Suppose that X and Y are continuous random variables with joint pdf f(x,y)=e−(x+y)0&lt;x&lt;∞ and 0&lt;y&lt;∞ and zero otherwise. Find P(X&gt;Y)

Step 1 Introduction: The joint density function of two random variables X and Y is given below: fXY(x,y)={c, x≥0,y≥0,(x2+y)≤1,0, otherwise The marginal density function of X is, f(x)=∫0∞e−(x+y)dy =e−x∫0∞e−ydy =−e−x[−e−y]0∞ =−e−x[e−∞−e−0] =−e−x[0,1] =e−x Step 2 The probability of P(X &gt; Y) is obtained as 0.5 from the calculation given below: P(X≥Y)=P(Y≤X) =∫0∞∫0xe−(x+y)dxdy =∫0∞∫0xe−xe−ydxdy =∫0∞e−xdx∫0xe−ydy =∫0∞e−xdx[−e−y]0x =−∫0∞e−xdx[e−x−e−0] =−∫0∞e−xdx[e−x−1] =∫0∞e−xdx[1−e−x] =∫0∞e−x−e−2xdx =

Expert answers to "Suppose that X and Y are continuous random..."

BenoguigoliB

Answered question

2021-10-25

Suppose that X and Y are continuous random variables with joint pdf

f (x, y) = e^{- (x + y)} 0 < x < \infty and 0 < y < \infty

and zero otherwise.
Find P(X>Y)

Answer & Explanation

i1ziZ

Skilled2021-10-26Added 92 answers

Step 1
Introduction:
The joint density function of two random variables X and Y is given below:
$f_{X Y} (x, y) = {\begin{cases} c, x \geq 0, y \geq 0, (x^{2} + y) \leq 1, \\ 0, o t h e r w i s e \end{cases}$
The marginal density function of X is,
$f (x) = \int_{0}^{\infty} e^{- (x + y)} d y$
$= e^{- x} \int_{0}^{\infty} e^{- y} d y$
$= - e^{- x} {[- e^{- y}]}_{0}^{\infty}$
$= - e^{- x} [e^{- \infty} - e^{- 0}]$
$= - e^{- x} [0, 1]$
$= e^{- x}$
Step 2
The probability of P(X > Y) is obtained as 0.5 from the calculation given below:
$P (X \geq Y) = P (Y \leq X)$
$= \int_{0}^{\infty} \int_{0}^{x} e^{- (x + y)} d x d y$
$= \int_{0}^{\infty} \int_{0}^{x} e^{- x} e^{- y} d x d y$
$= \int_{0}^{\infty} e^{- x} d x \int_{0}^{x} e^{- y} d y$
$= \int_{0}^{\infty} e^{- x} d x {[- e^{- y}]}_{0}^{x}$
$= - \int_{0}^{\infty} e^{- x} d x [e^{- x} - e^{- 0}]$
$= - \int_{0}^{\infty} e^{- x} d x [e^{- x} - 1]$
$= \int_{0}^{\infty} e^{- x} d x [1 - e^{- x}]$
$= \int_{0}^{\infty} e^{- x} - e^{- 2 x} d x$
$=$

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