An exercise about conditional expectation of Jean-François Le Gall's book Recently I've been studying Brownian Motion, Martingales, and Stochastic Calculus by Jean-François Le Gall. But I was stuck by this exercise (1.16 p.15):

Consider a sequence of random variables $(X_n)$ and $(Y_n)$ defined recursively by
  $$X_{n+1}=a_nX_n+\epsilon_{n+1}$$
  and
  $$Y_n=cX_n+\eta_n$$
where $a_n>0$, $c>0$ and $\epsilon_n\sim N(0,\sigma^2)$, $\eta_n\sim N(0,\delta^2)$ i.i.d.. Also, assume $(\epsilon_n)$ and $(\eta_n)$ is independent. Now define 
  $$\hat{X}_{n/m}=E[X_n|Y_0,\dots,Y_m].$$
  Show that for $n\geq 1$,
  $$\hat{X}_{n/n}=\hat{X}_{n/n-1}+\frac{E[X_nZ_n]}{E[Z_n^2]}Z_n.$$
  where $Z_n:=Y_n-c\hat{X}_{n/n-1}$.

I guess the solution involve some kind of inductive arguments, but I have no idea how to start... This would be nice for someone to offer me hints and ideas. Thanks!
 A: Here are some hints. 
When the random variables in question are square integrable (the Gaussianity is even not needed), you may understand conditional expectation as orthogonal projection. Let $V_m = \operatorname{span} (Y_0,Y_1,\dots,Y_m)$. Then $\hat X_{n/m}$ is the orthogonal projection of $X_n$ to $V_m$. Your claim reads,
$$
\hat X_{n/n}  =  \hat X_{n/n-1} + a Z_n =  \operatorname{pr}_{V_{n-1}} \hat X_{n/n} + a Z_n.
$$
We know that 
$$
\hat X_{n/n}  = \operatorname{pr}_{V_{n-1}^{\vphantom{\perp}}} \hat X_{n/n} + \operatorname{pr}_{V_{n-1}^{\perp}} \hat X_{n/n},
$$
where $V_{n-1}^{\perp}$ is the orthogonal complement of $V_{n-1}$. So you need to prove that $a Z_n = \operatorname{pr}_{V_{n-1}^{\perp}} \hat X_{n/n}$. Since $\hat X_{n/n}\in V_n\supset V_{n-1}$, this exactly amounts to proving that 


*

*$V_n = \operatorname{span}(V_{n-1}, Z_n)$; 

*$Z_n \perp V_{n-1}$;

*$aZ_n = \operatorname{pr}_{Z_n} \hat X_{n/n}$. 
Write if you have problems with one of these three points.
A: First note that we are working in the space of centered gaussian random variables. Since $\operatorname{span}(Y_0,\dots,Y_m)$ is a closed subspace, we have $$\hat{X}_{n/m}=P[X_n|Y_0,\dots,Y_m]$$ be the projection of $X_n$ on $\operatorname{span}(Y_0,\dots,Y_m)$.


*

*$V_n = \operatorname{span}(V_{n-1}, Z_n)$: just linear algebra.

*For $i=0,...,n-1$,
$$E[Y_iZ_n]=E[Y_iY_n]-cE[Y_iE[X_n|Y_0,\dots, Y_n]]=E[Y_iY_n]-cE[Y_iX_n]=E[Y_i\eta_{n+1}]=0.$$

*By our note and 2. we get $\hat{X}_{n/n}=\hat{X}_{n/n-1}+aZ_n$. By 2., $Z_n \perp V_{n-1}$, so we have $E[\hat{X}_{n/n-1}Z_n]=0$. Also, 
$$E[\hat{X}_{n/n}Z_n]=E[E[X_nZ_n|Y_0,\dots,Y_n]]=E[X_nZ_n].$$
Thus, $$a=\frac{E[X_nZ_n]}{E[Z_n^2]}.$$

