A sequence of rv $X_{n}$ i.i.d for $n \in \mathbb{N}$ with $S_{0}=0$ and $$S_{n} = \sum_{k=1}^{n}X_{k}.$$
with $k \leq n$ and $$\mathbb{E}[S_{k}\mid S_{n}] = \frac{k}{n}S_{n}$$
If $$M_{k} = \frac{S_{n-k}}{n-k},k=0,1,\dots,n-1$$ how can I show that $$\mathbb{E}[M_{k+1}\mid M_{0},\dots,M_{k}] =M_{k}$$
I am thinking that :
$$E[S_{n-k+1}\mid M_{0},\dots,M_{k}] = \mathbb{E}[X_{n-k+1}\mid M_{0},\dots,M_{k}]-S_{n-k}=a S_{n-k+1}-S_{n-k}$$
and find $a$. But is that right? And how can I find $a$?
Preliminaries
$E[X_i|S_n]=a$ does not depend on $i$, if $i \le n$. Than since:
$na=\sum_{i\le n} E[X_i|S_n]=E[S_n|S_n]=S_n$
We have therefore that $E[X_i|S_n]=S_n/n$ and
$$E[S_k|S_n]=\frac{k}{n} S_n, k \le n \ [1]$$
We also observe that:
$E[X_i|S_n,S_{n+1},..]=E[X_i|S_n,X_{n+1},...]=E[X_i|S_n] \ [2]$
because $X_{n+1}$ is independent of $X_i$. Analogously:
$$E[S_k|S_n,S_{n+1},..]=E[S_k|S_n], k\le n$$
The identity we want to show
Now we show:
$\mathbb{E}[M_{k+1}\mid M_{0},\dots,M_{k}] =M_{k}$
First observe:
$\mathbb{E}[M_{k+1}\mid M_{0},\dots,M_{k}]=1/(n-k-1)\mathbb{E}[S_{n-k-1}\mid S_{n},\dots,S_{n-k}]$
And now apply first [2] and than [1]. All factors are going to simplify :).
In detail thanks to [2]:
$1/(n-k-1)\mathbb{E}[S_{n-k-1}\mid S_{n},\dots,S_{n-k}]=1/(n-k-1)\mathbb{E}[S_{n-k-1}\mid S_{n-k}]$
and thanks to [1]:
$1/(n-k-1)\mathbb{E}[S_{n-k-1}\mid S_{n-k}]=1/(n-k-1) \times (n-k-1)/(n-k) \times S_{n-k}=M_k$