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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!

TNicky
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  • Implication c) holds for integrable random variables X and Y not necessarily square integrable, as shown in answers to several questions on this site. – Did Dec 17 '18 at 22:19
  • @Did I believe c) it to be true in general, but I failed to provide a proof. Do you have any of this links or a proof-sketch? – William M. Dec 17 '18 at 22:37
  • I thought, that my proof for c) (given above) is valid. Is there anything wrong with it? The only step I need square integrability is (at least I think so) $\mathbb{E}[XY|X] = X \mathbb{E}[Y|X]$. With Hölder's inequality I can conclude, that if X,Y are square-integrable then XY is integrable: $\int XY dP \leq \int |XY| dp \stackrel{Hölder}{=} \left( \int |X|^2 dP \right)^{\frac{1}{2}} \left( \int |Y|^2 dP \right)^{\frac{1}{2}} < \infty$. Or am I wrong? – TNicky Dec 17 '18 at 22:42
  • It is true that if $X$ and $Y$ are square integrable, then $XY$ is integrable. – William M. Dec 18 '18 at 01:36
  • @WillM. https://math.stackexchange.com/a/1520480/6179 – Did Dec 18 '18 at 07:20
  • @TNicky For the version of (c) that requires only integrability of $X$ and $Y$, please see https://math.stackexchange.com/q/34101/215011 – grand_chat Dec 19 '18 at 00:26

1 Answers1

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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.

William M.
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  • Hi and thanks for your quick respond! I have a question about the second equal-sign. Why is it $X^2$ and not $\mathbb{E}[X|\mathcal{A}]$? I think it doesn't make much difference in the result, due to if i write it this way I get zero as well, but maybe I missed something so my way might be faulty (?) – TNicky Dec 17 '18 at 22:46
  • You would need $\mathbf{E}(X^2 \mid \mathscr{A}) = X^2$ since $X$ is $\mathscr{A}$-measurable. – William M. Dec 18 '18 at 01:35
  • Ok! But I don't see, why $X$ is $\mathcal{A}$-measurable. We also need this for the step with the term $\mathbb{E}[XY|\mathcal{A}]$. – TNicky Dec 18 '18 at 07:52
  • If $X = \mathbf{E}(Y \mid \mathscr{A})$ then $X$ is $\mathscr{A}$-measurable. If this is not clear, you need to read again the definition of conditional expectation. – William M. Dec 18 '18 at 07:53
  • AAh sure it is clear! Stupid me!

    Thank you very much!

     – TNicky Dec 18 '18 at 07:55
  • You get $Y^2 + 2YX + X^2$ and then you take the conditional expectation, $X$ being measurable with respecto to $\mathscr{A},$ can "come out" of each one of the conditional expectations. – William M. Dec 18 '18 at 07:55
  • Ok - I wasn't sure whether you just left out this step or there is something I have overseen. I'm quite new to this topic, so I write out a few more steps. But I think now it's clear! Thanks again! – TNicky Dec 18 '18 at 07:57