5
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ what did you get for $\mathbb E[Y|X=x]$ ? $\endgroup$ – GWu Apr 20 '11 at 5:32
  • $\begingroup$ 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. $\endgroup$ – icobes Apr 20 '11 at 5:37
  • $\begingroup$ You said $\rho$ is $-\sqrt 3$? That's impossible. $\endgroup$ – GWu Apr 20 '11 at 5:45
  • $\begingroup$ My bad its -(root3)/2. $\endgroup$ – icobes Apr 20 '11 at 5:46
  • $\begingroup$ I got $\mathbb E[Y|X=x]=-\frac{\sqrt 3}4(x-2)+1$ not $-1$. $\endgroup$ – GWu Apr 20 '11 at 5:49
10
$\begingroup$

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

$\endgroup$
  • $\begingroup$ How can I calculate E[X^2|Y=y], E[Y^2|Y=y] or E[XY|Y=y] for a bivariate normal distribution? Can you share a link that does this? everything I see only has E[X|Y=y] Thanks $\endgroup$ – Mona Jalal Jan 27 '18 at 4:12
  • $\begingroup$ @MonaJalal Then you did no read carefully enough since a conditional variance is indeed computed above. Note also that the two latter conditional expectations in your comment are trivial, for example asking for $E(Y^2\mid Y=y)$ is strange, to say the least. $\endgroup$ – Did Jan 27 '18 at 8:32
  • $\begingroup$ I asked because I expanded E[(X-Y)^2 | Y=y]. Basically how would you find the value of E[(X-Y)^2 | Y=y] for a bivariate normal distribution? $\endgroup$ – Mona Jalal Jan 27 '18 at 21:05
  • $\begingroup$ @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. $\endgroup$ – Did Jan 27 '18 at 21:39
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – icobes Apr 20 '11 at 6:03
  • $\begingroup$ It is complicated. Maybe your approach is simpler. BTW, the conditional variance is $1/2$ according to Mathematica $\endgroup$ – GWu Apr 20 '11 at 6:09
  • $\begingroup$ hi, I am also working on bivariate, could you tell me how did you get $\frac{1}{12}$ in the end? Thanks much! $\endgroup$ – user13985 Nov 18 '13 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.