Explained: "Suppose that X and Y are continuous random..."

Wierzycaz

Answered question

2021-06-02

Suppose that X and Y are continuous random variables with joint pdf $f (x, y) = e^{- (x + y)} 0$ and zero otherwise.
Find $P (X > Y)$

Answer & Explanation

Liyana Mansell

Skilled2021-06-03Added 97 answers

Step 1
Introduction:
The joint density function of two random variables X and Y is given below:
$f (x, y) = {\begin{matrix} e^{- (x + y)} . 0 \leq x \leq \infty, 0 \leq y < \infty \\ 0. e l s e w h e r e \end{matrix}}$
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:
$P (X \geq Y) = P (Y \leq X)$
$= \int_{0}^{\infty} \int_{0}^{x} e^{- (x + y)} d x d y$
$= \int_{0}^{\infty} \int_{0}^{x} e^{- x} e^{- y} d x d y$
$= \int_{0}^{\infty} e^{- x} d x \int_{0}^{x} e^{- y} d y$
$= \int_{0}^{\infty} e^{- x} d x [- e^{- y}]_{0}^{x}$
$= - \int_{0}^{\infty} e^{- x} d x [e^{- x} - e^{- 0}]$
$= - \int_{0}^{\infty} e^{- x} d x [e^{- x} - 1]$
$= \int_{0}^{\infty} e^{- x} d x [1 - e^{- x}]$
$= \int_{0}^{\infty} e^{- x} - e^{- 2 x} d x$
$= [e^{- x}]_{0}^{\infty} + [\frac{e^{- 2 x}}{2}]_{0}^{\infty}$
$= - [e^{- \infty} - e^{- 0}] + [\frac{e^{- 2 (\infty)}}{2} - \frac{e^{- 2 (0)}}{2}]$
$= 1 - \frac{1}{2}$
=0.5
Thus, the probability of $P (X > Y)$ is 0.5.

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