LEGACY CONTENT. If you are looking for Voteview.com, PLEASE CLICK HERE

This site is an archived version of Voteview.com archived from University of Georgia on May 23, 2017. This point-in-time capture includes all files publicly linked on Voteview.com at that time. We provide access to this content as a service to ensure that past users of Voteview.com have access to historical files. This content will remain online until at least January 1st, 2018. UCLA provides no warranty or guarantee of access to these files.

45-733 PROBABILITY AND STATISTICS I Topic #4A


Sixth Lecture, 28 January 1999


Bivariate Distributions

  1. 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.

  2. Properties of Discrete Bivariate Probability Distributions
    1. f(xi, yj) ³ 0 for all i and j

    2. åi=1,n åj=1,m f(xi, yj) = 1

    3. P[a £ X £ b, c £ Y £ d] = P[a £ X £ b Ç c £ Y £ d] = åi=a,b åj=c,d f(xi, yj)

  3. 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

  4. Properties of Continuous Bivariate Probability Distributions
    1. f(x,y) ³ 0 over the real plane

    2. ò-¥+¥ ò-¥+¥ f(x,y)dxdy = 1 over the real plane

    3. P[a £ X £ b, c £ Y £ d] = P[a £ X £ b Ç c £ Y £ d] = òab òcd f(x,y)dxdy

  5. Example: 
    
                           æ cx2  0 < x < 2
                           ç      0 < y < 3
                  f(x,y) = ç
                           ç
                           è 0 otherwise

    1. 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

    2. 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

    3. 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