given the joint PDF:
$$f(x,y) = \frac{1}{4π} e^{-\frac{1}{2}(x^2-2xy+5y^2)}$$
a) show that $X \sim N(0, 5/4)$ and $Y \sim N(0, 1/4)$
b) find $\operatorname{corr}(X,Y)$ and show whether $X$ and $Y$ are independent.
c) what are expected values and how can you conclude $μX = μY = 0$ for this PDF
I understand that a) can be shown by utilizing the properties of the multivariate normal distribution but don't know how to arrive at an answer.
The joint PDF should be $$ f_{X,Y}(x,y)=\frac{1}{\pi}\exp\left(-\frac{1}{2}\left(x^2-2xy+5y^2\right)\right). $$ Otherwise, $$ \int_{\mathbb{R}^2}f_{X,Y}(x,y){\rm d}x{\rm d}y\ne 1, $$ making $f_{X,Y}(x,y)$ no longer a PDF.
According to the definition, provided the joint PDF $f_{X,Y}=f_{X,Y}(x,y)$, the marginal PDF of $X$ (i.e., $f_X$) reads \begin{align} f_X(x)&=\int_{\mathbb{R}}f_{X,Y}(x,y){\rm d}y\\ &=\int_{-\infty}^{\infty}\frac{1}{\pi}\exp\left(-\frac{1}{2}\left(x^2-2xy+5y^2\right)\right){\rm d}y\\ &=\frac{2}{\sqrt{10\pi}}\exp\left(-\frac{2}{5}x^2\right). \end{align} Compare this result with the PDF of normal distribution $\mathcal{N}(\mu,\sigma^2)$ $$ \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{1}{2}\frac{\left(x-\mu\right)^2}{\sigma^2}\right), $$ and it is clear that $$ X\sim\mathcal{N}\Bigl(0,\frac{5}{4}\Bigr). $$
Similarly, the marginal PDF of $Y$ (i.e., $f_Y$) reads \begin{align} f_Y(y)&=\int_{\mathbb{R}}f_{X,Y}(x,y){\rm d}x\\ &=\int_{-\infty}^{\infty}\frac{1}{\pi}\exp\left(-\frac{1}{2}\left(x^2-2xy+5y^2\right)\right){\rm d}x\\ &=\frac{2}{\sqrt{\pi}}\exp\left(-2y^2\right). \end{align} Again, compare this result to the PDF of normal distribution $\mathcal{N}(\mu,\sigma^2)$, and it is clear that $$ Y\sim\mathcal{N}\Bigl(0,\frac{1}{4}\Bigr). $$
According to the definition, \begin{align} \text{corr}\left(X,Y\right)&=\frac{\mathbb{E}\left(X-\mu_X\right)\left(Y-\mu_Y\right)}{\sigma_X\sigma_Y}=\frac{\mathbb{E}XY}{\sqrt{5/4}\sqrt{1/4}}=\frac{4}{\sqrt{5}}\mathbb{E}XY\\ &=\frac{4}{\sqrt{5}}\int_{\mathbb{R}^2}xyf_{X,Y}(x,y){\rm d}x{\rm d}y\\ &=\frac{4}{\sqrt{5}}\int_{\mathbb{R}^2}xy\frac{1}{\pi}\exp\left(-\frac{1}{2}\left(x^2-2xy+5y^2\right)\right){\rm d}x{\rm d}y\\ &=\frac{4}{\sqrt{5}}\cdot\frac{1}{4}=\frac{1}{\sqrt{5}}. \end{align} Since $X$ and $Y$ are Pearson-correlated, they must be dependent random variables.
(I did not get what part (c) was asking about...)