf of X1,X2, . . . ,Xr is called a multinomialdistribution.


Multinomial Distribution:1. The experiment consists of a sequence of n trials, where n isfixed in advance of the experiment;

2. Each trial can result in one of the r possible outcomes;3. The trials are independent;3. The trials are identical, which means the probabilities foroutcomes of each trial are the same. We use p1, p2, . . . , pr todenote them. (pi > 0 and

∑ri=1 pi = 1)



Multinomial Distribution:1. The experiment consists of a sequence of n trials, where n isfixed in advance of the experiment;2. Each trial can result in one of the r possible outcomes;

3. The trials are independent;3. The trials are identical, which means the probabilities foroutcomes of each trial are the same. We use p1, p2, . . . , pr todenote them. (pi > 0 and

∑ri=1 pi = 1)



Multinomial Distribution:1. The experiment consists of a sequence of n trials, where n isfixed in advance of the experiment;2. Each trial can result in one of the r possible outcomes;3. The trials are independent;

3. The trials are identical, which means the probabilities foroutcomes of each trial are the same. We use p1, p2, . . . , pr todenote them. (pi > 0 and

∑ri=1 pi = 1)




∑ri=1 pi = 1)




∑ri=1 pi = 1)



Remark:The joint pmf is

p(x1, x2, . . . , xr ) ={n!

(x1!)(x2!)···(xr !)px11 px2

2 · · · pxrr 0 ≤ xi ≤ n with

∑ri=1 xi = n

0 otherwise


Remark:The joint pmf is

p(x1, x2, . . . , xr ) ={n!

(x1!)(x2!)···(xr !)px11 px2

2 · · · pxrr 0 ≤ xi ≤ n with

∑ri=1 xi = n

0 otherwise


DefinitionThe random variables X1,X2, . . . ,Xn are said to be independent iffor every subset Xi1 ,Xi2 , . . . ,Xik of the variables (each pair, eachtriple, and so on), the joint pmf or pdf of the subset is equal to theproduct of the marginal pmf’s or pdf’s.

e.g. one way to construct a multinormal distribution is totake the product of pdf’s of n independent standard normal rv’s:

f (x1, x2, . . . , xn) =

(1√2π

e−x21 /2

)(1√2π

e−x22 /2

)· · ·(

1√2π

e−x2n/2

)=

1

(√

2π)ne−(x2

1+x22+···+x2

n )/2




f (x1, x2, . . . , xn) =

(1√2π

e−x21 /2

)(1√2π

e−x22 /2

)· · ·(

1√2π

e−x2n/2

)=

1

(√

2π)ne−(x2

1+x22+···+x2

n )/2




f (x1, x2, . . . , xn) =

(1√2π

e−x21 /2

)(1√2π

e−x22 /2

)· · ·(

1√2π

e−x2n/2

)=

1

(√

2π)ne−(x2

1+x22+···+x2

n )/2


Recall the following example (Problem 12):Two components of a minicomputer have the following joint pdffor their useful lifetimes X and Y :

f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

If we find out that the lifetime for the second component is 8(Y = 8), what is the probability for the first component to have alifetime more than 8, i.e. what is P(X ≥ 8 | Y = 8)?We can answer this question by studying conditional probabilitydistributions.



f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise




f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

If we find out that the lifetime for the second component is 8(Y = 8), what is the probability for the first component to have alifetime more than 8, i.e. what is P(X ≥ 8 | Y = 8)?

We can answer this question by studying conditional probabilitydistributions.



f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise



DefinitionLet X and Y be two continuous rv’s with joint pdf f (x , y) andmarginal Y pdf f

Y(y). Then for any Y value y for which

fY

(y) > 0, the conditional probability density function of Xgiven that Y = y is

fX |Y (x | y) =

f (x , y)

fY

(y)−∞ < x <∞

If X and Y are discrete, then conditional probability massfunction of X given that Y = y is

pX |Y (x | y) =

p(x , y)

pY

(y)−∞ < x <∞


DefinitionLet X and Y be two continuous rv’s with joint pdf f (x , y) andmarginal Y pdf f

Y(y). Then for any Y value y for which

fY

(y) > 0, the conditional probability density function of Xgiven that Y = y is

fX |Y (x | y) =

f (x , y)

fY

(y)−∞ < x <∞

If X and Y are discrete, then conditional probability massfunction of X given that Y = y is

pX |Y (x | y) =

p(x , y)

pY

(y)−∞ < x <∞


Example (Problem 12 revisit):Two components of a minicomputer have the following joint pdffor their useful lifetimes X and Y :

f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

What is P(X ≥ 8 | Y = 8)?

fX |Y (x | y) =

f (x , y)

fY

(y)=

{x(1 + y)2e−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

Then

P(X ≥ 8 | Y = 8) = 1−∫ 8

−∞fX |Y (x | 8)dx

= 1−∫ 8

081xe−9xdx = 73e−72

Jointly Distributed Random VariablesExample (Problem 12 revisit):Two components of a minicomputer have the following joint pdffor their useful lifetimes X and Y :

f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

What is P(X ≥ 8 | Y = 8)?

fX |Y (x | y) =

f (x , y)

fY

(y)=

{x(1 + y)2e−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

Then

P(X ≥ 8 | Y = 8) = 1−∫ 8

−∞fX |Y (x | 8)dx

= 1−∫ 8

081xe−9xdx = 73e−72


f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

What is P(X ≥ 8 | Y = 8)?

fX |Y (x | y) =

f (x , y)

fY

(y)=

{x(1 + y)2e−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

Then

P(X ≥ 8 | Y = 8) = 1−∫ 8

−∞fX |Y (x | 8)dx

= 1−∫ 8

081xe−9xdx = 73e−72


f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

What is P(X ≥ 8 | Y = 8)?

fX |Y (x | y) =

f (x , y)

fY

(y)=

{x(1 + y)2e−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

Then

P(X ≥ 8 | Y = 8) = 1−∫ 8

−∞fX |Y (x | 8)dx

= 1−∫ 8

081xe−9xdx = 73e−72


f (x , y) =

{xe−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

What is P(X ≥ 8 | Y = 8)?

fX |Y (x | y) =

f (x , y)

fY

(y)=

{x(1 + y)2e−x(1+y) x ≥ 0 and y ≥ 0

0 otherwise

Then

P(X ≥ 8 | Y = 8) = 1−∫ 8

−∞fX |Y (x | 8)dx

= 1−∫ 8

081xe−9xdx = 73e−72

Date post:	14-Sep-2020
Category:	Documents
Upload:	others
View:	2 times
Download:	0 times

Jointly Distributed Random Variableslzhang/teaching/3070spring2009... · Two components of a...

Documents