Marginal Density Function from Joint Probability Density Function

Question

I am in an introduction to probability class and we just covered joint probabilities. I came across a question which I cannot compute and would appreciate some help.

Suppose $X$, $Y$ are jointly continuous with joint probability density function $$f(x, y) = \frac{1}{2\pi}e^{-\frac{x^2}{2}-\frac{(x-y)^2}{2}}$$ with $x$, $y$ $\in$ (-$\infty$, $\infty$). Find the marginal density functions of $X$ and $Y$. Hint: you can do this without complicated integrals.

I am aware that $f(x, y)$ looks like two normal densities, but I am unable to figure out how to use this to my advantage when calculating the marginal densities without brute integration. Any help would be appreciated.

Answer 1

You are right that this looks like a bivariate normal distribution. Therefore let us try to express the given density in such a way. The density of a bivariate normal distribution is given by $g(z)= \frac{1}{2\pi \sqrt{\text{det} \Sigma}} e^{-\frac{1}{2}(z - \mu)^T \Sigma^{-1} (z-\mu)}, z\in \mathbb{R}^2 $, where $\Sigma \in \mathbb{R}^{2 \times 2}$ is the covariance matrix and $\mu \in \mathbb{R}^2$ is the expectation. The fact that the term in the exponent of your given density function does not contain any terms that are not depending on $x$ or $y$ suggests $\mu =0$. Now easy calculations give that \begin{align} \Sigma^{-1}= \left( \begin{matrix} 2& -1 \\ -1 & 1 \end{matrix} \right) \end{align} fulfills \begin{align} \left( \begin{matrix} x &y \end{matrix} \right) \Sigma^{-1} \left( \begin{matrix} x \\y \end{matrix} \right) = x^2 + (x-y)^2. \end{align} Now inverting gives $\Sigma= \left( \begin{matrix} 1& 1 \\ 1 & 2 \end{matrix} \right) $ and $\text{det}\Sigma=1$. Therefore the given density function is the density of a bivarite normal distribution with covariance matrix $\Sigma $ as above and zero expectation. It is well known that for a multivariate normal distribution $Z \sim N(\mu,\Sigma)$ the random variable $c^TZ$ has the distribution $N(c^T\mu, c^T \Sigma c)$ and thus in your case $X \sim N(0,1)$ and $Y \sim N(0,2)$.

Answer 2

$f_X(x)=\int_{-\infty}^\infty f(x,y)\, dy=\frac{1}{2\pi}e^{-x^2/2}\int_{-\infty}^\infty e^{-(x-y)^2/2}\, dy$. Notice that with $u=x-y,$ above is equivalent to $$ \frac{1}{2\pi}e^{-x^2/2}\int_{-\infty}^\infty e^{-u^2/2}\, du $$ In general, for a normal distribution $N(\mu,\sigma)$, the density is $$ \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}. $$ For the integral above involving $u$, suppose $\sigma=1$. Then it can be rewritten as $$ \left[\frac{1}{2\pi}\cdot\sqrt{2\pi\sigma^2} e^{-x^2/2}\right]\left[\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty e^{-u^2/2\sigma^2}\, du\right], $$ with the term on the right evaluating to $1$. See if you can apply a similar technique for $f_Y(y)$. Also, if you clean up the term on the left, you will see that $X\sim N(0,1)$.

Of course, this is a child's way of doing it. The grownup's method is to work backwards from a bivariate normal.