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 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 $$Pr(X>sqrt{Y})$$ 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:
$$P(X>\sqrt{Y})=\int_{\sqrt{y}}^{1}(x+\frac{1}{2})dx$$
$$=\int_{\sqrt{y}}^{1}xdx+\int_{\sqrt{y}}^{1}\frac{1}{2}dx$$
$$=[\frac{x^{2}}{2}]_{\sqrt{y}}^{1}+[\frac{x}{2}]_{\sqrt{y}}^{1}$$
$$=\frac{1}{2}-\frac{y}{2}+\frac{1}{2}-\frac{\sqrt{y}}{2}$$
$$=1-\frac{y}{2}-\frac{\sqrt{y}}{2}$$