Jointly Distributed Random Variables
Jointly Distributed Random Variables
Example (variant of Problem 12)
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −(x+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
Jointly Distributed Random Variables
Example (variant of Problem 12)
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −(x+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
a. What is the probability that the lifetimes of both components
excceed 3?
Jointly Distributed Random Variables
Example (variant of Problem 12)
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −(x+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
a. What is the probability that the lifetimes of both components
excceed 3?
b. What are the marginal pdf’s of X and Y ?
Jointly Distributed Random Variables
Example (variant of Problem 12)
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −(x+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
a. What is the probability that the lifetimes of both components
excceed 3?
b. What are the marginal pdf’s of X and Y ?
c. What is the probability that the lifetime X of the first
component excceeds 3?
Jointly Distributed Random Variables
Example (variant of Problem 12)
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −(x+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
a. What is the probability that the lifetimes of both components
excceed 3?
b. What are the marginal pdf’s of X and Y ?
c. What is the probability that the lifetime X of the first
component excceeds 3?
d. What is the probability that the lifetime of at least one
component excceeds 3?
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Recall the following example (variant of Problem 12):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −(x+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
Jointly Distributed Random Variables
Recall the following example (variant of Problem 12):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −(x+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
The marginal pdf’s of X and Y are
fX (x) = xe −x and fY (y ) = e −y ,
respectively.
Jointly Distributed Random Variables
Recall the following example (variant of Problem 12):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −(x+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
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 ).
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Definition
Two random variables X and Y are said to be independent if for
every 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 Y
are said to be dependent.
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Examples:
Problem 12. The joint pdf for X and Y is
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
Jointly Distributed Random Variables
Examples:
Problem 12. The joint pdf for X and Y is
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
The marginal pdf’s of X and Y are
fX (x) = −e −x and fY (y ) =
respectively.
1
,
(1 + y )2
Jointly Distributed Random Variables
Examples:
Problem 12. The joint pdf for X and Y is
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
The marginal pdf’s of X and Y are
fX (x) = −e −x and fY (y ) =
1
,
(1 + y )2
respectively. We see that
fX (x) · fY (y ) 6= f (x, y ).
Jointly Distributed Random Variables
Jointly Distributed Random Variables
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.
Jointly Distributed Random Variables
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
.50
20
0 .20 .10
.30
p(·, y ) .10 .35 .55
pX (12) · pY (12) 6= p(12, 12).
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Definition
If X1 , X2 , . . . , Xn are all discrete random variables, the joint pmf of
the variables is the function
p(x1 , x2 , . . . , xn ) = P(X1 = x1 , X2 = x2 , . . . , Xn = xn )
If the random variables are continuous, the joint pdf of
X1 , X2 , . . . , Xn is the function f (x1 , x2 , . . . , xn ) such that for any n
intervals [a1 , b1 ], . . . , [an , bn ],
P(a1 ≤ X1 ≤ b1 , . . . an ≤ Xn ≤ bn ) =
Z b1 Z bn
···
f (x1 , . . . , xn )dxn . . . dx1
a1
an
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Example:
Consider tossing a particular die six times. The probabilities for
outcomes of each toss are given as following:
1
2
3
4
5
6
x
p(x) .15 .20 .25 .20 .15 .05
If we are interested in obtaining exactly three “1’s”, then this
experiment can be modeled by the binomial distribution.
Jointly Distributed Random Variables
Example:
Consider tossing a particular die six times. The probabilities for
outcomes of each toss are given as following:
1
2
3
4
5
6
x
p(x) .15 .20 .25 .20 .15 .05
If we are interested in obtaining exactly three “1’s”, then this
experiment can be modeled by the binomial distribution.
However, if the question is “what is the probability for obtaining
exactly three 1’s, two 5’s and one 6”, the binomial
distribution can not do the job.
Jointly Distributed Random Variables
Example:
Consider tossing a particular die six times. The probabilities for
outcomes of each toss are given as following:
1
2
3
4
5
6
x
p(x) .15 .20 .25 .20 .15 .05
If we are interested in obtaining exactly three “1’s”, then this
experiment can be modeled by the binomial distribution.
However, if the question is “what is the probability for obtaining
exactly 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!
(.15)3 (.15)2 (.05)1
3!2!1!
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Multinomial Distribution:
1. The experiment consists of a sequence of n trials, where n is
fixed in advance of the experiment;
Jointly Distributed Random Variables
Multinomial Distribution:
1. The experiment consists of a sequence of n trials, where n is
fixed in advance of the experiment;
2. Each trial can result in one of the r possible outcomes;
Jointly Distributed Random Variables
Multinomial Distribution:
1. The experiment consists of a sequence of n trials, where n is
fixed in advance of the experiment;
2. Each trial can result in one of the r possible outcomes;
3. The trials are independent;
Jointly Distributed Random Variables
Multinomial Distribution:
1. The experiment consists of a sequence of n trials, where n is
fixed 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 for
outcomes of each trial are P
the same. We use p1 , p2 , . . . , pr to
denote them. (pi > 0 and ri=1 pi = 1)
Jointly Distributed Random Variables
Multinomial Distribution:
1. The experiment consists of a sequence of n trials, where n is
fixed 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 for
outcomes of each trial are P
the same. We use p1 , p2 , . . . , pr to
denote them. (pi > 0 and ri=1 pi = 1)
Definition
An experiment for which Conditions 1 — 4 are satisfied is called a
multinomial experiment.
Let Xi = the number of trials resulting in outcome i,
then the joint pmf of X1 , X2 , . . . , Xr is called a multinomial
distribution.
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Remark:
The joint pmf is
p(x1 , x2 , . . . , xr ) =
(
x1 x2
n!
xr
(x1 !)(x2 !)···(xr !) p1 p2 · · · pr
0
0 ≤ xi ≤ n with
otherwise
Pr
i=1 xi
=n
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Definition
The random variables X1 , X2 , . . . , Xn are said to be independent if
for every subset Xi1 , Xi2 , . . . , Xik of the variables (each pair, each
triple, and so on), the joint pmf or pdf of the subset is equal to the
product of the marginal pmf’s or pdf’s.
Jointly Distributed Random Variables
Definition
The random variables X1 , X2 , . . . , Xn are said to be independent if
for every subset Xi1 , Xi2 , . . . , Xik of the variables (each pair, each
triple, and so on), the joint pmf or pdf of the subset is equal to the
product of the marginal pmf’s or pdf’s.
e.g. one way to construct a multinormal distribution is to
take the product of pdf’s of n independent standard normal rv’s:
1 −x12 /2
1 −x22 /2
1 −xn2 /2
√ e
f (x1 , x2 , . . . , xn ) = √ e
··· √ e
2π
2π
2π
1
2
2
2
= √
e −(x1 +x2 +···+xn )/2
( 2π)n
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Recall the following example (Problem 12):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
Jointly Distributed Random Variables
Recall the following example (Problem 12):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
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 a
lifetime more than 8, i.e. what is P(X ≥ 8 | Y = 8)?
Jointly Distributed Random Variables
Recall the following example (Problem 12):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
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 a
lifetime more than 8, i.e. what is P(X ≥ 8 | Y = 8)?
We can answer this question by studying conditional probability
distributions.
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Definition
Let X and Y be two continuous rv’s with joint pdf f (x, y ) and
marginal Y pdf fY (y ). Then for any Y value y for which
fY (y ) > 0, the conditional probability density function of X
given that Y = y is
fX |Y (x | y ) =
f (x, y )
fY (y )
−∞<x <∞
If X and Y are discrete, then conditional probability mass
function of X given that Y = y is
pX |Y (x | y ) =
p(x, y )
pY (y )
−∞<x <∞
Jointly Distributed Random Variables
Jointly Distributed Random Variables
Example (Problem 12 revisit):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
Jointly Distributed Random Variables
Example (Problem 12 revisit):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
What is P(X ≥ 8 | Y = 8)?
Jointly Distributed Random Variables
Example (Problem 12 revisit):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
What is P(X ≥ 8 | Y = 8)?
(
x(1 + y )2 e −x(1+y )
f (x, y )
fX |Y (x | y ) =
=
fY (y )
0
x ≥ 0 and y ≥ 0
otherwise
Jointly Distributed Random Variables
Example (Problem 12 revisit):
Two components of a minicomputer have the following joint pdf
for their useful lifetimes X and Y :
(
xe −x(1+y ) x ≥ 0 and y ≥ 0
f (x, y ) =
0
otherwise
What is P(X ≥ 8 | Y = 8)?
(
x(1 + y )2 e −x(1+y )
f (x, y )
fX |Y (x | y ) =
=
fY (y )
0
x ≥ 0 and y ≥ 0
otherwise
Then
Z
8
P(X ≥ 8 | Y = 8) = 1 −
−∞
Z 8
fX |Y (x | 8)dx
81xe −9x dx = 73e −72
=1−
0