Suppose that the random variables X and Y have joint p.d.f. (f(x,y)={kx(x−y)0Find the marginal p.d.f. of the two random variables.

Given : (f(x,y)={kx(x−y)0 To find marginal P.D.F of x fx(x)=18∫−xxx(x−y)dy 18∫−xx(x2−xy)dy =18[x2y−xy22]−xx =18[x3−x32+x3+x32] =18(2x3) =x34 To find marginal P.D.F of y fy(y)=18∫02x(x−y)dx =18∫02(x2−xy)dx =18[x33−x2y2]02 =18[83−4y2−0−0] =18(83−y) =13−y8 =8−3y24

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babeeb0oL

Answered question

2021-05-31

Suppose that the random variables X and Y have joint p.d.f.
$(f (x, y) = {\begin{cases} k x (x - y) \\ 0 \end{cases}$ Find the marginal p.d.f. of the two random variables.

Answer & Explanation

Demi-Leigh Barrera

Skilled2021-06-01Added 97 answers

Given :
$(f (x, y) = {\begin{cases} k x (x - y) \\ 0 \end{cases}$
To find marginal P.D.F of x
$f_{x} (x) = \frac{1}{8} \int_{- x}^{x} x (x - y) d y$
$\frac{1}{8} \int_{- x}^{x} (x^{2} - x y) d y$
$= \frac{1}{8} [x^{2} y - \frac{x y^{2}}{2}]_{- x}^{x}$
$= \frac{1}{8} [x^{3} - \frac{x^{3}}{2} + x^{3} + \frac{x^{3}}{2}]$
$= \frac{1}{8} (2 x^{3})$
$= \frac{x^{3}}{4}$
To find marginal P.D.F of y
$f_{y} (y) = \frac{1}{8} \int_{0}^{2} x (x - y) d x$
$= \frac{1}{8} \int_{0}^{2} (x^{2} - x y) d x$
$= \frac{1}{8} [\frac{x^{3}}{3} - \frac{x^{2} y}{2}]_{0}^{2}$
$= \frac{1}{8} [\frac{8}{3} - \frac{4 y}{2} - 0 - 0]$
$= \frac{1}{8} (\frac{8}{3} - y)$
$= \frac{1}{3} - \frac{y}{8}$
$= \frac{8 - 3 y}{24}$

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