45-733 PROBABILITY AND STATISTICS I Notes #4C


February 2000



  1. Expected Values of Bivariate Distributions
    Continuous Distributions
    E(X) = ò-¥+¥ ò-¥+¥ xf(x,y)dydx = ò-¥+¥ xf1(x)dx
    E(Y) = ò-¥+¥ ò-¥+¥ yf(x,y)dxdy = ò-¥+¥ yf2(y)dy
    Discrete Distributions
    E(X) = åi=1,n åj=1,m xif(xi, yj) = åi=1,n xif1(xi)
    E(Y) = åj=1,m åi=1,n yjf(xi, yj) = åj=1,m yjf2(yj)

  2. Example Using (8):
    E(X) = ò02 x(3/8)(x2)dx = (3/32)(x4)|02 = 3/2
    E(Y) = ò03 y(1/3)dy = (y2/6)|03 = 3/2

  3. Independence
    Definition: Two random variables are Independent if and only if:
    f(x,y) = f1(x)f2(y)

  4. Example Using (8):
    It is obvious on expection that X and Y are independent random variables.

  5. Example:
    
                                æ 2xe-y  0 < x < 1
                                ç        y > 0
                       f(x,y) = ç
                                ç
                                è  0 otherwise
    

    Are X and Y independent?
    f1(x) = ò0+¥ 2xe-ydy = -2xe-y| 0+¥ = 2x[0 - -1] = 2x. Technically:
    
                               æ 2x  0 < x < 1
                       f1(x) = ç
                               è  0 otherwise
    

    f2(y) = ò01 2xe-ydx = 2[e-y](x2/2)|01 = e-y. Technically:
    
                               æ e-y   y > 0 
                       f2(y) = ç
                               è  0 otherwise
    

    Clearly, X and Y are independent.

  6. Example:
    
                                æ cxy  x = 1,2,3
                                ç      y = 1,2
                       f(x,y) = ç
                                ç
                                è  0 otherwise
    

    1. Find c.
      To find c construct the following table:
      
                                     y
                                 1       2
                               ------------
                             1 | 1       2 | 3
                               |           |
                          x  2 | 2       4 | 6
                               |           |
                             3 | 3       6 | 9
                               |           |
                               ---------------
                                 6      12 |18
      

      By inspection, c = 1/18

    2. Find f1(x) and f2(y)
      By inspection of the table (that is why they are called table marginals!):
      
                                  æ 3/18  x = 1   
                                  ç                æ x/6  x = 1,2,3
                                  ç 6/18  x = 2    ç
                          f1(x) = ç              = ç
                                  ç 9/18  x = 3    ç
                                  ç                è  0 otherwise
                                  è  0 otherwise
      

      
                                  æ  6/18 y = 1                    
                                  ç                æ y/3  y = 1,2
                          f2(y) = ç 12/18 y = 2  = ç
                                  ç                è  0 otherwise
                                  è  0 otherwise
      

    3. Are X and Y independent?
      Obvious on inspection.

  7. Example:
    
                                æ c(2x - y)  x = 2,4
                                ç            y = 1,2,3
                       f(x,y) = ç
                                ç
                                è  0 otherwise
    

    1. Find c.
      To find c construct the following table:
      
                                     y
                                 1   2   3
                               ------------
                             2 | 3   2   1 |  6
                          x    |           |
                             4 | 7   6   5 | 18
                               |           |
                               ---------------
                                10   8   6 | 24
      

      By inspection, c = 1/24

    2. Find f1(x) and f2(y)
      
                                  æ  6/24 x = 2                    
                                  ç                              
                          f1(x) = ç 18/24 x = 4     
                                  ç                                  
                                  è  0 otherwise
      

      
                                  æ 10/24  y = 1   
                                  ç                                   
                                  ç  8/24  y = 2     
                          f2(y) = ç                              
                                  ç  6/24  y = 3     
                                  ç                                
                                  è  0 otherwise
      

    3. Are X and Y independent?
      No. f(2,1) = 3/24 but f1(2)*f2(1) = (6/24)*(10/24) = 2.5/24

  8. Proof of (12B) Notes #5:
    E(X1 + X2 + ... + Xn) = E(X1) + E(X2) + ... + E(Xn)
    For two continuous random variables, X and Y:
    E(X + Y] = ò-¥+¥ ò-¥+¥ (x + y)f(x,y)dxdy =
    ò-¥+¥ ò-¥+¥ xf(x,y)dxdy + ò-¥+¥ ò-¥+¥ yf(x,y)dxdy =
    ò-¥+¥ xf1(x)dx + ò-¥+¥ yf2(y)dy =
    E(X) + E(Y) using (11) in Notes #6.

  9. If X and Y are Independent, then E(XY) = E(X)E(Y)

    To see this, recall that: f(x,y) = f1(x)f2(y)
    Hence:
    E(XY) = ò-¥+¥ ò-¥+¥ xyf(x, y)dxdy = ò-¥+¥ ò-¥+¥ xyf1(x)f2(y)dxdy =
    [ ò-¥+¥ xf1(x)dx] [ ò-¥+¥ yf2(y)dy] = E(X)E(Y)

  10. In (1B) of Notes #6, it was stated that:
    If X1, X2, ... , Xn, are independent random variables, then:
    VAR(a1X1 + a2X2 + ... + anXn + b) = åi=1,n (ai)2VAR(Xi)

    This can be proven using (1) and (2).

  11. Example: Binomial Distribution
    Let X1, X2, X3, ..., Xn be random variables corresponding to n Bernoulli trials (experiments), and let Z be the random variable:
    Z = åi=1,n Xi
    The distribution of Z is the binomial
    
                             æ ænöpz(1-p)(n-z)
                             ç ç ÷
                             ç èzø
                      f(z) = ç            z=0,1,2,3,...,n
                             ç
                             è 0 otherwise
    

    E(Z) = E[X1 + X2 + ... + Xn] = E(X1) + E(X2) + ... + E(Xn) =
    p + p + ... + p = np
    Using the fact that E(X) = p for the Bernoulli distribution.

  12. Variance of the Binomial Distribution

    VAR(Z) = VAR[X1 + X2 + ... + Xn] = VAR(X1) + VAR(X2) + ... + VAR(Xn) =
    p(1 - p) + p(1 - p) + ... + p(1 - p) = np(1 - p)
    Using the fact that VAR(X) = s2 = p(1 - p) for the Bernoulli distribution. [See Notes #5 (17).]