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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.

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  • $\begingroup$ To find the marginal of X, you want to integrate the joint density with respect to y. Rather than solving the resulting integral, try to manipulate it a little to make it look like a known density function, which you know integrates to 1. $\endgroup$
    – Nasenhaar
    Mar 28, 2020 at 19:40
  • $\begingroup$ @Nasenhaar I tried to manipulate it to look like the standard bivariate normal distribution, but wasn't able to. Is this the correct density function (which I know integrates to 1)? $\endgroup$
    – aslconwnb
    Mar 28, 2020 at 19:46

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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 $Y \sim N(0,2)$.

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$f_X(x)=\int_{-\infty}^\infty f(x,y)\, dy=\frac{1}{2\pi}e^{-x^2/2}\int_{-\infty}^\infty e^{-(x-y)^2/2}\, dy$. 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.

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  • $\begingroup$ Thanks! One small question: wouldn't du = -dy, so therefore when we substitute u in we should change the sign of the integrand? I've seen problems like this done before and others have left out this step, so I'm pretty sure I'm missing something. Could you clarify what I'm missing? $\endgroup$
    – aslconwnb
    Mar 28, 2020 at 22:53
  • $\begingroup$ Suppose $u=x-y$. Then you're right that $du=-dy$. But when $y \to -\infty$, we see that $u \to \infty$ and when $y \to \infty$, we see that $u \to -\infty$. Hence you get an integral like $-\int_{\infty}^{-\infty} ...$. But in general, $-\int_a^b=\int_b^a$. Thus, the sign doesn't really change. $\endgroup$
    – ProfOak
    Mar 28, 2020 at 22:56
  • $\begingroup$ That makes sense thank you so much. $\endgroup$
    – aslconwnb
    Mar 28, 2020 at 22:58

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