<s>[INST] Find the marginal probability density function The random vector $[\,X \,\,\, Y \,]'$ has probability density function 
$f_{X,Y} (x,y) = ke^{-2x^2-3xy-\frac{9}{2}y^2}$, where $k$ is some constant
Find $k.$
Find the marginal probability density functions of $X$ and $Y.$
I know for it to be a valid pdf its integral from negative to positive infinity must be equal to one, and that it must be greater than $0$ for all $x.$ But for starters I'm not sure on the integration.
 [/INST] A standard form of the bivariate normal density with expected value $(0,0)$ is this:
$$
f(x,y) = \frac{1}{2 \pi  \sigma_X \sigma_Y \sqrt{1-\rho^2}}
\exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{x^2}{\sigma_X^2} + \frac{y^2}{\sigma_Y^2} - \frac{2\rho xy}{\sigma_X \sigma_Y} \right] \right)
$$
The exponent is this:
$$
-\frac 1 {2(1-\rho^2)}\left[ \frac{x^2}{\sigma_X^2} + \frac{y^2}{\sigma_Y^2} - \frac{2\rho xy}{\sigma_X \sigma_Y} \right] = -2x^2-3xy-\frac{9}{2}y^2
$$
So
$$
\frac 1 {1-\rho^2} \left[ \left( \frac x {\sigma_X}\right)^2 + \left( \frac y {\sigma_Y} \right)^2 - 2\rho\left( \frac x {\sigma_X} \right)\left( \frac y {\sigma_Y} \right) \right] = 4x^2 + 6xy +9y^2 \tag 1
$$
We can see that this is positive-definite by completing the square and writing it as
$$
\left(2x + \frac 3 2 y \right)^2 + \frac {27} 4 y^2,
$$
so it's always positive.
Equating coefficients, we get
\begin{align}
& \frac 1 {(1-\rho^2)\sigma_X^2} = 4 \\[10pt]
& \frac 1 {(1-\rho^2)\sigma_Y^2} = 9 \\[10pt]
& \frac{-2\rho}{(1-\rho^2)\sigma_X\sigma_Y} = 6
\end{align}
From this it follows that
\begin{align}
\sigma_X^2 & = 1/3, \\[8pt]
\sigma_Y^2 & = 4/27, \\[8pt]
\sigma_X \sigma_Y & = 2/9, \\[8pt]
\rho & = -1/2.
\end{align}
The marginal density for $X$ is
$$
\frac 1 {\sigma_X\sqrt{2\pi}} \exp\left( \frac {-1} 2 \left( \frac x {\sigma_X} \right)^2 \right)
$$
and similarly for $Y.$ (If we had $\mu_X\ne0$ then instead of $\dfrac x {\sigma_X}$ we'd have $\dfrac {x-\mu_X}{\sigma_X}.$)
The constant $k$ should be $\dfrac{1}{2\pi\sigma_X\sigma_Y(1-\rho^2)}.$
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