# Suppose the random variables X and Y have a joint pdff_{XY}(x,y)=\begin{cases}x+y & 0\leq x,y \leq 1\\0 & otherwise\end{cases}Find Pr(X>sqrt{Y})

Suppose the random variables X and Y have a joint pdf
${f}_{XY}\left(x,y\right)=\left\{\begin{array}{ll}x+y& 0\le x,y\le 1\\ 0& otherwise\end{array}$
Find $Pr\left(X>\sqrt{Y}\right)$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Derrick
Step 1
Given information:
The joint probability density function of the random variables X and Y is as given below:

Step 2
Compute the value of $P\left(X>\sqrt{Y}\right)$:
The marginal probability density function of X is obtained as given below:
${f}_{X}\left(x\right)={\int }_{y}{f}_{XY}\left(x,y\right)dy$
$={\int }_{0}^{1}\left(x+y\right)dy$
$=x\left[y{\right]}_{0}^{1}+\left[\frac{{y}^{2}}{2}{\right]}_{0}^{1}$
${f}_{X}\left(x\right)=x+\frac{1}{2}$
The value of $P\left(X>\sqrt{Y}\right)$ is obtained as given below:
$P\left(X>\sqrt{Y}\right)={\int }_{\sqrt{y}}^{1}\left(x+\frac{1}{2}\right)dx$
$={\int }_{\sqrt{y}}^{1}xdx+{\int }_{\sqrt{y}}^{1}\frac{1}{2}dx$
$=\left[\frac{{x}^{2}}{2}{\right]}_{\sqrt{y}}^{1}+\left[\frac{x}{2}{\right]}_{\sqrt{y}}^{1}$
$=\frac{1}{2}-\frac{y}{2}+\frac{1}{2}-\frac{\sqrt{y}}{2}$
$=1-\frac{y}{2}-\frac{\sqrt{y}}{2}$