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I am to show that $$f_{XY}(x,y)= \begin{cases} c(y^2-x^2)e^{-y}, & (-y \le x \le y),\quad y\gt0 \\ 0, & \text{otherwise} \end{cases}$$ has gamma density and show that $c=\frac18$. I know I need to set up the double integral, and I got to solving for $dx$ first but and after that it is just a complicated function. I feel it should be a little shorter since it's a homework problem but at the same time it wouldn't surprise me that it's this long. Here is what I started with maybe my bounds are incorrect? $$\int_0^\infty \int_{-y}^y cy^2e^{-y}-cx^2e^{-y}dxdy$$ Is this set up correctly? Or is this really just a long problem?

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Your problem is indeed framed correctly. Merely equate RHS to 1 after integration and you are done. If necessary, change the order of integration. You might end up with a simpler form. Also what bounds are you talking about? –  Gautam Shenoy Nov 19 '12 at 19:50
    
@GautamShenoy In my setup I have x ranging from -y to y, and y 0 to infinity, is that the right bounds? –  TheHopefulActuary Nov 19 '12 at 23:34
    
The correct term is "limits of integration", not bounds. –  Gautam Shenoy Nov 20 '12 at 7:50
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up vote 1 down vote accepted

I am not sure what you mean by the joint distribution having Gamma density (can anyone comment on this?), but you can use it to help simplify the integral of the joint prob. density function:

$$\int_0^\infty \int_{-y}^y c (y^2-x^2) e^{-y} dx dy$$ $$=c\int_0^\infty e^{-y} \left(\int_{-y}^y (y^2-x^2)dx\right)dy$$ $$=c\int_0^\infty e^{-y} \left(\frac{4}{3} y^3\right)dy$$ $$=\frac{4}{3}c\int_0^\infty e^{-y} y^3dy$$ $$=(3!)\frac{4}{3} c\int_0^\infty \frac{e^{-y} y^3}{3!}dy$$ $$=8c$$

In the fourth line of my work above, I used the fact that the integrand looked suspiciously close to the Gamma distribution. Dividing by $3!$ would make it a pdf of a Gamma distribution with shape parameter 4 and rate parameter 1, whose integral over all $x\ge 0$ would equal 1.

Hopefully you see why $c=\frac18$ and how the Gamma distribution is used to help solve this problem.

Remark: this is a pattern in many probability homework problems: rarely do you have to actually compute integrals; you manipulate them until the integrand resembles a known p.d.f., and then the integral evaluates to 1.

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Holy crap wow... This is why math math never fails to amaze me. Thank you so much. –  TheHopefulActuary Nov 20 '12 at 4:51
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The set up, including the bounds, appears to be correct, assuming you equate that integral times $c$ to 1 and solve for c. It doesn't seem too long — one normal integration and then three integrations by parts, which can be done very quickly with tabular integration.

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