The exponent $e$ of a bivariate normal density is given as follows : $$ -\dfrac{1}{102} [ (x+2)^{2} - 2.8(x+2)(y-1) + 4 (y-1)^{2} ]$$
We need to find : E(X) , E(Y) , Var(X) , Var(y) and $\rho$ (correlation coefficient).
I tried to compare that $102$ term with $ 2 ( \sqrt{1 - \rho^{2}})^{2}$ , but that gives $ 1 - \rho^{2} = 51$ , but that's not possible..
What am I doing wrong ?
Am I correct in comparing the coefficients straightaway ??
The bivariate Normal density for $(X,Y)$ is given by
\begin{align} 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)\\ \end{align}
By comparison with your formula it follows that
\begin{align} \frac{1}{2(1-\rho^2)\sigma_X^2} &= \frac{1}{102}\\[2ex] \frac{1}{2(1-\rho^2)\sigma_Y^2} &= \frac{4}{102}\\[2ex] \frac{2\rho}{2(1-\rho^2)\sigma_X\sigma_Y} &= \frac{2.8}{102}\\[2ex] \mu_X &= -2\\[2ex] \mu_Y &= 1 \\[2ex] \end{align}
The first $3$ equations form a system for $3$ unknowns with solution
$$ \sigma_X^2 =10, \quad \sigma_Y^2 =5, \quad \text{and} \quad \rho =0.7 $$
