How to find a marginal distribution from a joint distribution?

Question

So I have this joint dist

$$f(x,y) = \frac{1}{2\pi}\exp(-\frac{x^2}{2} - \frac{y^2}{2} + x^2y - \frac{x^4}{2})$$

I'd like to find $f_x(x)$. So I know that means I need to integrate function wrt y. So my strategy is to pull out whatever terms I can so some of it will look like the normal dist then it'll integrate to one and I'll be left with the marginal. So I have $$f_x(x) = \frac{1}{\sqrt{2\pi}}\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2} - \frac{y^2}{2} + x^2y - \frac{x^4}{2})dy$$

This issue then is I'm not sure how I can modify $\exp(-\frac{x^2}{2} - \frac{y^2}{2} + x^2y - \frac{x^4}{2})$ so it will take the form $\exp(-\frac{(y-\mu)^2}{2\sigma^2})$.

Any help would be great, thanks. Sorry for the uberitalicized tex, kinda hacked it together from what I know.

Also I know this is a lot to ask, but this is homework so if you could provide hints or tips as opposed to a complete solution that would be better.

Edited for clarity.

Answer 1

First re-write the joint density as

$$f(x,y) = \frac{1}{2\pi} {\rm exp} ({-x^{2}/2}) \cdot {\rm exp} \Big( -\frac{1}{2} \left( y^2 -2x^{2}y + x^4 \right) \Big) $$

Note that $y^2 -2x^{2}y + x^4 = (y-x^{2})^2$. Now the marginal density of $x$ is

$$ f(x) = \int_{-\infty}^{\infty} f(x,y) dy = \frac{1}{\sqrt{2 \pi}} {\rm exp} ({-x^{2}/2}) \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} {\rm exp} \Big( -\frac{(y-x^{2})^2}{2} \Big) dy $$

The integrand is a $N(x^{2},1)$ density and therefore it integrates to 1. Therefore,

$$f(x) = \frac{1}{\sqrt{2 \pi}} {\rm exp} ({-x^{2}/2})$$

In other words, the marginal distribution of $X$ is standard normal.

Answer 2

Look back in your college algebra or pre-calc text book for the concept of "completing the square" and remember that since you are integrating out 'y' that 'x' and the expressions including 'x' but not 'y' are constants.