3
$\begingroup$

Suppose we have two random variables $(X,Y)$ according to a joint pdf $f_{X,Y}(x,y)$.

We can, of course, talk about the conditional expectation of $E[X\mid Y]$. Note that the conditional expectation is a random variable too.

My question is: Given $f_{X,Y}(x,y)$ can we find the distribution of $E[X\mid Y]$ in general ? Was this ever attempted?

Clearly, for specific examples, we can do this. For example, if $(X,Y)$ are jointly Gaussian then $E[X\mid Y]= \frac{E[XY]}{E[Y^2]} Y$ which is a Gaussian random variable.

My initial guess is that this is very difficult to do. This is because if we let $f(Y)=E[X\mid Y]$ then \begin{align} \mathbb{P}[ E[X\mid Y]\le y]= \mathbb{P}[ f(Y)\le y]= \mathbb{P}[ Y \le f^{-1}(y)], \end{align} but I doubt that we can find the inverse of $E[X\mid Y=y]$.

Anyway, I am looking forward to your thoughts.

$\endgroup$
3
$\begingroup$

If you have a joint density $f_{X,Y}(x,y)$ for $(X,Y)$ then you can obtain $E(X|Y)=:h(Y)$ via the expression $$ h(y) := \int x\,f_{X|Y}(x\mid y)\,dy = \int x\,\frac{f_{X,Y}(x,y)}{f_Y(y)}\,dy\tag1 $$ If the integrals involved in (1) [ including the integral that defines the marginal density $f_Y(y)$ ] are tractable you get an explicit form for $h$. One case where this is possible is if $f_{X,Y}$, when viewed as a function of $x$ alone, is seen to be (a constant times) the density of some random variable; then $h(y)$ is precisely the mean of that random variable. For example in the bivariate Gaussian case the joint density $f_{X,Y}(x,y)$, when viewed as a function of $x$, is, after some algebra, proportional to the density of a univariate Gaussian having mean $\mu_X +\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)$. This is one way to derive the conditional expectation: $$E(X\mid Y=y) = \mu_X +\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y).$$ Having obtained $h$, it's a separate matter to determine the distribution of $h(Y)$. If you're lucky, $h$ is invertible, or $h(Y)$ is a transformation that yields a familiar distribution.

$\endgroup$

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.