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Let $X_i$ for $i=1,\cdots,n$ random variables i.i.d., such that $P(X_1=1)=\frac{1}{2}$ and $P(X_1=-1)=\frac{1}{2}$. Let $S_n=X_1+\cdots+X_n$ Let $Y_n=S_n^3-3nS_n$, is $Y_n$ a martingale or isn't?

Attempt: We have $\mathbb{E}(X_1)= 1\cdot \frac{1}{2}-1\cdot\frac{1}{2}= 0$. I also have verified that $\mathbb{E}(|Y_n|)<\infty$ and that taking $h_n(X_1,\cdots,X_n)=(X_1+\cdots+X_n)^3-3n(X_1+\cdots+X_n)=Y_n$ so is medible. If we can prove that $\mathbb{E}(S_n^3-3nS_n|X_1,\cdots, X_{n-1})=Y_{n-1}=S_{n-1}^3-3(n-1)S_{n-1}$ then we are done, but not sure how to proceed, because $Y_n$ is unknow at filtration at time $X_{n-1}$.

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  • $\begingroup$ What exactly is $S_{n}$? $\endgroup$
    – B_B
    Commented May 8, 2021 at 17:03
  • $\begingroup$ @B_B Is a random walk sum. $\endgroup$ Commented May 8, 2021 at 17:04
  • $\begingroup$ OK, you have edited it. $\endgroup$
    – B_B
    Commented May 8, 2021 at 17:05
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    $\begingroup$ It might help to split $S_n$ into what is known at time $n-1$ and what is unknown. I.e. write $S_n = S_{n-1} + X_n$. $\endgroup$ Commented May 8, 2021 at 17:08
  • $\begingroup$ Yes, Jamie is right. $\endgroup$
    – B_B
    Commented May 8, 2021 at 17:36

1 Answer 1

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If $Y_{n}=S_{n}^{3}-3nS_{n}$, then

$Y_{n+1}=(S_{n}+X_{n+1})^{3}-3(n+1)(S_{n}+X_{n+1})$

so after multiplication and re-arranging, we get

$Y_{n+1}=Y_{n}+3S_{n}(X_{n+1}^{2}-1)+3S_{n}^{2}X_{n+1}-3(n+1)X_{n+1}+X_{n+1}^{3}$

We need to check if

$\mathbb{E}[Y_{n+1}|\mathcal{F}_{n}]=Y_{n}$

so in our case, if

$\mathbb{E}\Big[3S_{n}(X_{n+1}^{2}-1)+3S_{n}^{2}X_{n+1}-3(n+1)X_{n+1}+X_{n+1}^{3}|\mathcal{F}_{n}\Big]=0$

Note that:

1.

$\mathbb{E}\Big[3S_{n}(X_{n+1}^{2}-1)|\mathcal{F}_{n}\Big]=3S_{n}\Big(\mathbb{E}\Big[X_{n+1}^{2}|\mathcal{F}_{n}\Big]-1\Big)=3S_{n}\Big(\mathbb{E}\Big[X_{n+1}^{2}\Big]-1\Big)$

2.

$\mathbb{E}\Big[3S_{n}^{2}X_{n+1}|\mathcal{F}_{n}\Big]=3S_{n}^{2}\mathbb{E}\Big[X_{n+1}|\mathcal{F}_{n}\Big]=3S_{n}^{2}\mathbb{E}\Big[X_{n+1}\Big]$

3.

$\mathbb{E}\Big[3(n+1)X_{n+1}|\mathcal{F}_{n}\Big]=3(n+1)\mathbb{E}\Big[X_{n+1}|\mathcal{F}_{n}\Big]=3(n+1)\mathbb{E}\Big[X_{n+1}\Big]$

4.

$\mathbb{E}\Big[X_{n+1}^{3}|\mathcal{F}_{n}\Big]=\mathbb{E}\Big[X_{n+1}^{3}\Big]$

Because $X_{n+1}^{3}$, $X_{n+1}^{2}$, $X_{n+1}$ and $X_{1},\ldots,X_{n}$ are independent random variables.

Now

$\mathbb{E}\Big[X_{n+1}^{3}\Big]=\mathbb{E}\Big[X_{n+1}\Big]=0$

$\mathbb{E}\Big[X_{n+1}^{2}\Big]=1$

so

$\mathbb{E}\Big[3S_{n}(X_{n+1}^{2}-1)+3S_{n}^{2}X_{n+1}-3(n+1)X_{n+1}+X_{n+1}^{3}|\mathcal{F}_{n}\Big]=3S_{n}(1-1)+3S_{n}^{2}\cdot 0-3(n+1)\cdot 0+ 0=0$

It means, that

$\mathbb{E}[Y_{n+1}|\mathcal{F}_{n}]=Y_{n}$

and $Y_{n}=S_{n}^{3}-3nS_{n}$ is a martingale.

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  • $\begingroup$ Alternatively, we can use $$Y_{n+1}=Y_{n}+3S_{n}(X_{n+1}^{2}-1)+3S_{n}^{2}X_{n+1}-3(n+1)X_{n+1}+X_{n+1}^{3}$$ and the distribution of $X_i$ to write $$Y_{n+1}=Y_{n}+(3S_{n}^{2}-3n-2)X_{n+1}$$ almost surely. $\endgroup$ Commented May 8, 2021 at 18:04

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