Suppose the random variables X and Y have a joint pdf

Find

Ava-May Nelson
2021-05-01
Answered

Suppose the random variables X and Y have a joint pdf

Find

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Derrick

Answered 2021-05-02
Author has **94** answers

Step 1

Given information:

The joint probability density function of the random variables X and Y is as given below:

${f}_{XY}(x,y)=\{\begin{array}{l}x+y.\text{}0\le x,y\le 1\\ 0.otherwise\end{array}$

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:

$P(X>\sqrt{Y})={\int}_{\sqrt{y}}^{1}(x+\frac{1}{2})dx$

$={\int}_{\sqrt{y}}^{1}xdx+{\int}_{\sqrt{y}}^{1}\frac{1}{2}dx$

$=[\frac{{x}^{2}}{2}{]}_{\sqrt{y}}^{1}+[\frac{x}{2}{]}_{\sqrt{y}}^{1}$

$=\frac{1}{2}-\frac{y}{2}+\frac{1}{2}-\frac{\sqrt{y}}{2}$

$=1-\frac{y}{2}-\frac{\sqrt{y}}{2}$

Given information:

The joint probability density function of the random variables X and Y is as given below:

Step 2

Compute the value of

The marginal probability density function of X is obtained as given below:

The value of

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