Question 1: I am interested in finding out, why the following is or is not the case:
$\mathbb{E}\left[ \mathbb{E} \left[( X_k|(Y_0,Y_1,...,Y_{k-1}) )|( Y_k|(Y_0,Y_1,...,Y_{k-1}) )\right]|(Y_0,Y_1,...,Y_{k-1}) \right] = \mathbb{E} \left[X_k|(Y_0,Y_1,...,Y_{k}) \right] $
where $X_i, Y_j$ are random variables in $\mathbb{R}^n$? Or, is there way to break up the expectation on the right which is conditioned on measurements up to $Y_k$, into terms which involve the condiotional expectation with respect to $Y_1,...,Y_{k-1}$
Background: This question arises in probabilistic Kalman-Filtering derivation, where we are trying to simplify the conditional expectation $\mathbb{E} \left[X_k|(Y_0,Y_1,...,Y_{k}) \right]$ (Minimum-Variance Estimate of State $X_k$ based on measurements $(Y_0,Y_1,...,Y_{k})$ up until time $k$) by trying to compute it iteratively based on on our already computed estimates based on the information of $(Y_0,Y_1,...,Y_{k-1})$. This part of the derivation is a crucial part, as it motivates and derives the iterative approach of computing the estimates of the Kalman Filter. Reading through literature, the explanation at this part, which connects the computation of estimates of one time-step to another, is never detailed enough for me. Backwards engineering suggests to me though, that the presented conditional expectation formula must have been used to complete the derivation. That's why I am asking for a proof or counterexample of the above statement.
Question 2: In case, the above formula is not true, is there any related formula, which would suite the purpose of the described derivation?
EDIT:
I think I found my question and answer at the same time:
More proper question:
Q: Given the probability density function $f\left( \left.x_k,y_k\right|y_0,y_1,...,y_{k-1}\right)$, how do I find $E(x_k|y_0,...,y_k)$?
A: \begin{align} E(x_k|y_0,...,y_k) &= \int\limits^{\infty}_{-\infty} x_k f\left( \left.x_k\right|y_0,y_1,...,y_{k-1},y_{k}\right) dx_k = \int\limits^{\infty}_{-\infty} x_k \frac{f\left(x_k,y_0,y_1,...,y_{k-1},y_{k}\right)}{f\left(y_0,y_1,...,y_{k-1},y_{k}\right)} dx_k\\ &=\int\limits^{\infty}_{-\infty} x_k \frac{f\left(x_k,y_0,y_1,...,y_{k-1},y_{k}\right)}{f\left(y_0,y_1,...,y_{k-1},y_{k}\right)}\frac{f\left(y_0,y_1,...,y_{k-1}\right)}{f\left(y_0,y_1,...,y_{k-1}\right)} dx_k\\ &=\int\limits^{\infty}_{-\infty} x_k \frac{f\left(\left.x_k,y_k\right|y_0,...,y_{k-1}\right)}{f\left(\left.y_k\right|y_0,...,y_{k-1}\right)} dx_k \end{align}
We basically obtain $E(\left.x_k\right|y_0,...,y_k)$ by building the conditional expectation of $x_k$ given $y_k$, both conditioned on $(y_0,...,y_k-1)$. Not knowing how to otherwise express this operation, I formulated this calculation as
$\mathbb{E}\left[ \mathbb{E} \left[( x_k|(y_0,y_1,...,y_{k-1}) )|( y_k|(y_0,y_1,...,y_{k-1}) )\right]|(y_0,y_1,...,y_{k-1}) \right] = \mathbb{E} \left[x_k|(y_0,y_1,...,y_{k}) \right] $
which received some bad resonance from our online audience.
My final question or help for advice would be though, how would you otherwise write down this computation. Speaking non-mathematically/inproper, it is just the building of the conditional expectation of one random variable given another random variable, were both random variables each have distributions conditioned on the same set of other ($y_0,...y_{k-1}$) random variables? Please, give me some insight how to express this more mathematically rigorous.