Is $Y_n=S_n^3-3nS_n$ a martingale? 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 $X_{n-1}$.
 A: 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.
