91Ó°ÊÓ

The joint distribution of X and Y is uniform.

Let,

\(f\left( {x,y} \right) = c\,;\, - 1 < x < 1,x - 1 < y < x + 1\)

First, we have to find the value of c

\(\begin{array}{l}\int_{x - 1}^{x + 1} {\int_{ - 1}^1 {f\left( {x,y} \right)dxdy} } = 1\\\int_{x - 1}^{x + 1} {\int_{ - 1}^1 {cdxdy} } = 1\end{array}\)

\(\begin{array}{l}c\int_{x - 1}^{x + 1} {\left[ x \right]_{ - 1}^1} dy = 1\\c\int_{x - 1}^{x + 1} {2dy = 1} \\2c\left[ y \right]_{x - 1}^{x + 1} = 1\\2c\left[ {\left( {x + 1} \right) - \left( {x - 1} \right)} \right] = 1\\4c = 1\\c = \frac{1}{4}\end{array}\)

The joint p.d.f. of X and Y is,

\(f\left( {x,y} \right) = \frac{1}{4};\, - 1 < x < 1,x - 1 < y < x + 1\)

The marginal pdf of X is,

\(\begin{aligned}{}f\left( x \right) &= \int_{x - 1}^{x + 1} {\frac{1}{4}dy} \\ &= \frac{1}{4}\left[ y \right]_{x - 1}^{x + 1}\\ &= \frac{1}{4}\left[ {x + 1 - x + 1} \right]\\ = \frac{2}{4}\\ &= \frac{1}{2}\end{aligned}\)

The marginal p.d.f. of Y is,

\(\begin{aligned}{}f\left( y \right) &= \int_{ - 1}^1 {\frac{1}{4}dx} \\ &= \frac{1}{4}\left[ x \right]_{ - 1}^1\\ &= \frac{1}{4}\left[ {1 + 1} \right]\\ &= \frac{2}{4}\\ &= \frac{1}{2}\end{aligned}\)

The, area in the second plus the fourth quadrants is 1.

Therefore, the area in the first plus the third quadrants is 3,

\(\Pr \left( {XY > 0} \right) = \frac{3}{4}\)

The conditional p.d.f. of Y given that\(X = x\)is,

\(\begin{aligned}{}h\left( {y\left| {X = x} \right.} \right) &= \frac{{f\left( {x,y} \right)}}{{f\left( x \right)}}\\ &= \frac{{\frac{1}{4}}}{{\frac{1}{2}}}\\ &= \frac{1}{2}\end{aligned}\)

Therefore, the conditional distribution of of Y given that\(X = x\)is\(\frac{1}{2}\)