Question

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

Random variables
ANSWERED
asked 2021-05-01
Suppose the random variables X and Y have a joint pdf
\(f_{XY}(x,y)=\begin{cases}x+y & 0\leq x,y \leq 1\\0 & otherwise\end{cases}\)
Find \(Pr(X>sqrt{Y})\)

Expert Answers (1)

2021-05-02
Step 1
Given information:
The joint probability density function of the random variables X and Y is as given below:
\(f_{XY}(x,y)=\begin{cases}x+y.\ 0\leq x,y \leq 1\\0. otherwise\end{cases}\)
Step 2
Compute the value of \(P(X>\sqrt{Y})\):
The marginal probability density function of X is obtained as given below:
\(f_{X}(x)=\int_{y}f_{XY}(x,y)dy\)
\(=\int_{0}^{1}(x+y)dy\)
\(=x[y]_{0}^{1}+[\frac{y^{2}}{2}]_{0}^{1}\)
\(f_{X}(x)=x+\frac{1}{2}\)
The value of \(P(X>\sqrt{Y})\) is obtained as given below:
\(P(X>\sqrt{Y})=\int_{\sqrt{y}}^{1}(x+\frac{1}{2})dx\)
\(=\int_{\sqrt{y}}^{1}xdx+\int_{\sqrt{y}}^{1}\frac{1}{2}dx\)
\(=[\frac{x^{2}}{2}]_{\sqrt{y}}^{1}+[\frac{x}{2}]_{\sqrt{y}}^{1}\)
\(=\frac{1}{2}-\frac{y}{2}+\frac{1}{2}-\frac{\sqrt{y}}{2}\)
\(=1-\frac{y}{2}-\frac{\sqrt{y}}{2}\)
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