Prove that a succession of random variables is a martingale

Question

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?

Answer 1

$Y_{i}$ having finite variance implies that $E[|Y_{i}|^{2}]<\infty$, since $\text{Var}(Y_{i})=E[|Y_{i}|^{2}]-E[Y_{i}]^{2}$.

Answer 2

I think the complete answer is the following.

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

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 $Y_1,...,Y_n$)

$=n\sigma^2-(\mu^n)^2<\infty$

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

$=E[(S_n+Y_{n+1})^2-(n+1)\sigma^2|Y_1,...,Y_n]$

$=E[S_n^2+2S_nY_{n+1}+Y_{n+1}^2-n\sigma^2-\sigma^2|Y_1,...,Y_n]$

$=S_n^2-n\sigma^2+2SnE[Y_{n+1}]+E[Y_{n+1}^2]-\sigma^2$

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}