Recursive relation between conditional expectations Let $\epsilon _n$ and $\eta _n$ iid random variables (and the sequences are independent of each other) such that $\epsilon _n \sim \mathcal{N}(0, \sigma ^2)$ and $\eta _n \sim \mathcal{N}(0, \delta ^2)$. Let $X_0=0$, $X_{n+1}=a_nX_n+\epsilon_{n+1}$ and $Y_n=cX_n+\eta_n$ where $c$ and $a_n$ are positive constants. Put $\widehat{X_{n/n}}=E(X_n|Y_0,\ldots ,Y_n$) and $\widehat{X_{n/n-1}}=E(X_n|Y_0,\ldots ,Y_{n-1}$). I want to show that 
$$\widehat{X_{n/n}}=\widehat{X_{n/n-1}}+\frac{E(X_nZ_n)}{E(Z_n^2)}Z_n $$ where $Z_n=Y_n-c\widehat{X_{n/n-1}}$, but my attempts didn't suceed.
 A: Let $\mathcal{W}_{n-1}\subset\mathcal{W}_{n}$ be the two vector subspaces spanned by $Y_0,\ldots,Y_k$ for $k\in\{n, n-1\}$.  Since $(X_n,Y_0,\ldots,Y_n)$ are joint normal and $X_n$ is mean zero: $$\hat{X}_{n/n} \in\mathcal{W}_n\text{ and }\hat{X}_{n/n-1} \in\mathcal{W}_{n-1}.$$ Furthermore conditional expectation is the orthogonal projection with respect to the inner product $E[UV]$ which means 
$$ \hat{X}_{n/n}  - \hat{X}_{n/n-1} \perp \mathcal{W}_{n-1}.$$ 
But the orthogonal complement to $\mathcal{W}_{n-1}$ in $\mathcal{W}_{n}$ is spanned by 
\begin{align} 
E[Y_n \mid Y_0,\ldots,Y_n] - E[Y_n \mid Y_0,\ldots,Y_{n-1}] 
&= Y_n - E[cX_n-\eta_n \mid Y_0, \ldots, Y_{n-1}]\\
&= Y_n - c \hat{X}_{n/n-1}\\
&= Z_n
\end{align}
where the second step uses the assumption that $\eta_n$ is independent and mean zero.
Hence 
$$ \hat{X}_{n/n} - \hat{X}_{n/n-1} = \alpha Z_n.$$
Multiplication by $Z_n$, taking expectation on both sides and using that $E[\hat{X}_{n/n-1}Z_n]=0$ due to orthogonality, shows that
$$ \alpha = \frac{E[\hat{X}_{n/n} Z_n]}{E[Z_n^2]}.$$ 
The conclusion follows since $X_n - \hat{X}_{n/n}$ is orthogonal to $\mathcal{W}_n$ which means $E[\hat{X}_{n/n} Z_n] = E[X_n Z_n]$.
