Calculate $E[XY]$ for $(X,Y)\sim N(\mu_{1},\mu_{2},\sigma_{1}^{2},\sigma_{2}^{2}, \rho)$ I need to calculate $E[XY]$ for $(X,Y) \sim N(\mu_{1},\mu_{2},\sigma_{1}^{2}, \sigma_{2}^{2}, \rho)$ by using integration and then determine the correlation coefficient afterwards.
Now, when $X \sim N(\mu_{1},\sigma_{1}^{2})$ and $Y \sim N(\mu_{2}, \sigma_{2}^{2})$, the probability density function is given by $$f(x,y)= \\ \frac{1}{2\pi \sigma_{1} \sigma_{2} \sqrt{1-\rho^{2}}}\exp \left( - \frac{1}{2(1-\rho^{2})} \left[ \frac{(x-\mu_{1})^{2}}{\sigma_{1}^{2}}+ \frac{(y-\mu_{2})^{2}}{\sigma_{2}^{2}} - \frac{2 \rho (x-\mu_{1})(y-\mu_{2})}{\sigma_{1}\sigma_{2}}\right]\right) $$
But, I don't know how to calculate $E[XY]$ as an integral, short of setting up $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xyf(x,y)dydx. \,\,\,(*)$$
Somebody told me that this integral is equal to  $$ \int_{-\infty}^{\infty} x \left[\int_{-\infty}^{\infty} y f(y \vert x) dy\right]f(x)dx,$$
where $f(y|x)$ is normal density with mean $=\int_{-\infty}^{\infty}yf(y\vert x)dy = \mu_{2}$, but I don't know what this means or how to get from $(*)$ to this last integral above.
Could someone please explain this to me in an answer?   
 A: The hint you suggest in your question is to use condition expectations. although not the only way to solve this problem, it is a quick method if you are comfortable with conditioning arguments. 
$\mathbb{E}(XY) = \mathbb{E}_X(\mathbb{E}_y(XY|X))$
where the subscripts denote expectation with respect to which variable (for clarity to the reader. Then 
$\mathbb{E}_X(\mathbb{E}_Y(XY|X))= \mathbb{E}_X(X\mathbb{E}_Y(Y|X)).$
Now if you know the distribution of $Y|X$ this is easy, 
$Y|X \sim N(\mu_y+\rho\frac{\sigma_y}{\sigma_x}(X-\mu_x), \sigma_y^2(1-\rho)).$
This means the mean of $Y|X$ is the first quantity, 
$\mathbb{E}_X(X(\mu_y+\rho\frac{\sigma_y}{\sigma_x}(X-\mu_x))).$
Now multiply the $X$ through and take expectations to get
$\mathbb{E}_X(X)\mu_y+\rho\frac{\sigma_y}{\sigma_x}\left[\mathbb{E}_X(X^2)-\mu_x\mathbb{E}_X(X)\right].$ 
Now using the variance breakdown of the second and the first moment gives:
$\mathbb{E}_X(X)\mu_y+\rho\frac{\sigma_y}{\sigma_x}\left[\sigma_x^2 +\mu_x^2 -\mu_x^2\right].$ And a little simplification yields:
$\mu_X\mu_y+\rho\sigma_y\sigma_x.$
You can check this against the correlation by substractions of $\mu_x\mu_y$ to get the co-variance, verifying the result is correct. 
