Although this link said that $$\int_{0}^{+\infty} {e^{-x^2}} dx = \frac{\sqrt{\pi}}{2}$$ I wasn't unable to leverage the information to derive marginal density functions from this joint density function $$f_{X,Y}(x, y) = \begin{cases}\frac{e^{-y(x^2+1)}}{\pi}, \quad -\infty<x<+\infty, \quad y>0\\ 0, \quad \textrm{otherwise} \end{cases} $$ That is, $$f_X(x) = ? $$ $$f_Y(y) = -\frac{e^{-y(x^2+1)}}{\pi(x^2+1)} $$ Could you please show me:

  • How to calculate $f_X(x)$?

  • Is $f_Y(y)$ summed to 1?

Thanks in advance.


1 Answer 1


It is not the case that $$ f_Y(y) = \frac{1}{\pi} e^{-y(x^2+1)}.$$ That is what they gave you for $f_{X,Y}(x,y).$ Furthermore, $f_Y(y)$ should only depend on $y$.

It is given by integrating $f_{X,Y}$ over $x,$ so $$ f_Y(y) = \int_{-\infty}^\infty \frac{1}{\pi} e^{-y(x^2+1)}dx.$$

Simplifying a little, $$ \int_{-\infty}^\infty \frac{1}{\pi} e^{-y(x^2+1)}dx = \frac{1}{\pi}e^{-y} \int_{-\infty}^\infty e^{-yx^2}dx = \frac{2}{\pi}e^{-y} \int_{0}^\infty e^{-yx^2}dx.$$

Then we can use the integral you were given. Substituting $z=\sqrt{y}x$ gives $$ \frac{2}{\pi}e^{-y} \int_{0}^\infty e^{-yx^2}dx=\frac{2}{\pi}e^{-y} \frac{1}{\sqrt{y}}\int_{0}^\infty e^{-z^2}dz=\frac{1}{\sqrt{\pi y}}e^{-y}$$ where $y$ ranges over $(0,\infty).$

Similarly, $$ f_X(x) = \int_0^\infty \frac{1}{\pi}e^{-y(x^2+1)}dy.$$ This is an integral in $y$ so you can just treat the $x^2+1$ as a constant. A run of the mill u substitution gives, $$ \int_0^\infty e^{-ay}dy = \frac{1}{a}$$ for $a>0.$ Thus we have $$ f_X(x)=\frac{1}{\pi}\int_0^\infty e^{-y(x^2+1)}dy = \frac{1}{\pi}\frac{1}{x^2+1}.$$

  • 1
    $\begingroup$ @Nemo It's a definite integral, not an indefinite integral. Also you should be integrating with respect to x rather than y for that one (so treating $y$ as a constant, not $x$). $\endgroup$ Feb 26, 2020 at 4:43
  • $\begingroup$ Thanks for your quick response. I made a typo in my OP regarding $f_X(y)$. It should have been $\frac{e^{-y(x^2+1)}}{\pi(x^2+1)}$ instead of $frac{e^{-y(x^2+1)}}{\pi}$. This was how I did: set $u = -y(x^2+1)$ where $(x^2+1)$ was a constant. Then $\int{\frac{e^{-y(x^2+1)}}\{pi}} = \int{\frac{e^u}{\pi} = -\frac{e^u}{\pi(x^2+1} = \frac{e^{-y(x^2+1)}{\pi(x^2+1)}$. Could you please tell me what I did wrong? $\endgroup$
    – Nemo
    Feb 26, 2020 at 4:47
  • $\begingroup$ I appreciate your effort to try to fix the tex problems in your first comment, but I read it fine and my previous comment is a response to it. (Also, in case you don't know, you can edit comments within 5 minutes) $\endgroup$ Feb 26, 2020 at 4:53
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    $\begingroup$ @Nemo For instance, $\int_0^\infty e^{-x}dx=\left.-e^{-x}\right |_0^\infty = 0-(-1)=1.$ Not $\int_0^\infty e^{-x}dx=-e^{-x}.$ $\endgroup$ Feb 26, 2020 at 5:34
  • 1
    $\begingroup$ @Nemo Yes, that's right. $\endgroup$ Feb 28, 2020 at 2:55

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