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Question . The joint probability density function of X and Y is given by $f(x, y) = (1/8)(y^2 − x^2)e^{-y} , -y\leq x\leq y, 0\leq y \leq \infty $
Find the marginal density of x.

So i know that we need to integrate out the Y from the joint PDF,and thus i came up with this integral $f_x(x)=\int_0^\infty (1/8)(y^2 − x^2)e^{-y} dy$. i evaluated this to get $(1/8)(2-x^2)$

However, the solutions provided is different. Is there something wrong with my bounds? Help will be much appreciated.

hazard
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1 Answers1

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There is two ways to address this question. But both rely on the fact that $\mathbb{P} \left(-y \leq x \leq y \,\big| \, y\geq 0 \right)=1$, so you can see it as:

$$f_X(x) = \frac{1}{8} \int_x^{\infty} \left(y^2 - x^2 \right) e^{-y} \mathrm{d}y = \frac{1}{8} \int_0^{\infty} \left(y^2 - x^2 \right)\mathbb{1}_{\left\{-y \leq x \leq y \right\}} e^{-y} \mathrm{d}y = \frac{x+1}{4e^{x}} \mathbb{1}_{\left\{x \geq 0 \right\}}.$$

Guilherme Thompson
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  • how did your limits of integration changed from infinity-x to infinity-0? – hazard Apr 20 '16 at 10:21
  • In fact, both have the same result, but they are "different" integrations. In any case, think about the measure $(y^2-x^2)e^{-y}$ that cannot be negative, for $x \geq 0$. It implies $x \leq y$. I even have let $- y \leq x \leq y$ like this, but you can simplify it to $x \geq y$. – Guilherme Thompson Apr 20 '16 at 11:29