Find vector of expected values ​​and covariance matrix

Question

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?

Answer 1

$$f(x,y)=C \exp(-4x^2-6xy-9y^2)$$

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:

$$C=\frac{1}{2 \pi \sigma_x \sigma_y \sqrt{1-\rho^2}}$$

Answer 2

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.