Prove that a succession of random variables is a martingale I've been working on the following problem: 
Let $\{{Y_n:n\in \mathbb{N}}\}$ be independent identically distributed random variables with mean $\mu$ and variance $\sigma^2>0$. Define $S_n=Y_1+...+Y_n$ and $Z_n=S_n^2-n\sigma^2$.
Is $\{{Z_n:n\in \mathbb{N}}\}$ a martingale with respect to $\{{Y_n:n\in \mathbb{N}}\}$ if $\mu=0$? What if $\mu>0$?
So we need to use the definition of martingale. The second part was easy
$E[Z_{n+1} |Y_0,...,Y_n]=E[S_n^2-n\sigma^2|Y_0,...,Y_n]
\\=E[S_n^2|Y_0,...,Y_n]+E[-n\sigma^2|Y_0,...,Y_n]
\\=S_n^2E[1|Y_0,...,Y_n]-n\sigma^2
\\=S_n^2*1-n\sigma^2
\\=Z_n$. 
But what about the first part? (prove that $E[Z_n]$ is finite). We know that
$|Z_n|=|S_n^2-n\sigma^2| \leq |S_n|^2+n\sigma^2 =(|Y_1+...+Y_n|)^2+n\sigma^2\leq(\sum\limits_{i=1}^n |Y_i|)^2 +n\sigma^2$
so
$E[Z_n]\leq E[(\sum\limits_{i=1}^n |Y_i|)^2 +n\sigma^2]$
$=E[(\sum\limits_{i=1}^n |Y_i|)^2] +n\sigma^2$ 
$=\sum\limits_{i=1}^n E[|Y_i|^2]+n\sigma^2$ (by independence)
$=nE[|Y_i|^2]+n\sigma^2$ (because the rv are identically distributed)
How can I conclude that $E[Z_n]$ is indeed finite?
 A: I think the complete answer is the following. 


*

*$|Z_n|=|S_n^2-n\sigma^2|\leq|S_n|^2+n\sigma^2$, 


so 
$E[|Z_n|]\leq E[|S_n|^2+n\sigma^2]=E[|S_n|^2]+n\sigma^2<\infty$
because 
$E[|S_n|^2]=Var(S_n)-E[S_n]^2$
$=Var(Y_1,...,Y_n)-E[Y_1,...,Y_n]^2$
$=\sum_{i=1}^nVar(Y_i)-(\prod_{i=1}^nE[Y_i])^2$ (by independence of $Y_1,...,Y_n$)
$=n\sigma^2-(\mu^n)^2<\infty$


*$E[Z_{n+1}|Y1,...Y_n]=E[S_{n+1}^2-(n+1)\sigma^2|Y_1,...,Y_n]$


$=E[(S_n+Y_{n+1})^2-(n+1)\sigma^2|Y_1,...,Y_n]$
$=E[S_n^2+2S_nY_{n+1}+Y_{n+1}^2-n\sigma^2-\sigma^2|Y_1,...,Y_n]$
$=S_n^2-n\sigma^2+2SnE[Y_{n+1}]+E[Y_{n+1}^2]-\sigma^2$
So if $\mu=0$, $E[Y_{n+1}]=0$ and $E[Y_{n+1}^2]=Var(Y_{n+1})=\sigma^2$, therefore 
$=S_n^2-n\sigma^2+2SnE[Y_{n+1}]+E[Y_{n+1}^2]-\sigma^2=S_n^2-n\sigma^2=Z_n$. 
In that case, ${Z_n}$ is a martingale with respect to {Y_n}. 
Furthermore, if $\mu>0$, ${Z_n}$ is a submartingale with respect to {Y_n}
A: $Y_{i}$ having finite variance implies that $E[|Y_{i}|^{2}]<\infty$, since $\text{Var}(Y_{i})=E[|Y_{i}|^{2}]-E[Y_{i}]^{2}$.
