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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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    Sorry, you are right. Your $\mu_x=-1$ and I misread $1$.2011-04-20

2 Answers 2

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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 $\mathrm P(A)=1/12$

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    @MonaJalal To compute $E((X-Y)^2\mid X)$, I would use the $(U,V)$-representation in my post. To compute $E((X-Y)^2\mid Y)$, I would first compute a similar $(U,V)$-representation with $X$ and $Y$ exchanged.2018-01-27
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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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    hi, I am also working on bivariate, could you tell me how did you get $\frac{1}{12}$ in the end? Thanks much!2013-11-18