The conditional expectation of iid random variables 
Let $Y_1,Y_2,\dots,Y_n$ be a sequence of iid random variables. Each of them is integrable.
Let $X_1=(Y_1+Y_2+\cdots+Y_n)/n,X_2=(Y_1+Y_2+\cdots+Y_{n-1})/(n-1),\dots,X_{n-1}=(Y_1+Y_2)/2,X_n=Y_1$
Show that $X_1,X_2,\dots,X_n$ is a martingale relative to natural filtration $\mathcal F_n=\sigma(X_1,X_2,\dots,X_n)$.

I am stuck in a step where I need to prove that
$$\mathbf E(Y_{n-i+2}|\mathcal F_{i-1})=X_{i-1}.$$
However, I don't know why it holds. Is it because the conditional expectations of iid RVs given the same $\sigma$-algebra are equal? If so, why is that? Thanks.
 A: First, observe that $\mathcal F_i$ is the $\sigma$-algebra generated by the random variables $Y_1+\dots+Y_{n-i+1}, Y_{n-i+2},\dots,Y_n$. This is because $\sigma(c_i X_i)=\sigma(X_i)$ for constants $c_i\neq 0$ and the property $\sigma(X,X+Y)=\sigma(X,Y)$ used several times.
As a consequence of the previous observation with $i$ replaced by $i-1$,  we derive that for all $1\leqslant k\leqslant n-i+2$,
$$\tag{*}
\mathbf E(Y_{k}|\mathcal F_{i-1})= \mathbf E(Y_k|\sigma\left(Y_1+\dots+Y_{n-i+2}, Y_{n-i+3},\dots,Y_n\right)).
$$
Now use the fact that the vectors $(Y_k,Y_1+\dots+Y_{n-i+2}, Y_{n-i+3},\dots,Y_n)$ and $(Y_{n-i+2},Y_1+\dots+Y_{n-i+2}, Y_{n-i+3},\dots,Y_n)$ have the same distribution (due to the i.i.d. assumption) to derive that
$$
\mathbf E(Y_{k}|\mathcal F_{i-1})= \mathbf E(Y_{n-i+2}|\sigma\left(Y_1+\dots+Y_{n-i+2}, Y_{n-i+3},\dots,Y_n\right))=\mathbf E(Y_{n-i+2}|\mathcal F_{i-1}).
$$
Summing the equality $\mathbf E(Y_{k}|\mathcal F_{i-1})=\mathbf E(Y_{n-i+2}|\mathcal F_{i-1}) $ over $1\leqslant k\leqslant n-i+2$ gives
$$
(n-i+2)\mathbb E\left(X_{i-1}\mid \mathcal F_{i-1}\right)=(n-i+2)\mathbf E(Y_{n-i+2}|\mathcal F_{i-1}).
$$
Since $X_{i-1}$ is $\mathcal F_{i-1}$-measurable, we derive the wanted conclusion.
