Answer to Question #248496 in Statistics and Probability for Jac

Question

Given the joint density function

f(x,y) = {6-x-y / 8 ; 0<x<2, 2<y<4,

{0, elsewhere,

find P(1<Y<3 | X=1).

Answer 1

Solution.

Such as a range of Y is (2,4), then

$P(1<Y<3|X=1)=P(1<Y<2|X=1) +P(2<Y<3|X=1) = 0+ P(2<Y<3|X=1) \newline \text{}$

It is known that the conditional probability can be calculated by the formula

$$f(y|x)=\frac{f(x,y)}{g(x)}.$$

Find $g(x)=\int_2^4\frac{6-x-y}{8}dy=\frac{3-x}{4},$ where 0<x<2.

Find f(y|x), where x=1:$P(2<Y<3|X=1)=\int_2^3(\frac{6-x-y}{8}:\frac{3-x}{4})dy|_{x=1}=(\frac{5}{4}y-\frac{y^2}{8})|_2^3=\frac{15}{4}-\frac{9}{8}-\frac{10}{4}+\frac{4}{8}=\frac{5}{4}-\frac{5}{8}=\frac{5}{8}.$ Answer. $\frac{5}{8}.$