Download STAT 430/510 Probability Lecture 14: Joint Probability Distribution

STAT 430/510 Lecture 14
STAT 430/510 Probability
Lecture 14: Joint Probability Distribution,
Continuous Case
Pengyuan (Penelope) Wang
June 20, 2011
STAT 430/510 Lecture 14
Joint density function of continuous Random Variable
When X and Y are two continuous random variables, the
joint density function f (x, y ) is a function defined for
each pair of numbers (x,y) by
f (x, y )
Since the total probability of the all possible pairs of (x,y) is
1, the joint density function must satisfy
Z Z
f (x, y )dxdy = 1
x,y
Also
f (x, y ) ≥ 0
STAT 430/510 Lecture 14
Marginal pdf of Continuous Random Variable
The marginal pdf’s of X and Y , denoted by fX (x) and
fY (y ), respectively, are given by
Z
fX (x) =
f (x, y )dy
y
Z
fY (y ) =
f (x, y )dx
x
STAT 430/510 Lecture 14
Example
The joint density
function of X and Y is given by
2e−x e−2y , 0 < x < ∞, 0 < y < ∞
f (x, y ) =
0, otherwise
Check that f (x, y ) is a joint density function .
Check 1:
f (x, y ) ≥ 0
Check 2:
Z Z
f (x, y )dxdy = 1
x,y
STAT 430/510 Lecture 14
Example-continue
The joint density
function of X and Y is given by
2e−x e−2y , 0 < x < ∞, 0 < y < ∞
f (x, y ) =
0, otherwise
Compute the marginal density of X and Y .
STAT 430/510 Lecture 14
Example-continue
The joint density
function of X and Y is given by
2e−x e−2y , 0 < x < ∞, 0 < y < ∞
f (x, y ) =
0, otherwise
Compute the marginal density of X and Y .
R∞
fX (x) = 0 2e−x e−2y dy = e−x
R∞
fY (y ) = 0 2e−x e−2y dx = 2e−2y
STAT 430/510 Lecture 14
Usage 1: compute probability
X and Y are two continuous r.v.’s.
Z
P[(X , Y ) ∈ A] =
f (x, y )dxdy
(x,y )∈A
In the last example, compute (a) P(X > 1, Y < 1)
(b) P(X < Y )
STAT 430/510 Lecture 14
Usage 1: compute probability
X and Y are two continuous r.v.’s.
Z
P[(X , Y ) ∈ A] =
f (x, y )dxdy
(x,y )∈A
In the last example, compute (a) P(X > 1, Y < 1)
(b) P(X < Y )
R1R∞
P(X > 1, Y < 1) = 0 1 2e−x e−2y dxdy = e−1 − e−3
R∞Ry
P(X < Y ) = 0 0 2e−x e−2y dxdy = 1/3
STAT 430/510 Lecture 14
Usage 2: compute marginal Expected Value
X and Y are two continuous r.v.’s and they have marginal
distribution fX (x) and fY (y ).
Z
E[X ] = xfX (x)dx
x
Z
E[Y ] =
yfY (y )dy
y
In the last
R ∞ example, what is the expected value of X ?
EX = 0 xe−x dx = 1.
STAT 430/510 Lecture 14
Usage 3: compute Expected Value of a function of X and Y
If X and Y have a joint probability mass function p(x, y ),
then
XX
E[g(X , Y )] =
g(x, y )p(x, y )
y
x
X and Y are two continuous r.v.’s.
Z
E[g(X , Y )] =
g(X , Y )f (x, y )dxdy
x,y
1
In the last example, what is the expected value of e 2 X +Y ?
R∞Ry 1
1
E[e 2 X +Y ] = 0 0 e 2 X +Y ∗ 2e−x e−2y dxdy =
R ∞ R y − 1 X −Y
2
dxdy = 4.
0
0 2e
STAT 430/510 Lecture 14
Usage 4: compute Conditional probability
X and Y are two continuous r.v.’s and they have marginal
distribution fX (x) and fY (y ).
fX |Y (x|y ) =
f (x, y )
fY (y )
fY |X (y |x) =
f (x, y )
fX (x)
What is fX |Y (x|y )? Given that y = 2, what is the
distribution of x?
fX |Y (x|y ) = e−x , for any y .
STAT 430/510 Lecture 14
Conditional Expectation
E[X |Y = y ] =
R
xfX |Y (x|y )dx.
What is E[X |Y R= 2]?
E[X |Y = y ] = xe−x dx = 1.
STAT 430/510 Lecture 14
Comments
Again, conditional expectation satisfies all of the properties
of ordinary expectation, for example
Pn
Pn
E[ i=1 Xi |Y = y ] = i=1 E[Xi |Y = y ]
STAT 430/510 Lecture 14
Usage 5: Check independence
When X and Y are continuous, X and Y are independent if
and only if f (x, y ) = fX (x)fY (y ), for all x, y
Are X and Y independent?
They are, since f (x, y ) = 2e−x e−2y = fX (x)fY (y )
STAT 430/510 Lecture 14
Example
The joint density of X and Y is given by
12
5 x(2 − x − y ), 0 < x < 1, 0 < y < 1
f (x, y ) =
0, otherwise
Compute the conditional density of X given that Y = y ,
where 0 < y < 1.
STAT 430/510 Lecture 14
Example
The joint density of X and Y is given by
12
5 x(2 − x − y ), 0 < x < 1, 0 < y < 1
f (x, y ) =
0, otherwise
Compute the conditional density of X given that Y = y ,
where 0 < y < 1.
fX |Y (x|y ) =
=
=
f (x, y )
fY (y )
12
5 x(2 − x
R 1 12
0 5 x(2 − x
− y)
− y )dx
6x(2 − x − y )
4 − 3y
STAT 430/510 Lecture 14
Example 1:
f (x, y ) = 24xy , where 0 < x < 1, 0 < y < 1, 0 < x + y < 1,
and it equals to 0 otherwise.
Show that f (x, y ) is a joint probability density function.
Find out fX (x).
Given X = 0.5, find fY |X (y |x = 0.5) and E[Y |X = 0.5].
STAT 430/510 Lecture 14
Example:
f (x, y ) = 24xy , where 0 < x < 1, 0 < y < 1, 0 < x + y < 1,
and it equals to 0 otherwise. Show that f (x, y ) is a joint
probability density function.
STAT 430/510 Lecture 14
Example:
f (x, y ) = 24xy , where 0 < x < 1, 0 < y < 1, 0 < x + y < 1,
and it equals to 0 otherwise. Show that f (x, y ) is a joint
probability density function.
R 1 R 1−x
x=0 y =0 f (x, y )dydx = 1.
STAT 430/510 Lecture 14
Example:
f (x, y ) = 24xy , where 0 < x < 1, 0 < y < 1, 0 < x + y < 1,
and it equals to 0 otherwise. Show that f (x, y ) is a joint
probability density function.
R 1 R 1−x
x=0 y =0 f (x, y )dydx = 1.
Find out fX (x).
STAT 430/510 Lecture 14
Example:
f (x, y ) = 24xy , where 0 < x < 1, 0 < y < 1, 0 < x + y < 1,
and it equals to 0 otherwise. Show that f (x, y ) is a joint
probability density function.
R 1 R 1−x
x=0 y =0 f (x, y )dydx = 1.
Find out fX (x).
R 1−x
fX (x) = y =0 f (x, y )dy = 12x(1 − x)2 .
STAT 430/510 Lecture 14
Example:
f (x, y ) = 24xy , where 0 < x < 1, 0 < y < 1, 0 < x + y < 1,
and it equals to 0 otherwise. Show that f (x, y ) is a joint
probability density function.
R 1 R 1−x
x=0 y =0 f (x, y )dydx = 1.
Find out fX (x).
R 1−x
fX (x) = y =0 f (x, y )dy = 12x(1 − x)2 .
Given X = 0.5, find fY |X (y |x = 0.5) and E[Y |X = 0.5].
STAT 430/510 Lecture 14
Example:
f (x, y ) = 24xy , where 0 < x < 1, 0 < y < 1, 0 < x + y < 1,
and it equals to 0 otherwise. Show that f (x, y ) is a joint
probability density function.
R 1 R 1−x
x=0 y =0 f (x, y )dydx = 1.
Find out fX (x).
R 1−x
fX (x) = y =0 f (x, y )dy = 12x(1 − x)2 .
Given X = 0.5, find fY |X (y |x = 0.5) and E[Y |X = 0.5].
24∗0.5y
fY |X (y |x = 0.5) = f (X = 0.5, y )/fX (0.5) = 12∗0.5(1−0.5)
2 =
8y , for y ∈ (0, 1 − 0.5).
R 1−0.5
E[Y |X = 0.5] = y =0 yfY |X (y |x = 0.5)dy =
R 1−0.5 2
y =0 8y dy = 1/3.
STAT 430/510 Lecture 14
Example 2
A man and a woman decide to meet at a certain location. If
each of them independently arrives at a time uniformly
distributed between 12 noon and 1 P.M.
Then f (x, y ) = (1/60)2 , 0 < x < 60, 0 < y < 60, where x
represents the number of minutes after 12 noon when the
man arrives, and y is for the woman. (why?)
find the probability that the first to arrive has to wait longer
than 10 minutes.
STAT 430/510 Lecture 14
Example: Solution
X and Y denote, respectively, the time past 12 that the
man and the woman arrive.
X and Y are independent uniform random variables over
(0,60).
P(X + 10 < Y ) + P(Y + 10 < X )
Z Z
Z Z
=
f (x, y )dxdy +
Z
{x+10<y }
60 Z y −10
2
Z
60 Z
(1/60) dxdy
=
10
0
= 25/36
10
f (x, y )dxdy
{y +10<x}
x−10
0
(1/60)2 dydx
STAT 430/510 Lecture 14
Example 3
An accident occurs at a point X that is uniformly distributed
on a road of length L. At the time of the accident, an
ambulance is at a location Y that is also uniformly
distributed on the road. Assuming that X and Y are
independent, find the expected distance between the
ambulance and the point of the accident.
STAT 430/510 Lecture 14
Example: Continued
Need to compute E|X − Y |
The joint density function of X and Y is
f (x, y ) = L12 , 0 < x < L, 0 < y < L
Z
E[|X − Y |] =
=
=
=
LZ L
1
|x − y |dydx
2
L
0
0
Z L Z x
Z L
1
(
(x
−
y
)dy
+
(y − x)dy )dx
L2 0 0
x
Z L 2
L
1
( + x 2 − x)dx
2
L 0 2
L
3