I struggle to calculate a conditional expectation:
Let $X$ be a random variable on $\mathbb{R}^n$ with density function $\alpha$. Further, let $\eta$ be a unit vector and $g:\mathbb{R}^n \rightarrow \mathbb{R}$ a measurable function.
I already know that the density function of the random variable $\langle \eta, X \rangle$ is $\rho(y) := \int_{\{x:\langle \eta,x\rangle = y\}} \alpha(x) \;d\lambda_{n-1}(x)$.
Now I need to prove the following statement:
\begin{equation}\mathbb{E}(g(X) \vert \langle \eta, X \rangle) = \frac{\int_{\{x:\langle \eta,x\rangle =y\}} g(x)\alpha(x) \;d\lambda_{n-1}(x)}{\rho(y)}.
\end{equation}
This is what I've tried:
If we denote $Z:= g(X)$,$Y:= \langle \eta, X \rangle$ and $f_{Z,Y}$ as the joint density function of $Z$ and $Y$ then according to wikipedia,
\begin{equation}
f_{Z\vert Y} (z,y) = \frac{f_{Z,Y}(z,y)}{\int_? f_{Z,Y}(u,y) du}
\end{equation}
and
\begin{equation}
\mathbb{E}(Z \vert Y) = \int_? z \cdot f_{Z\vert Y} (z,Y)\;dz.
\end{equation}
Comparing this to the solution we are tying to get, it looks a like $\alpha$ was treated as the joint PDF of $Z$ and $Y$. But I don't understand why we can do that. Shouldn't $f_{Z,Y}$ be a function $\mathbb{R} \times\mathbb{R} \rightarrow \mathbb{R}$, while $\alpha:\mathbb{R}^n \rightarrow \mathbb{R}$?
I am very confused about this.
Thanks a lot for any help!
We know the density $\alpha(x)$ of $X$ and the density $\rho(y)$ of $Y=\langle\eta,X\rangle$ and want to calculate the function $$ h(y):=\mathbb E[\,g(X)|Y=y\,]. $$ The trick is to choose a Borel set $A\subset\mathbb R$ and just use the properties of the conditional expectation: \begin{eqnarray*} \int_A\int_{\{x:\langle \eta,x\rangle\, =\, y\}}g(x)\,\alpha(x)\,dx\,dy&=&\mathbb E[\,g(X)1_{\{Y\in A\}}\,]=\mathbb E[\,\mathbb E[\,g(X)\,1_{\{Y\in A\}}|Y\,]\,]\\&=& E[\,\mathbb E[\,g(X)|Y\,]\,1_{\{Y\in A\}}\,]=\int_Ah(y)\,\rho(y)\,dy\\ &=&\int_Ah(y)\int_{\{x:\langle \eta,x\rangle\, =\, y\}} \alpha(x)\,dx\,dy. \end{eqnarray*} Using the fact that $A$ was an arbitrary Borel set and comparing the first and last term in this chain of equations shows $$ h(y)=\frac{\int_{\{x:\langle \eta,x\rangle\, =\, y\}}g(x)\,\alpha(x)\,dx}{\int_{\{x:\langle \eta,x\rangle\, =\, y\}} \alpha(x)\,dx}=\frac{\int_{\{x:\langle \eta,x\rangle\, =\, y\}}g(x)\,\alpha(x)\,dx}{\rho(y)}. $$