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### 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).]