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So I've got the density function for the $2$-dimensional random variable $(X,Y)$: $$p(x,y) = \frac{4}{3}xe^{-x-y} $$ when $ 0 < y < x$. Otherwise, it's $0$.

I am now interested in the density of the random variable $W = X + Y$. This is given by:

$$\int_{-\infty}^\infty p(x,w-x)dx$$

Fine enough, but now I run into a problem I always have... what are all the limits concerning this problem? Not just for the integral, but for $W$ as a random variable as well; i.e. when does its density function equal $0$, and when does it equal a function $d(w)$?

recursion.ninja
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Stan fesn
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1 Answers1

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As usual, the solution is to write the PDFs correctly, that is, as functions defined on their full domain, including indicator functions. Here, the density $p$ is defined on the whole plane $\mathbb R^2$ by $$p(x,y)=\tfrac43x\mathrm e^{-x-y}\mathbf 1_{0\lt y\lt x},$$ hence the density $q$ of $X+Y$ is defined on the whole line $\mathbb R$ by $$q(w)=\int_\mathbb R p(x,w-x)\mathrm dx=\int_\mathbb R \tfrac43x\mathrm e^{-w}\mathbf 1_{0\lt w-x\lt x}\mathrm dx=\tfrac43\mathrm e^{-w}\mathbf 1_{w\gt0}\int_\mathbb R x\mathbf 1_{w/2\lt x\lt w}\mathrm dx,$$ that is, $$q(w)=\tfrac43\mathrm e^{-w}\mathbf 1_{w\gt0}\int_{w/2}^w x\mathrm dx=\tfrac23\mathrm e^{-w}\mathbf 1_{w\gt0}\left.\ x^2\right|_{w/2}^w=\tfrac12w^2\mathrm e^{-w}\mathbf 1_{w\gt0}.$$

Did
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  • This is what I don't understand; $$\int_\mathbb R \tfrac43x\mathrm e^{-w}\mathbf 1_{0\lt w-x\lt x}\mathrm dx=\tfrac43\mathrm e^{-w}\mathbf 1_{w\gt0}\int_\mathbb R x\mathbf 1_{w/2\lt x\lt w}\mathrm dx,$$ How are those two equal? I understand how we reach the expression on the left, but not the one on the right – Stan fesn Jan 03 '15 at 19:54
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    $$\mathbf 1_{0\lt w-x\lt x}=\mathbf 1_{x\lt w\lt2x}=\mathbf 1_{w\gt0}\mathbf 1_{w/2\lt x\lt w}$$ – Did Jan 03 '15 at 20:07
  • @Did how did you come up with limits $w$ and $w\over 2$ ? – Andy Jan 08 '15 at 15:53
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    @Andy See my previous comment. – Did Jan 08 '15 at 15:55