# The conditional expectation of i.i.d. random variables

Let $$Y_1,Y_2,\dots,Y_n$$ be a sequence of i.i.d. 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 i.i.d. random variables given the same $$\sigma$$-algebra are equal? If so, why is that? Thanks.

• It would probably help if you state what your filtration is. Commented May 28, 2020 at 11:41

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.