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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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  • $\begingroup$ 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
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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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  • $\begingroup$ 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
  • $\begingroup$ can you elaborate on the P(๐‘‹>๐œ‡๐‘ฅ,๐‘Œ>๐œ‡๐‘ฆ)=14+12๐œ‹arcsin๐œŒ part? how did you get that analytical form? trying to integrate by myself, didn't quite get the results $\endgroup$ – peng yu Apr 26 '20 at 7:28
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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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  • $\begingroup$ 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

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