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I'm having trouble seeing why the bounds of integration used to calculate the marginal density of $X$ aren't $0 < y < \infty$.

Here's the problem:

$f(x,y) = \frac{1}{8}(y^2 + x^2)e^{-y}$ where $-y \leq x \leq y$, $0 < y < \infty$

Find the marginal densities of $X$ and $Y$.

To find $f_Y(y)$, I simply integrated away the "x" component of the joint probability density function:

$f_Y(y) = \frac{1}{8}\int_{-y}^y (y^2 + x^2)e^{-y} \, dx = \frac{1}{6}y^3e^{-y}$

Then to find $f_X(x)$,

$f_X(x) = \frac{1}{8}\int_0^\infty (y^2 + x^2)e^{-y} \, dy = \frac{-(x^2-2)}{8}$

However, the solutions I have say that the marginal density of $X$ above is wrong. Instead, it says that $f_X(x)$ is

$f_X(x) = \frac{1}{8}\int_{|x|}^\infty (y^2 + x^2)e^{-y} \, dy = \frac{1}{4}e^{-|x|}(1+|x|)$

Unfortunately, there is no explanation as to why the lower bound is $|x|$. The only thing that stands out to me are the bounds of $x$: $-y \leq x \leq y$.

Any constructive input is appreciated.

Mlagma
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    Perhaps drawing the domain where $f(x,y)$ is not zero would help. – Guest Mar 30 '14 at 05:03
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    Well, that certainly explains it. I suppose I will never forget to plot the domain again... – Mlagma Mar 30 '14 at 05:18
  • Usually this is the first thing one does in this kind of problems. – Guest Mar 30 '14 at 05:20
  • Yes, this is true, but for some reason it slipped my mind. It's been a while since I worked these sorts of problems, and yes, plotting the domain is quite basic. – Mlagma Mar 30 '14 at 05:26

2 Answers2

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The constraints are $-y \leq x \leq y$ and $0 < y < \infty$. That first constraint is equivalent to

$$ |x| \leq y $$

which, when combined with the second constraint, naturally yields

$$ |x| \leq y < \infty $$

Whence the integration limits.

Brian Tung
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If you graph $x = y$ and $x = -y$ ; and only look at $0 < y < \infty$ ; you will only use quadrants I and IV.

Thus $x$ can only be positive.

$0 < y < \infty $ aren't x-bounds.

Nizar
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Javier Gutierrez
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