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45-733 PROBABILITY AND STATISTICS I Topic #4A
Sixth Lecture, 28 January 1999
Bivariate Distributions
Example: Discrete Uniform Bivariate Distribution
We toss two dice. Let X = "the number of dots on the first die", and
let Y = "the number of dots on the second die". Hence:
æ 1/36 x = 1,2,3,4,5,6
ç y = 1,2,3,4,5,6
f(x,y) = ç
ç
è 0 otherwise
Note that this is a bivariate discrete uniform distribution.
Properties of Discrete Bivariate Probability Distributions
f(xi, yj) ³ 0 for all i and j
åi=1,n
åj=1,m f(xi, yj) = 1
P[a £ X £
b, c £ Y £ d] =
P[a £ X £ b
Ç
c £ Y £ d] =
åi=a,b
åj=c,d f(xi, yj)
Example: With respect to (4):
P(X ³ 5, Y ³ 5) =
åi=5,6
åj=5,6
f(xi, yj) = 4/36 = 1/9
Properties of Continuous Bivariate Probability Distributions
f(x,y) ³ 0 over the real plane
ò-¥+¥
ò-¥+¥
f(x,y)dxdy = 1 over the real plane
P[a £ X £ b,
c £ Y £ d] =
P[a £ X £ b
Ç
c £ Y £ d] =
òab
òcd f(x,y)dxdy
Example:
æ cx2 0 < x < 2
ç 0 < y < 3
f(x,y) = ç
ç
è 0 otherwise
Find c
ò03
ò02 cx2 dxdy =
c{ò03
[x3/3 |20]dy} =
c[ò03 (8/3)dy] =
c[(8/3)y]|03 = 8c
Hence, c = 1/8
Find P(X < 1, Y < 1] =
P(X < 1 Ç Y < 1)
P(X < 1, Y < 1] =
ò01
ò01
(x2/8)dydx =
ò01
{[(x2/8)y]|01}dx =
ò01
(x2/8)dx = (x3/24)|01 = 1/24
Find P(X < 1/2)
Note that: P(X < 1/2) =
P(X < 1/2 Ç 0 < Y < 3). Hence:
P(X < 1/2) =
ò01/2
ò03
(x2/8)dydx =
ò01/2
{[(x2/8)y]|03}dx =
ò01/2
(3/8)(x2)dx = (3/8)(x3/3)|01/2 =
1/64