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Let $\{X_n:n=1,2,\ldots\}$ be a sequence of independent random variables with common distribution $$\mathbb P(X_1=1) = \mathbb P(X_1=-1) = \frac12. $$ Set $S_0:=0$ and $S_n = \sum_{j=1}^n X_j$ for $n=1,2,\ldots$. Show that $M_n:= S_n^2 - n$ is an adapted process with respect to the filtration $\mathcal F_n = \sigma(X_1,\ldots, X_n)$, and that $M_n$ is a martingale with respect to this filtration.

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    What process would $X_t$ be? – 2017-01-29
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    toss a fair coin, if it shows up head, then $X_j = 1$, otherwise $X_j = -1$, j represents the number of times of the experiment – 2017-01-29
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    so all $X_i$ are $\textit{iid}$? – 2017-01-29
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    yes, they are iid – 2017-01-29
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    @HaoQuan I rewrote your question using the information provided in your comments. – 2017-01-29

3 Answers 3

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It is clear from inspection that for each positive integer $n$ the map $x\mapsto x^2-n$ is a measurable function, and so $M_n$ is adapted with respect to $\mathcal F_n$. Moreover, we see that \begin{align} \mathbb E[|M_n|] &= \mathbb E[|S_n^2-n|] \leqslant \mathbb E[S_n^2] + n \leqslant n^2 + n <\infty, \end{align} so the sequence of random variables $M_n$ is integrable. To check the martingale condition, we compute \begin{align} \mathbb E[M_{n+1}\mid \mathcal F_n] &= \mathbb E\left[S_{n+1}^2 -(n+1)\mid \mathcal F_n\right]\\ &= \mathbb E\left[S_n^2 + X_{n+1}^2(1-2S_n) - (n+1)\mid\mathcal F_n \right]\\ &= \mathbb E\left[S_n^2 - n \mid\mathcal F_n \right] + \mathbb E\left[X_{n+1} -2S_nX_{n+1}- 1\mid\mathcal F_n \right]\\ &= S_n^2-n + \mathbb E\left[ X_{n+1}^2\mid\mathcal F_n \right] -2S_n\mathbb E\left[X_{n+1}\mid\mathcal F_n\right] - 1\\ &= M_n + 1 - 2\cdot0 -1\\ &= M_n. \end{align}

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    how to prove $\mathbb{E} [S_n^2 - n| F_n] = S_n^2 - n$? – 2017-01-29
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    $S_n^2-n$ is $\mathcal F_n$-measurable, so this follows directly from basic properties of conditional expectation. – 2017-01-29
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    how to use $\mathbb{E}[M_{n+1} | F_n] = M_n $ to derive $\mathbb{E} [M_t|F_s] = M_s$, for $s\le t$? – 2017-01-29
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    That has nothing to do with your previous question, so I am not sure what you are really asking... – 2017-01-29
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    Because $\mathbb{E} [M_t|F_s] = M_s$, for $s\le t$ is the definition that my class introduces about martingale. – 2017-01-29
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    @HaoQuan You just use the tower property. For example, $E[M_{n+2} \mid F_n] = E[E[M_{n+2} \mid F_{n+1}] \mid F_n]=E[M_{n+1} \mid F_n]=M_n$. With induction you can replace $M_{n+2}$ here with $M_{n+k}$ for arbitrary $k$. – 2017-01-30
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$$\mathbb{E}(S_n^2-n|\mathcal F_{n-1}) = \mathbb{E}(S_{n-1}^2+X_n^2-2S_{n-1}X_n-n|\mathcal F_{n-1}) $$

$$ = S_{n-1}^2-n-2S_{n-1}\mathbb{E}(X_n|\mathcal F_{n-1})+\mathbb{E}(X_n^2|\mathcal F_{n-1})$$

$$=S_{n-1}^2-n-2S_{n-1}\mathbb{E}(X_n)+\mathbb{E}(X_n^2)$$ $$=S_{n-1}^2-n-0+1=S_{n-1}^2-(n-1).$$

Update:

Note that $\mathbb{E}(M_t|F_{t-1})=M_{t-1}$ suffices to show that $M_n$ is a Martingale, since for $t>s$, $$\mathbb{E}(M_t|\mathcal F_s)=\mathbb{E}(\mathbb{E}(M_t|\mathcal F_{t-1})|\mathcal F_s)=\mathbb{E}(M_{t-1}|F_{s}).$$

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    is there a way other than using induction? – 2017-01-29
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    If I'm not wrong, the definition of Martingale involves induction. – 2017-01-29
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    @HaoQuan Is the update what you are looking for? – 2017-01-30
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You said, that you are primarily interested in the measurability part of the now clarified question. You can see, that $$ M_n:= S_n^2 - n $$
is $\mathcal{F}_n$ measurable, while conducting the following steps:

First, we note, that $S_n = \sum_{j=1}^n X_j$ is of course $\mathcal F_n = \sigma(X_1,\ldots, X_n)$ (as a finite sum of $\mathcal F_n$-measurable random variables). Then we use the fact, that the composition of measurable functions is indeed once again measurable.

With $f:x\mapsto x^2$ we have a continuous function which is therefore measurable. Now we see $$ M_n=f(S_n)-n $$ and conclude, since the first part is just a composition of measurable functions, is once again measurable, and $n$ is as a deterministic function again $\mathcal F_n$ measurable. We conclude, that $\forall n\in\mathbb N: M_n$ is $\mathcal F_n = \sigma(X_1,\ldots, X_n)$ measurable ($\equiv$ adpated),so in particular we have (as also noted by Math1000) $$ {E} [M_n|\mathcal F_n]= M_n $$