Suppose the random variables X and Y have a pdf given by f(x,y)=x+y on 0&lt;x&lt;1, 0&lt;y&lt;1 Find P(X&gt;Y).

Given that, X and Y are two random variables The joint pdf of them is given by, f(x,y)={x+y 0&lt;x&lt;1,0&lt;y&lt;10 otherwise P(X&gt;Y) is to be obtained. It can be written as P(X&gt;Y)=1-P(X Marginal pdf of X is given by ∫01(x+y)dy =∫01xdy+∫01ydy =x(y)01+(y22)01 =x+12 Now, (X&lt;Y)=∫0yx+12dx=(x22)0y+(x2)0y =y22+y2 Therefore, P(X&gt;Y)=1−y22+y2

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Brennan Flores

Answered question

2021-05-05

Suppose the random variables X and Y have a pdf given by
f(x,y)=x+y on 0<x<1, 0<y<1
Find P(X>Y).

Answer & Explanation

Mayme

Skilled2021-05-06Added 103 answers

Given that,
X and Y are two random variables
The joint pdf of them is given by,
$f (x, y) = {\begin{cases} x + y 0 < x < 1, 0 < y < 1 \\ 0 o t h e r w i s e \end{cases}$
P(X>Y) is to be obtained.
It can be written as P(X>Y)=1-P(X Marginal pdf of X is given by
$\int_{0}^{1} (x + y) d y$
$= \int_{0}^{1} x d y + \int_{0}^{1} y d y$
$= x (y)_{0}^{1} + (\frac{y^{2}}{2})_{0}^{1}$
$= x + \frac{1}{2}$
Now,
$(X < Y) = \int_{0}^{y} x + \frac{1}{2} d x$ $= (\frac{x^{2}}{2})_{0}^{y} + (\frac{x}{2})_{0}^{y}$
$= \frac{y^{2}}{2} + \frac{y}{2}$
Therefore, $P (X > Y) = 1 - \frac{y^{2}}{2} + \frac{y}{2}$

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