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### 45-733 PROBABILITY AND STATISTICS I Notes #2B

January 2000

#### Continuous Random Variables

1. Continuous Distributions
P[X Î A] = òA f(x)dx
2. Properties of Continuous Probability Distributions
a) f(x) ³ 0 over the real line
b) òf(x)dx = 1 over the real line
c) P[a £ X £ b] = òab f(x)dx = F(b) - F(a)
3. P(X = x) = 0. To see this:
P(X = a) = òaa f(x)dx = F(a) - F(a) = 0
4. Continuous Uniform Distribution
```
æ c  a < x < b
f(x) = ç
è 0 otherwise
```
This is what the function looks like: 5. To solve for the constant, c, use (b) in (22). That is:
òab cdx = cx|ab = c[b-a] = 1; hence, c = 1/(b-a)
6. Example: Let f(x) be:
```
æ c(1 - x3)  0 < x < 1
f(x) = ç
è   0 otherwise
```
Find c.
ò01 c(1 - x3)dx = c[x - x4/4]|01 = c[1 - 1/4] = c(3/4) = 1
Hence, c = 4/3

Find F(1/2).
F(1/2) = P(X < 1/2) = ò01/2 (4/3)(1 - x3)dx = (4/3)[x - x4/4]|01/2 =
(4/3)[1/2 - 1/64] = 31/48
7. Example: Let f(x) be:
```
æ c/x  0 < x < 1
f(x) = ç
è 0 otherwise
```
Find c.
ò01 cxdx = clnx|01 = c[0 - - ¥] = ??
So this is not a probability distribution.

8. Problem 4.8 p.145
```
æ c(2 - y)  0 < y < 2
f(y) = ç
è 0 otherwise
```
1. Find c
ò02 c(2 - y)dy = c[2y - y2/2]|02 = c(4 - 4/2) = 2c
Hence, c = 1/2

2. Find F(y)
F(y) = P(Y < y) = ò0y 1/2(2 - t)dt =
1/2[2t - t2/2]|0y = 1/2(2y - y2/2) = y - y2/4
. Hence:
```
æ   0   y < 0
ç
F(y) = ç  y - y2/4  0 £ y £ 2
ç
è   1   y > 2
```