Question

Suppose that the joint distribution of X and Y is uniform over the region in the\({\bf{xy}}\)plane bounded by the four lines\({\bf{x = - 1,x = 1,y = x + 1}}\)and\({\bf{y = x - 1}}\). Determine (a)\({\bf{Pr}}\left( {{\bf{XY > 0}}} \right)\)and (b) the conditional p.d.f. of Y given that\({\bf{X = x}}\).

Short answer

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

b. The conditional distribution of Y given that \(X = x\) is \(\frac{1}{2}\)

Step-by-step solution

Given information

The joint distribution of X and Y is uniform.

Finding the form of joint density function

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,

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

Finding the form of marginal density functions

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,

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

Calculating the probability for part (a)

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

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

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

Calculating the probability for part (b)

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\(\frac{1}{2}\)