Bivariate Normal Conditional Variance I am given the parameters for a bivariate normal distribution ($\mu_x, \mu_y, \sigma_x, \sigma_y,$ and $\rho$). How would I go about finding the Var($Y|X=x$)? I was able to find E[$Y|X=x$] by writing $X$ and $Y$ in terms of two standard normal variables and finding the expectation in such a manner. I am unsure how to do this for the variance.
Also, how do I find the probability that both $X$ and $Y$ exceed their mean values (i.e., $P(X>\mu_x, Y > \mu_y)$)?
Thanks for the help!
 A: First, the joint PDF $f(x,y)$ is obvious, just plug in your parameters. Bivariate Normal.
Then you can find the marginal density for $X$, which gives you the conditional density of $Y$ given $X=x$:
$$f_{Y|X}(y|x)=\frac{f(x,y)}{f_X(x)}.$$
Now use the conditional density you can evaluate both conditional expectation and conditional variance :
$$\mathbb{E} (Y|X=x)=\int_{-\infty}^\infty y f_{Y|X}(y|x)dy,$$
and 
$$\text{Var} (Y|X=x)=\int_{-\infty}^\infty (y-h(x))^2 f_{Y|X}(y|x)dy=\frac14,$$
where $h(x)=\mathbb{E} (Y|X=x)=-\frac{\sqrt 3}4(x-2)-1$.
And with the joint PDF, $P(X>\mu_x, Y > \mu_y)$ is just an integration:
$$P(X>\mu_x, Y > \mu_y)=\int_{\mu_x}^\infty\int_{\mu_y}^\infty f(x,y)dydx=\frac1{12},$$
though I guess there's an easier way to compute.
A: Rather than embarking on some pretty involved computations of conditional distributions, one should rely on one of the main assets of Gaussian families, namely, the... 

Key feature: In Gaussian families, conditioning acts as a linear projection.

Hence, as the OP suggested, one could do worse than to start from a representation of $(X,Y)$ by standard i.i.d. Gaussian random variables $U$ and $V$, for example,
$$
X=\mu_x+\sigma_xU\qquad
Y=\mu_y+\sigma_y(\rho U+\tau V)$$ where the parameter $\tau$ is $$\tau=\sqrt{1-\rho^2}
$$
Since $\sigma_x\ne0$, the sigma-algebra generated by $X$ is also the sigma-algebra generated by $U$ hence conditioning by $X$ or by $U$ is the same. Furthermore, constants and functions of $X$ or $U$ are all $U$-measurable while functions of $V$ are independent on $U$, thus,
$$
\mathrm E(Y\mid X)=\mu_y+\sigma_y(\rho U+\tau \mathrm E(V))=\mu_y+\sigma_y \rho U
$$
which is equivalent to
$$
\color{red}{\mathrm E(Y\mid X)=\mu_y+\rho\frac{\sigma_y}{\sigma_x}(X-\mu_x)}
$$
Likewise, when computing conditional variances conditionally on $X$, deterministic functions of $X$ or $U$ should be considered as constants, hence their conditional variance is zero, and functions of $V$ are independent on $X$, hence their conditional variance is their variance. Thus,
$$
\mbox{Var}(Y\mid X)=\mbox{Var}(\sigma_y\tau V\mid X)=\sigma_y^2\tau^2\mbox{Var}(V\mid X)=\sigma_y^2\tau^2\mbox{Var}(V)
$$
that is,
$$
\color{red}{\mbox{Var}(Y\mid X)=\sigma_y^2(1-\rho^2)}
$$
Finally, the event $$A=[X>\mu_x,Y>\mu_y]$$ is also
$$
A=[U>0,\rho U+\tau V>0].
$$
To evaluate $\mathrm P(A)$, one can turn to the planar representation of couples of independent standard Gaussian random variables, which says in particular that the distribution of $(U,V)$ is invariant by rotations. The event $A$ means that the direction of the vector $(U,V)$ is between the angle $\vartheta$ in $(-\pi/2,\pi/2)$ such that $$\tan(\vartheta)=-\rho/\tau$$ and the angle $\pi/2$. Thus,
$$\mathrm P(A)=\frac{\pi/2-\vartheta}{2\pi}$$
that is,
$$
\color{red}{\mathrm P(X>\mu_x,Y>\mu_y)=\frac14+\frac1{2\pi}\arcsin\rho}
$$
Numerical application: If $\mu_x=2$, $\mu_y=-1$, $\sigma_x=2$, $\sigma_y=1$ and $\rho=-\sqrt3/2$, then
$$
\mathrm E(Y\mid X)=-1+\sqrt3/2-(\sqrt3/4)X\qquad
\mbox{Var}(Y\mid X)=1/4
$$
and $\tau=1/2$, hence $\vartheta=\pi/3$ and $$\mathrm P(A)=1/12$$
