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This is an abbreviated version of Exercise 4.14 from Stocastic Calculus for Finance II by Steven Shreve but the context of the question should be sufficient to answer the question.

Define $Z_j = f''(W(t_j))[(W(t_{j+1}) - W(t_j))^2 - (t_{j+1} - t_{j})]$

where $W(t)$ is a Brownian motion , $t \geq 0.$ Show that $\mathbb{E}\sum_{j=0}^{n-1}Z_j = 0.$

My attempt:

\begin{align} \mathbb{E}\sum_{j=0}^{n-1}Z_j &= \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))[W(t_{j+1}) - W(t_j)]^2 - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\ &= \sum_{j=0}^{n-1}\mathbb{E}f''(W(t_j))\mathbb{E}[W(t_{j+1}) - W(t_j)]^2 - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\ &= \sum_{j=0}^{n-1}\mathbb{E}f''(W(t_j))(t_{j+1}-t_j) - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\ &= 0. \end{align} where in the second equality, I have used the fact that $f''(W(t_j))$ is $\mathcal{F}(t_j)$- measurable while $[W(t_{j+1}) - W(t_j)]^2$ is independent of $\mathcal{F}(t_j)$ and in order to get the third equality, I have simply plugged in the value of $\mathbb{E}[W(t_{j+1}) - W(t_j)]^2$ which is $(t_{j+1}-t_{j}).$ Please comment if I'm right.

The official solution goes as follows:

$Z_j$ is $\mathcal{F}(t_{j+1})$- measurable and $\mathbb{E}[Z_j|\mathcal{F}(t_j)] = 0.$ I'm okay with both these steps and have easily proved it. But then they do the following working which I'm not clear with:

$\mathbb{E}\big[\sum_{j=0}^{n-1}Z_j\big] = \mathbb{E}\big[\sum_{j=0}^{n-1}\mathbb{E}[Z_j|\mathcal{F}(t_j)]\big] = 0.$

My question regarding their solution is the following:

$Z_j$ is $\mathcal{F}(t_{j+1})$- measurable so $\mathbb{E}[Z_j|\mathcal{F}(t_{j+1})] = Z_j$ so how can they replace $Z_j$ with $\mathbb{E}[Z_j|\mathcal{F}(t_{j})]$ ?

I know that $\mathbb{E}[Z_j|\mathcal{F}(t_{j})]$ is $\mathcal{F}(t_{j})$- measurable but can't understand the logic used in their solution. Could be a fundamental misunderstanding so any help highly appreciated.

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2 Answers 2

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To you question: "Shouldn't the iterated conditioning work this way : $E[E[Z_j|\mathcal F_{t_{j+1}}|\mathcal F_{t_j}]$ since $\mathcal F_{t_j} \subset \mathcal F_{t_{j+1}}$?"

Yes, you are right, but the relevant $\sigma$-algebras here are $\mathcal F_0$ and $\mathcal F_{t_j},$(not $\mathcal F_{t_{j+1}}$). We have

$$\mathbb{E}\big[\sum_{j=0}^{n-1}Z_j\big] = \mathbb{E}\big[\sum_{j=0}^{n-1}Z_j\big | \mathcal F_0]= \mathbb{E}\big[\sum_{j=0}^{n-1}\underbrace{\mathbb{E}[Z_j|\mathcal{F}(t_j)]}_{=0}\big| \mathcal F_0] = 0,$$ since $\mathcal F_0 \subset \mathcal F_{t_j}$ and $\mathcal F_0$ is the trivial $\sigma$-algebra.

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  • $\begingroup$ Got it now. Thanks. Do you mind having a look at my attempt? $\endgroup$
    – user715112
    Commented Jan 1, 2021 at 17:55
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    $\begingroup$ I think your attempt is correct. $\endgroup$
    – UBM
    Commented Jan 1, 2021 at 17:59
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They are using the Law of Iterated Expectations.

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  • $\begingroup$ Shouldn't the iterated conditioning work this way : $ \mathbb{E} \bigg[\mathbb{E}[Z_j|\mathcal{F}(t_{j+1})] \bigg| \mathcal{F}(t_j)\bigg] = \mathbb{E}[Z_j|\mathcal{F}(t_{j})]$ since $\mathcal{F}(t_{j}))$ is a sub-$\sigma$ algebra of $\mathcal{F}(t_{j+1})$. Do you mind elaborating your answer ? $\endgroup$
    – user715112
    Commented Jan 1, 2021 at 11:14

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