$\newcommand{\Cov}{\operatorname{Cov}}$and thank you for taking the time to read this. :)
My question is about evaluating $\Bbb E[f(X) \mid Y]$ (a random variable in $Y$). There's plenty online (and in my first year lecture notes) about $\Bbb E[X\mid Y]$ and $\Bbb E[f(X)]$, but not $\Bbb E[f(X)\mid Y]$. I appreciated that if you know the pdf of $f(X)$, then this is easy (set $X^* = f(X)$ and just do normally). Is there some special way that I can do this (similar to the $\Bbb E[f(X)] = \int_\Bbb R f(x)p(x) \, dx$, where $p$ is the pdf - if the pdf exists, of course!)?
For a bit of background, I was trying to show that, for $(X,Y)$ jointly Gaussian, $$\Cov(f(X),Y) = \Bbb E[f'(X)] \Cov(X,Y)$$ for any "nice" function $f$. I was attempting this using "iterated expectation": $$\Cov(f(X),Y) = \Bbb E[(f(X) - \mu)(Y - \nu)] = \Bbb E[\Bbb E[(f(X)-\mu)(Y-\nu)\mid Y]] \\ = \Bbb E[(Y-\nu)\Bbb E[f(X) - \mu \mid Y]],$$ then using the fact that we know that $X\mid Y\sim N(\mu + \rho\sigma(Y-\nu)/\tau, \sigma^2(1-\rho^2))$ (with the usual symbol conventions). Unfortunately, I couldn't get anywhere with this. I'm not even sure if this is the right method, as I didn't know how to convert this to using $f(X)$ instead of $X$.
Any help would be most appreciated! Thanks! :)
[PS - if someone with lots of SE-experience thinks that I should make this two questions, please just let me know - they are directly related though! :)]
Update 1: Consider $Y = aX + bZ$ where $X$ and $Y$ have mean 0 and variance 1 and $Z$ ~ $N(0,1)$ is independent of $X$; so $\Cov(X,Y) = a$, by independences. So $$\Cov(f(X),Y) = \Cov(f(X),aX + bZ) = a\Cov(f(X),X)$$ as $X$ and $Z$ are independent, so $f(X)$ and $Z$ are also. Further, earlier in the question we showed that $$\Bbb E[g(R)(R-\mu)] = \sigma^2 \Bbb E[g'(R)]$$ where $g$ is a "nice" function and $R$ ~ $N(\mu,\sigma)$. Applying this, we have, for $g(X) = f(X) - \Bbb E[f(X)]$ and using the fact that X has mean 0 and variance 1, that $$\Cov(f(X),X) = \Bbb E[g(X)X] = \Bbb E[g'(X)] = \Bbb E[g'(X)],$$ as required. However, this is only for $Y$ in the form given. I need to consider general $(X,Y)$ jointly Gaussian.
I'm not 100% clear on what 'jointly Gaussian' is - I just used it as having pdf given by $$p_{X,Y}(x,y) = \exp \left( -{1 \over {2(1-\rho^2)}} \left( \left( {{x - \mu} \over \sigma} \right)^2 ... \right) \right).$$ Is this the right definition?