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45-733 PROBABILITY AND STATISTICS I Topic #4A

Sixth Lecture, 28 January 1999

- 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. -
**Properties of Discrete Bivariate Probability Distributions**

- f(x
_{i}, y_{j}) ³ 0 for all i and j - å
_{i=1,n}å_{j=1,m}f(x_{i}, y_{j}) = 1 -
**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})

- f(x
- Example: With respect to (4):

**P(X ³ 5, Y ³ 5) = å**= 4/36 = 1/9_{i=5,6}å_{j=5,6}f(x_{i}, y_{j}) -
**Properties of Continuous Bivariate Probability Distributions**

- f(x,y) ³ 0 over the real plane
- ò
_{-¥}^{+¥}ò_{-¥}^{+¥}f(x,y)dxdy = 1 over the real plane -
**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

- Example:
**æ cx**^{2}0 < x < 2 ç 0 < y < 3 f(x,y) = ç ç è 0 otherwise

- Find c

ò_{0}^{3}ò_{0}^{2}cx^{2}dxdy = c{ò_{0}^{3}[x^{3}/3 |_{2}^{0}]dy} = c[ò_{0}^{3}(8/3)dy] = c[(8/3)y]|_{0}^{3}= 8c

Hence, c = 1/8 - Find
**P(X < 1, Y < 1] = P(X < 1 Ç Y < 1)**

**P(X < 1, Y < 1]**= ò_{0}^{1}ò_{0}^{1}(x^{2}/8)dydx = ò_{0}^{1}{[(x^{2}/8)y]|_{0}^{1}}dx =

ò_{0}^{1}(x^{2}/8)dx = (x^{3}/24)|_{0}^{1}= 1/24 - Find
**P(X < 1/2)**

Note that:**P(X < 1/2) = P(X < 1/2 Ç 0 < Y < 3)**. Hence:

**P(X < 1/2)**= ò_{0}^{1/2}ò_{0}^{3}(x^{2}/8)dydx = ò_{0}^{1/2}{[(x^{2}/8)y]|_{0}^{3}}dx =

ò_{0}^{1/2}(3/8)(x^{2})dx = (3/8)(x^{3}/3)|_{0}^{1/2}= 1/64

- Find c