Martingales: how can I prove the third condition? I have the process $(Y_m, F_m)$ defined in this way: 
$$Y_m=\frac{S_{n-m+1}}{n-m+1}$$
$$S_n=\sum_{i=1}^n X_i$$ with $X_i$ i.i.d. and $X_i \in L^1$ for each $i$. 
$$F_1=\sigma(S_n)$$
$$F_2=\sigma(S_n,S_{n-1})$$
$$...$$
$$F_m=\sigma(S_n,S_{n-1},...,S_{n-m+1})$$
$$...$$
$$F_n=\sigma(S_n,S_{n-1},...,S_1)$$
I have to prove that $$E(Y_m \mid F_{m-1})=Y_{m-1}, \forall m$$ that is the third condition of martingales.
I have this hint: $Y_n=X_1$ so $E(X_1 \mid F_n)=Y_n$ (maybe I have to prove that  $E(X_1 \mid F_m)=Y_m, \forall m=1,...,n$?).
 A: Fix $m \leq n$. First of all, note that
$$\mathcal{F}_{m-1} = \sigma(S_{n-m+2},X_{n-m+3},\ldots,X_{n-1},X_n). \tag{1}$$
Since $X_1,\ldots,X_{n-m+1}$ and $X_{n-m+2},\ldots,X_n$ are independent, this gives
$$\begin{align*} \mathbb{E}(S_{n-m+1} \mid \mathcal{F}_{m-1}) &= \mathbb{E}(S_{n-m+1} \mid \sigma(S_{n-m+2},X_{n-m+3},\ldots,X_{n-m+3},\ldots,X_n))\\ &= \mathbb{E}(S_{n-m+1} \mid \sigma(S_{n-m+2})). \end{align*}$$
Thus, by the linearity of conditional expectation,
$$\mathbb{E}(S_{n-m+1} \mid \mathcal{F}_{m-1}) = \sum_{j=1}^{n-m+1} \mathbb{E}(X_j \mid \sigma(S_{n-m+2})). \tag{2}$$
Using that the random variables $X_1,\ldots,X_{n-m+1}$ are independent and identically distributed, it follows that
$$\mathbb{E}(X_j \mid \sigma(S_{n-m+2})) = \mathbb{E}(X_1 \mid \sigma(S_{n-m+2})) \tag{3}$$
for all $j=1,\ldots,n-m+1$. As
$$S_{n-m+2} = \mathbb{E}(S_{n-m+2} \mid \sigma(S_{n-m+2})) \stackrel{(3)}{=} (n-m+2) \mathbb{E}(X_1 \mid \sigma(S_{n-m+2}))$$
we get
$$\mathbb{E}(X_1 \mid \sigma(S_{n-m+2})) = \frac{1}{n-m+2} S_{n-m+2} = Y_{m-1}. \tag{4}$$
Moreover, $(3)$ shows
$$\mathbb{E}(S_{n-m+1} \mid \mathcal{F}_{m-1}) = (n-m+1) \mathbb{E}(X_1 \mid \sigma(S_{n-m+2})).$$
Dividing both sides by $n-m+1$, we get
$$\mathbb{E}(Y_m \mid \mathcal{F}_{m-1}) = \mathbb{E}(X_1 \mid \sigma(S_{n-m+2})) \stackrel{(4)}{=} Y_{m-1} .$$
