Two random variables X and Y with joint density function given by: f(x,y)={13(2x+4y)0≤x,≤10elsewhere Find the marginal density of Y.

The joint density function of random variables X and Y is : f(x,y)={13(2x+4y)0≤x,≤10otherwise We have to find : marginal density of y fY(y)=∫−∞∞f(x,y)dy =∫0113(2x+3y)dy =13∫01(2x+3y)dy =13[2xy+3y22]01 =13[(2x+32)−0] =4x+36

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Albarellak

Answered question

2021-10-21

Two random variables X and Y with joint density function given by:

f (x, y) = {\begin{cases} \frac{1}{3} (2 x + 4 y) & 0 \leq x, \leq 1 \\ 0 & e l s e w h e r e \end{cases}

Find the marginal density of Y.

Answer & Explanation

SchulzD

Skilled2021-10-22Added 83 answers

The joint density function of random variables X and Y is :
$f (x, y) = {\begin{cases} \frac{1}{3} (2 x + 4 y) & 0 \leq x, \leq 1 \\ 0 & o t h e r w i s e \end{cases}$
We have to find :
marginal density of y
$f_{Y} (y) = \int_{- \infty}^{\infty} f (x, y) d y$
$= \int_{0}^{1} \frac{1}{3} (2 x + 3 y) d y$
$= \frac{1}{3} \int_{0}^{1} (2 x + 3 y) d y$
$= \frac{1}{3} {[2 x y + \frac{3 y^{2}}{2}]}_{0}^{1}$
$= \frac{1}{3} [(2 x + \frac{3}{2}) - 0]$
$= \frac{4 x + 3}{6}$

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