In my textbook, it asks us to find E[X], E[Y], Var[X], Var[Y] and Cov(X,Y) of the following jointly Gaussian random variable:
$$ f_{x,y}(X,Y) = \frac1{2\pi} \exp\Bigg\{-\frac{1}{2}(x^2+4y^2-3xy+3y-2x+1)\Bigg\} $$
I've spent an unacceptable amount of time finding the marginal pdf of x via integration:
$$f_x(X) = \frac1{\sqrt{2\pi}}\exp\Bigg\{ -\frac{41}{64} \big(x-1 \big)^2 \Bigg\}$$
and finding $E[X]$ by solving $\int_x xf_x(X) dx$ seems like an insane amount of work on top of what I've already done for just the solution to the first subquestion.
Am I missing something? Is there a simpler way of finding these things, or is my textbook just especially sadistic?
There is a simpler way to find the pdf $f(x)$ without resorting to integration, instead, by just comparing the standard joint pdf,
$$\frac1{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}} \exp\left[-\frac1{2(1-\rho^2)} \left( \frac{(x-\mu_x)^2}{\sigma_x^2} + \frac{(y-\mu_y)^2}{\sigma_y^2} - \frac{2\rho(x-\mu_x)(y-\mu_y)}{\sigma_x\sigma_y} \right) \right]$$
with the given pdf of the form
$$\frac1{2\pi A} e^{-\left( ax^2+by^2 + c xy + dx+ ey +g \right)}$$
Note $\sigma_x\sigma_y\sqrt{1-\rho^2}=A$. Then, we obtain $\rho$ from matching the coefficient $c$, $$\frac{\rho}{A\sqrt{1-\rho^2}} = - c$$
Next, with known $\rho$, obtain $\sigma_x$ and $\sigma_y$ from
$$a = \frac1{2(1-\rho^2)\sigma_x^2},\>\>\>\>\> b = \frac1{2(1-\rho^2)\sigma_y^2}$$
In the end, obtain $\mu_x$ and $\mu_y$ from matching the coefficients $d$ and $e$. Thus, from the deduced $\mu_x$ and $\sigma_x$, the pdf of $x$ is
$$f(x) = \frac1{\sqrt{2\pi\sigma_x}} e^{-\frac{(x-\mu_x)^2}{2\sigma_x^2} }$$