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Let $\{Y_n\}$ be an independent and identically distributed sequence of random variables, with: \begin{equation*} P(Y_n = a) = \frac{1}{2} = P(Y = -a) \end{equation*} for some $a > 0$, and let $S_n$ be the stochastic process: \begin{equation*} S_n = S_{n-1}(1+Y_n^3), n \geq 1, S_0 = 1. \end{equation*} I am trying to show this process is a martingale, and have shown that $E[S_n|\mathcal{F}_m] = S_m$ for all $n > m$, but am struggling to prove that $E[|S_n|] < \infty$ for all $n$.

So far I have that $|S_n| \leq \prod_{i = 1}^n (|1| + |Y_i^3|) = (1 + a^3)^n$, but I'm not sure if I can then conclude anything about $E[|S_n|]$.

From my bound on $|S_n|$ can I conclude that $E[|S_n|]$ is also bounded, even though my bound tends to $\infty$ with $n$?

2 Answers 2

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Yes, you have shown what you need for it to be a martingale --- assuming you are taking the martingale to be with respect to some appropriate increasing sequence of $\sigma$-fields.

You just need that $S_n$ is integrable for each fixed $n$.

So your bound is enough -- assuming it is correct. I didn't check it myself.

The adaptedness condition for martingales would be automatic if the filtration you are working with is $\sigma(Y_1 \ldots Y_n)$ or some other filtration to which the sequence $\{Y_n\}$ is adapted.

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Let's do it with mathematical induction.

$$E[\mid S_1 \mid]=1+E[\mid Y_1^3\mid]=1+a^3.$$

Assume now that

$$E[\mid S_{n-1}\mid]<\infty.$$

By the definition of the sequence, we have

$$E[\mid S_n\mid]=E[\mid S_{n-1}(1+Y_{n}^3)\mid]\le E[\mid S_{n-1}\mid]E[\mid 1+Y_{n}^3\mid]<\infty.$$

(Because of the independence of $Y_n$ and $S_{n-1}$). This proves the claim.