1999 Flex-Time and Flex-Mode Prob-Stat I Notes #6 Page

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45-733 PROBABILITY AND STATISTICS I Notes #4C

February 2000

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) = å_i=1,n å_j=1,m x_if(x_i, y_j) = å_i=1,n x_if₁(x_i)
E(Y) = å_j=1,m å_i=1,n y_jf(x_i, y_j) = å_j=1,m y_jf₂(y_j)

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

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

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

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.

Example:


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

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

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

Are X and Y independent?
Obvious on inspection.

Example:


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

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

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

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

Proof of (12B) Notes #5:
E(X₁ + X₂ + ... + X_n) = E(X₁) + E(X₂) + ... + E(X_n)
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.

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)

In (1B) of Notes #6, it was stated that:
If X₁, X₂, ... , X_n, are independent random variables, then:
VAR(a₁X₁ + a₂X₂ + ... + a_nX_n + b) = å_i=1,n (a_i)²VAR(X_i)
This can be proven using (1) and (2).

Example: Binomial Distribution
Let X₁, X₂, X₃, ..., X_n be random variables corresponding to n Bernoulli trials (experiments), and let Z be the random variable:
Z = å_i=1,n X_i
The distribution of Z is the binomial


                         æ ænöp^z(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 + ... + X_n] = E(X₁) + E(X₂) + ... + E(X_n) =
p + p + ... + p = np
Using the fact that E(X) = p for the Bernoulli distribution.

Variance of the Binomial Distribution

VAR(Z) = VAR[X₁ + X₂ + ... + X_n] = VAR(X₁) + VAR(X₂) + ... + VAR(X_n) =
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).]