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I'm given a joint probability density function of X and Y.

$$f(x, y)=c \mathrm{e}^{-y}|y-x|, \quad y>0 \text { and }-1 \leq x \leq 1$$

After normalizing I obtain $c = \frac{1}{\frac{-2}{e}+3}$. Now I am interested in the marginal densities of X and Y. How do I obtain these? I know to integrate with respect to each of the variables.

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  • $\begingroup$ it seems that you know what to do, where are you stuck? $\endgroup$ Nov 8, 2019 at 3:07
  • $\begingroup$ In particular, what are the bounds on each of the integrals? $\endgroup$
    – user581882
    Nov 8, 2019 at 3:37

2 Answers 2

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Guide:

If we want to compute $f_X(x)$, we are fixing the $x$ and we want to integrate over the $y$. Hence we should integrate from $0$ to $\infty$.

If we want to compute $f_Y(y)$, we are fixing the $y$ and we want to integrate over the $x$. Hence we should integrate from $-1$ to $1$.

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  • $\begingroup$ The problem is that when trying to compute the first one, there is an issue somewhere, due to the absolute value. $\endgroup$
    – user581882
    Nov 8, 2019 at 4:09
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    $\begingroup$ The second one is well-defined, but the first has a problem. $\endgroup$
    – user581882
    Nov 8, 2019 at 4:10
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    $\begingroup$ Split the integral into two parts when $y> x$ and when $y<x$ then. $\endgroup$ Nov 8, 2019 at 4:10
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You have found $c$ by using: $$\begin{align}1&=\iint_{y>0, -1\leq x\leq 1} f_{X,Y}(x,y)~\mathrm d (x,y)\\[1ex]&= \int_{-1}^1\int_0^\infty c\, \mathrm e^{−y} \lvert y−x\rvert~\mathrm d y~\mathrm d x\\[1ex]&=\int_{-1}^1\left(\int_0^x c\mathrm e^{-y}(x-y)\mathrm d y+\int_x^\infty c\mathrm e^{-y}(y-x)\mathrm d y\right)~\mathrm d x\end{align}$$

Surely you can then see that:

$$f_X(x)=\int_{y>0} f_{X,Y}(x,y)~\mathrm d y\\f_Y(y)=\int_{-1\leq x\leq 1} f_{X,Y}(x,y)~\mathrm d x$$

Brought to you by the fact that the supports are seperable.

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  • $\begingroup$ Thank you, the problem is with the first integral though, given the limits and the absolute value. I'm not sure how to break it up appropriately to make it solvable. $\endgroup$
    – user581882
    Nov 8, 2019 at 4:50
  • $\begingroup$ Partition the integrals on $y\leq x$ and $y> x$. $\endgroup$ Nov 8, 2019 at 4:52

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