Suppose that the random variables X and Y have joint p.d.f. f(x,y)={kx(x−y),0&lt;x&lt;2,−x&lt;y&lt;x0, elsewhere Evaluate the constant k.

Given : f(x,y)={kx(x−y),0&lt;x&lt;2,−x&lt;y&lt;x0, elsewhere To find k ∫−∞∞∫−∞∞f(x,y)dxdy=1 =∫02∫−xxkx(x−y)dxdy=1 =∫02∫−xxk(x2−xy)dydx=1 =∫02k[x2y−xy22]−xxdx=1 =∫02k[x3−x32+x3+x32]dx=1 =∫02k[2x3]dx=1 =2k[x44]02=1 =2k(4)=1 =k=18

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Caelan

Answered question

2021-10-25

Suppose that the random variables X and Y have joint p.d.f.
$f (x, y) = {\begin{cases} k x (x - y), 0 < x < 2, - x < y < x \\ 0, e l s e w h e r e \end{cases}$
Evaluate the constant k.

Answer & Explanation

pivonie8

Skilled2021-10-26Added 91 answers

Given :
$f (x, y) = {\begin{cases} k x (x - y), 0 < x < 2, - x < y < x \\ 0, e l s e w h e r e \end{cases}$
To find k
$\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} f (x, y) d x d y = 1$
$= \int_{0}^{2} \int_{- x}^{x} k x (x - y) d x d y = 1$
$= \int_{0}^{2} \int_{- x}^{x} k (x^{2} - x y) d y d x = 1$
$= \int_{0}^{2} k {[x^{2} y - \frac{x y^{2}}{2}]}_{- x}^{x} d x = 1$
$= \int_{0}^{2} k [x^{3} - \frac{x^{3}}{2} + x^{3} + \frac{x^{3}}{2}] d x = 1$
$= \int_{0}^{2} k [2 x^{3}] d x = 1$
$= 2 k {[\frac{x^{4}}{4}]}_{0}^{2} = 1$
=2k(4)=1
$= k = \frac{1}{8}$

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