Can the mean and variance of X be extracted from this joint probability?

Question

If we have the joint distribution function:
$\Large f_{X,Y}(x,y) = \frac{\exp{\frac{(36x²-24xy+ 25y^2 − 120x − 2y + 121)}{-1512}}}{55\pi}$

Can we know the mean and variance of X? How?

Answer 1

What you have is a multivariate normal distribution, more specifically yours is bivariate. In that case the density funcction is of the following form: $$ \begin{aligned} f(x,y) = \frac{1}{2\pi \sigma_X \sigma_Y\sqrt{1 - \rho^2}} \exp(-\frac{1}{2(1 - \rho^2)}[ \frac{(x - \mu_X)^2}{\sigma_X^2} + \frac{(y - \mu_Y)^2}{\sigma_Y^2} - \frac{(x - \mu_X)(y - \mu_Y)}{\sigma_X \sigma_Y} ]) \end{aligned} $$ We have $\rho$ as the correlation between the RVs $X$ and $Y$ where $\sigma_X, \sigma_Y > 0$ and $$ \begin{aligned} \textbf{mu} = \begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix}, \ \Sigma = \begin{bmatrix} \sigma_X^2 & \rho \sigma_X \sigma_Y \\ \rho \sigma_X \sigma_Y & \sigma_Y^2 \end{bmatrix} \end{aligned} $$ The rest is just a bit of algebra and factorization etc.