# Marginal density functions from joint density function $\int_{-\infty}^{\infty} {e^{-y(x^2+ 1)}} dx$

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 $$f_{X,Y}(x, y) = \begin{cases}\frac{e^{-y(x^2+1)}}{\pi}, \quad -\infty0\\ 0, \quad \textrm{otherwise} \end{cases}$$ 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?

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

• @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 $x$). – spaceisdarkgreen Feb 26 at 4:43
• 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? – Nemo Feb 26 at 4:47
• 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) – spaceisdarkgreen Feb 26 at 4:53
• @Nemo For instance, $\int_0^\infty e^{-x}dx=\left.-e^{-x}\right |_0^\infty = 0-(-1)=1.$ Not $\int_0^\infty e^{-x}dx=-e^{-x}.$ – spaceisdarkgreen Feb 26 at 5:34
• @Nemo Yes, that's right. – spaceisdarkgreen Feb 28 at 2:55