Suppose the random variables X and Y have a joint pdf fXY(x,y)={x+y0≤x,y≤10otherwise Find Pr(X&gt;Y)

Answered: "Suppose the random variables X and Y have..."

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Ava-May Nelson

Answered question

2021-05-01

Suppose the random variables X and Y have a joint pdf
$f_{X Y} (x, y) = {\begin{cases} x + y & 0 \leq x, y \leq 1 \\ 0 & o t h e r w i s e \end{cases}$
Find $P r (X > \sqrt{Y})$

Answer & Explanation

Derrick

Skilled2021-05-02Added 94 answers

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

f_{X Y} (x, y) = {\begin{cases} x + y . 0 \leq x, y \leq 1 \\ 0. o t h e r w i s e \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_{X Y} (x, y) d y

= \int_{0}^{1} (x + y) d y

= 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}) d x

= \int_{\sqrt{y}}^{1} x d x + \int_{\sqrt{y}}^{1} \frac{1}{2} d x

= [\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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