Random variables $X$ and $Y$ have joint PDF
$$f(x,y) = \begin{cases} \frac{3}{8} x^2 e^{-2y}, & -2<x<2, \, y>0\\[2ex] 0,&\text{otherwise} \end{cases} $$

a) Find the marginal PDF of $Y$:

This is my work, I am not sure if it is correct:

$$f(y)=\int_{-2}^2 (3/8) x^2 e^{-2y} dy = (3/8) x^2 (-1/2) e^{-2y} = (-3/16) x^2 e^{-2y}$$

b) Find $P(X>0)$:

I am unsure how to start this one. Would I need to find the marginal PDF of $X$ and then take the integral from $0$ to infinity?

c) Find $E(XY)$: I know that $E(XY)=E(X)E(Y)$. To find $E(X)$ would I take the integral on the range of $x$ of $x*fy(y)dy$?

  • $\begingroup$ Typically, $E(XY) = \displaystyle\int_0^\infty \int_{-2}^2 xy \cdot f(x,y) \, dxdy $. $\endgroup$ Apr 26 '16 at 4:44
  • $\begingroup$ Using that formula, I got 0, would that be correct? $\endgroup$
    – Lcheck
    Apr 26 '16 at 17:53
  • $\begingroup$ After the first integral I get 0, so would the integral of 0 be 1? Thats how I ended up with 0 $\endgroup$
    – Lcheck
    Apr 26 '16 at 19:53
  • $\begingroup$ Yes, you are right. My mistake. $\endgroup$ Apr 26 '16 at 19:58

$f(y)=\int_{-2}^2 \frac{3}{8}x^2e^{-2y}dx=\frac{3}{8}e^{-2y}\left(\left. \frac{x^3}{3} \right|_{-2}^2\right)=\frac{3}{8}e^{-2y}\frac{16}{3}=2e^{-2y}$

$f(x)=\frac{3}{8}x^2\int_0^\infty e^{-2y}dy=\frac{3}{8}x^2 \left( -\frac{1}{2}e^{-2y}|_0^\infty\right) = \frac{3}{16}x^2$

$P(X>0)=\int_0^2 \frac{3}{16}x^2 dx=\frac{3}{16}\frac{x^3}{3}|_0^2=\frac{1}{2}$


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