Find vector of expected values ​​and covariance matrix For vector (X,Y) with density
$f(x,y)=C exp \{ -4x^2-6xy-9y^2 \}$
find constans C, vector of expected values ​​and covariance matrix.
How to do this kind of exercises?
 A: Note that 
$$
4x^2+6xy+9y^2=9\left(y+\tfrac13x\right)^2+3x^2,
$$
and that, for every positive $a$,
$$
\int_\mathbb R\mathrm e^{-az^2}\,\mathrm dz=\sqrt{\frac{\pi}a},
$$
hence
$$
\iint\mathrm e^{-4x^2-6xy-9y^2}\,\mathrm dx\mathrm dy=\int\mathrm e^{-3x^2}\int\int\mathrm e^{-9\left(y+\tfrac13x\right)^2}\mathrm dy\mathrm dx
$$
is
$$
\int\mathrm e^{-3x^2}\sqrt{\frac{\pi}9}\mathrm dx=\sqrt{\frac{\pi}3}\,\sqrt{\frac{\pi}9}=\frac\pi{3\sqrt3}.
$$
This shows that
$$
C=\frac{3\sqrt3}\pi.
$$
The decomposition of the binomial we started with also shows that, if $$
Z=Y+\frac13X,
$$
then $(X,Z)$ is a centered independent normal vector with variances 
$$
\sigma^2_X=\frac16,\qquad\sigma^2_Z=\frac1{18}.
$$ 
To see this, simply note that the change of variable formula shows that the density of $(X,Z)$ is proportional to
$$
\mathrm e^{-3x^2-9z^2}=\mathrm e^{-x^2/(2\sigma_X^2)}\,\mathrm e^{-z^2/(2\sigma_Z^2)}.
$$
Finally, $Y=Z-\frac13X$ hence $\sigma^2_Y=\sigma^2_Z+\frac19\sigma^2_X$ and $\mathrm{Cov}(X,Y)=-\frac13\sigma^2_X$, hence you are done.
A: $$f(x,y)=C \exp(-4x^2-6xy-9y^2)$$
The form is clearly a normal, the bivariate case is the next:
   $$ 
     f(x,y) = \frac{1}{2 \pi  \sigma_x \sigma_y \sqrt{1-\rho^2}}
      \exp\left(
        -\frac{1}{2(1-\rho^2)}\left[
          \frac{(x-\mu_x)^2}{\sigma_x^2} +
          \frac{(y-\mu_y)^2}{\sigma_y^2} -
          \frac{2\rho(x-\mu_x)(y-\mu_y)}{\sigma_x \sigma_y}
        \right]
      \right)$$
How there is no constant in the expression the guess is $\mu_x=\mu_y=0$
$$ 
     f(x,y) = \frac{1}{2 \pi  \sigma_x \sigma_y \sqrt{1-\rho^2}}
      \exp\left(
        -\frac{1}{2(1-\rho^2)}\left[
          \frac{x^2}{\sigma_x^2} +
          \frac{y^2}{\sigma_y^2} -
          \frac{2\rho xy}{\sigma_x \sigma_y}
        \right]
      \right)$$
Then looking in your density:
$$2(1-\rho^2)\sigma_x^2=\frac{1}{4}$$
$$2(1-\rho^2)\sigma_y^2=\frac{1}{6}$$
$$\frac{(1-\rho^2)}{\rho}\sigma_x \sigma_y=-\frac{1}{9}$$
Once you have those values:
$$C=\frac{1}{2 \pi  \sigma_x \sigma_y \sqrt{1-\rho^2}}$$
