Expert Answer to "The area enclosed by the x and y..."

Kyran Hudson

Answered question

2021-05-25

The area enclosed by the x and y axes has a uniform distribution of the random variables X and Y, and the curve $y = 1 - x^{2}$ .
Compute E. (XY)

Answer & Explanation

SchulzD

Skilled2021-05-26Added 83 answers

So,
$X \sim U (0, 1)$
$Y \sim U (0, 1)$
Given curve $y = 1 - x^{2}$
$f (x) = {\begin{cases} 1 & 0 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$
$f (y) = {\begin{cases} 1 & 0 < y < 1 \\ 0 & o t h e r w i s e \end{cases}$
$f (x, y) = \int_{0}^{x} (1 - x^{2}) d x = x - \frac{x^{3}}{3}$
$E (X Y) = \int_{0}^{1} \int_{0}^{1} x y f (x, y) d y d x$
$= \int_{0}^{1} \int_{0}^{1} x y (x - \frac{x^{3}}{3}) d y d x$
$= \int_{0}^{1} x (x - \frac{x^{3}}{3}) d x \int_{0}^{1} y d y$
$= \int_{0}^{1} x (x - \frac{x^{3}}{3}) d x (\frac{y^{2}}{2})_{0}^{1}$
$= \int_{0}^{1} x (x - \frac{x^{3}}{3}) d x (\frac{1}{2} - 0)$
$= \frac{1}{2} [\int_{0}^{1} x (x - \frac{x^{4}}{3}) d x]$
$= \frac{1}{2} [\frac{x^{3}}{3} - \frac{x^{5}}{3 \times 5}]_{0}^{1}$
$= \frac{1}{2} [\frac{(1 - 0)}{3} - \frac{(1 - 0)}{15}]$
$= \frac{1}{2} [\frac{1}{3} - \frac{1}{15}]$
$= \frac{1}{2} [\frac{5 - 1}{15}]$
$= \frac{2}{15}$

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