<s>[INST] 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?
 [/INST] $$f(x,y)=C \exp(-4x^2-6xy-9y^2)$$
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:
$$C=\frac{1}{2 \pi  \sigma_x \sigma_y \sqrt{1-\rho^2}}$$
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