# 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!

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what did you get for $\mathbb E[Y|X=x]$ ? –  GWu Apr 20 '11 at 5:32
Well I was given numbers where mew x and mew y are 2 and negative 1 respectively. Variance of X is 4, variance of y is 1, and rho is negative root 3. I then expressed X as X = 2Z1+ 2 and Y as -root3/2*Z1 + 1/2*Z2 - 1 where Z1 and Z2 are standard normals and found E[Y|X=x] to be - root 3 / 2 * (x-2)/2 - 1. –  icobes Apr 20 '11 at 5:37
You said $\rho$ is $-\sqrt 3$? That's impossible. –  GWu Apr 20 '11 at 5:45
My bad its -(root3)/2. –  icobes Apr 20 '11 at 5:46
I got $\mathbb E[Y|X=x]=-\frac{\sqrt 3}4(x-2)+1$ not $-1$. –  GWu Apr 20 '11 at 5:49

Rather than embarking on some involved computations of conditional distributions, one can rely on one of the main assets of Gaussian families:

In a Gaussian family, conditioning acts as a linear projection.

Hence, as the OP suggested, one should start from a representation of $(X,Y)$ by standard i.i.d. Gaussian random variables $U$ and $V$, for example $$X=\mu_x+\sigma_xU,\quad Y=\mu_y+\sigma_y(\rho U+\tau V),\quad \tau=\sqrt{1-\rho^2}.$$ 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 $U$ 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, in 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, $$\mbox{Var}(Y\mid X)=\mbox{Var}(\sigma_y\tau V\mid X)=\sigma_y^2\tau^2\mbox{Var}(V\mid X)=\sigma_y^2\tau^2\mbox{Var}(V),$$ hence $$\color{red}{\mbox{Var}(Y\mid X)=\sigma_y^2(1-\rho^2)}.$$ 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, $\mathrm P(A)=(\pi/2-\vartheta)/(2\pi)$, hence $$\color{red}{\mathrm P(X>\mu_x,Y>\mu_y)=\frac14+\frac1{2\pi}\arcsin\rho}.$$ Numerical application: $\mu_x=2$, $\mu_y=-1$, $\sigma_x=2$, $\sigma_y=1$ and $\rho=-\sqrt3/2$, hence $$\mathrm E(Y\mid X)=-1+\sqrt3/2-(\sqrt3/4)X,\quad \mbox{Var}(Y\mid X)=1/4,$$ and $\tau=1/2$, hence $\vartheta=\pi/3$ and $\mathrm P(A)=1/12$.

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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.

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Doesn't this seem a bit too tedious? The integration is quite nasty given the horrific looking density... Is there no way to neatly solve Var(Y=-root(3)/2*Z1 + 1/2Z2 - 1 | Z1 = (x-2)/2)? With expectations you are allowed to split up the above statement due to linearity of it. Is there any way to do this with variance? –  icobes Apr 20 '11 at 6:03
It is complicated. Maybe your approach is simpler. BTW, the conditional variance is $1/2$ according to Mathematica –  GWu Apr 20 '11 at 6:09
hi, I am also working on bivariate, could you tell me how did you get $\frac{1}{12}$ in the end? Thanks much! –  user13985 Nov 18 '13 at 0:32