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(x_i, y_j) ³ 0 for all i and j
   
   
   2. å_i=1,n å_j=1,m f(x_i, y_j) = 1
   
   
   3. P[a £ X £ b, c £ Y £ d] = P[a £ X £ b Ç c £ Y £ d] = å_i=a,b å_j=c,d f(x_i, y_j)
   
   
3. Example: With respect to (4):
   P(X ³ 5, Y ³ 5) = å_i=5,6 å_j=5,6 f(x_i, y_j) = 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] = ò_a^b ò_c^d 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) = ò₀^1/2 ò₀³ (x²/8)dydx = ò₀^1/2 {[(x²/8)y]|₀³}dx =
      ò₀^1/2 (3/8)(x²)dx = (3/8)(x³/3)|₀^1/2 = 1/64