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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