I have a question and hope you can help me. The problem is about stochastic variables $X,Y$, which are square integrable, independent and identically distributed on $(\Omega, \mathcal{S}, P)$. Furthermore $\mathcal{A} \subset\mathcal{S}$ is a sub-sigma-algebra.
Now I have to show three statements:
a) $\mathbb{V}(X) = \mathbb{E}(X|\mathcal{A}) + \mathbb{E}[X-\mathbb{E}(X|\mathcal{A})]^2 \Rightarrow \mathbb{V}(\mathbb{E}(X|\mathcal{A})) \leq \mathbb{V}(X)$
b) $\mathbb{E}X^2 = \mathbb{E}Y^2$ and $\mathbb{E}(Y|\mathcal{A}) = X$ P-a.s. $\Rightarrow X=Y$ P.-a.s.
c) $\mathbb{E}(X|Y) = Y$ P-a.s. and $\mathbb{E}(Y|X) = X$ P-a.s. $\Rightarrow X=Y$ P-a.s.
I already managed to proof a) by using variance-decomposition-theorem and a little bit of calculation.
Furthermore I think I have an idea how to solve c):
$\mathbb{E}[XY|X] = X \mathbb{E}[Y|X] = X X = X^2$ P.-almost sure, where the second equal-sign uses the assumption. Similarly there one can conclude $\mathbb{E}[XY|Y] = Y^2$ P-a.s.
Applying $\mathbb{E}$ on both sides gives me:
$\mathbb{E}[X^2]=\mathbb{E}[\mathbb{E}[XY|X]] = \mathbb{E}[XY] = \mathbb{E}[\mathbb{E}[XY|Y]] = \mathbb{E}[Y^2]$ P-a.s. Now, by using this result, I can conclude $\mathbb{E}[(X-Y)^2] = 0$ to get the required result.
Sadly I don't have a clue how to prove b) to finish this task.
I would be very glad if anyone could help me solving b) too!
Thanks in advance for your help!
Best Answer
Partial solution.
Since you have a), I give a proof of b). Observe the hypothesis $\mathbf{E}(Y \mid \mathscr{A}) = X$ implies $X$ to be measurable with respect to $\mathscr{A}.$ Bearing this in mind, $$\begin{align*} \mathbf{E}((Y-X)^2) &= \mathbf{E}\big( \mathbf{E}((Y-X)^2 \mid \mathscr{A}) \big) \\ &= \mathbf{E}(\mathbf{E}(Y^2 \mid \mathscr{A}) - 2\mathbf{E}(Y \mid \mathscr{A}) X+ X^2) \\ &= \mathbf{E}(\mathbf{E}(Y^2 \mid \mathscr{A}) - X^2) = 0. \end{align*}$$
As for c), if $X$ and $Y$ were square integrable, you have $\mathbf{E}(XY) = E(X^2) = E(Y^2),$ so b) finishes the proof upon applying it with $\mathscr{A} = \sigma(X).$ For the general case, I am not sure how to tackle it.