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Let $X_t$ be a stochastic process. $\mathcal{F}_t:=\sigma(X_s:0\leq s\leq t)$.

Suppose $\mathrm{E}[e^{ik(X_t-X_s)}|X_s]=e^{-\frac{1}{2}k^2(t-s)}$.

My idea

  1. By the tower property of conditional expectation: if $\mathcal{H}$ is a sub $\sigma$-algebra of $\mathcal{G}$, $\mathrm{E}[\mathrm{E}[Y|\mathcal{G}]|\mathcal{H}]=\mathrm{E}[Y|\mathcal{H}]$, I get \begin{equation} \mathrm{E}[\mathrm{E}[e^{ik(X_t-X_s)}|\mathcal{F}_s]|X_s]=e^{-\frac{1}{2}k^2(t-s)}. \end{equation}

  2. That is, $\mathrm{E}[\mathrm{E}[e^{ik(X_t-X_s)}|\mathcal{F}_s]-e^{-\frac{1}{2}k^2(t-s)}|X_s]=0.$ Consequently, by uniqueness of conditional expectation, I get $\mathrm{E}[e^{ik(X_t-X_s)}|\mathcal{F}_s]=e^{-\frac{1}{2}k^2(t-s)}$

But, clearly this is absurd because assumption say only about $\sigma(X_s)$ which included in $\mathcal{F}_s$ . I don't know what is wrong. Thank you for your cooperation.

sate
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1 Answers1

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There is no such 'uniquness of conditional expectation' result. You are saying that if $E(Y|X_s)=0$ then $Y=0$. This is false. For example $Y$ could be any random variable with mean $0$ independent of $X_s$.

From $E(Y|X)=0$ we can only conclude that $EY=0$.

Kavi Rama Murthy
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  • Thank you for your answer, I get it why my idea is wrong. Then, from $\mathrm{E}[X|F]=0$ we can also only conclude that $\mathrm{E}[X]=0$? I think it is not correct because this is imply $X=0$ a.s. So, my question arise from sigma-algebra is $\sigma(X_s)$ not $\mathcal{F}_s$? – sate Jan 04 '20 at 10:23
  • @sate $EX=0$ does not imply that $X=0$ a.s.. This argument works only for positive $X$. If $X$ has standard normal distribution then $EX=0$ but it is not true that $X=0$ a.s.. – Kavi Rama Murthy Jan 04 '20 at 11:23
  • Sorry, before my comment does not make sense. I agree counterexample in your answer, but my understanding is not enough, why uniqueness does not hold in this case? Suppose $E[Y_1|\sigma(X)]=E[Y_2|\sigma(X)]=0$, then for any $A\in\sigma(X)$, $E[Y_1-Y_2,A]=0$. And this derive $Y_1=Y_2$, like uniqueness of general conditional expectation $E[Z|\mathcal{G}]$? – sate Jan 04 '20 at 12:01
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    When you are dealing with conditional expectations you have to be careful with measurability assumptions. If $E(X|Y)=0$ and $X$ is already measurable w.r.t. $\sigma (Y)$ then and only then you can use uniqueness of conditional expectations to say that $X=0$. Not in general. – Kavi Rama Murthy Jan 04 '20 at 12:05
  • Ah, I see. Thank you for your cooperation. I'll study more. – sate Jan 04 '20 at 12:49