Let $\xi_{n}$ be a symmetric random walk, i.e, $\xi_{n}=\eta_{1}+\eta_{2}+\ldots+\eta_{n}$ where $\eta_{1},\ldots$ is a sequence of independent identically distributed random variables such that $P\{\eta_{n}=1\}=P\{\eta_{n}=-1\}=\frac{1}{2}$. Show that $\xi_{n}^2-n$ is a martingale with respect to the filtration $\mathcal{F_{n}}=\sigma(\eta_{1},\eta_{2},\ldots,\eta_{n})$ .

I came upon this problem while I was trying to self study Martingale.

By the definition of Martingale given in the book that I'm following, I need to show that

$i)$ $\xi_{n}^2-n$ is integrable for each $n=1,2,\ldots$

$ii)$ $\xi_{n}^2-n$ is adapted to $\mathcal{F_{n}}$

$iii)$ $E(\xi_{n+1}^2-(n+1))|\mathcal{F_{n}})=\xi_{n^2}-n$ for each $n=1,2,\ldots$

The second point is pretty simple to prove as $\xi_{n}^2-n$ being a function of $\eta_{1},\ldots,\eta_{n}$ is $\mathcal{F_{n}}$ measurable.

I have some doubts is proving first and third point. For first point,I started off by

$E(|\xi_{n}^2-n|)=E(|(\eta_{1}+\ldots+\eta_{n})^{2}-n|)\leq E(|\xi_{n}|^{2})+n$

Then the book proceeded to show that it is $\leq n^{2}+n<\infty$ because $|\xi_{n}|=|\eta_{1}+\ldots+\eta_{n}|\leq|\eta_{1}|+\ldots+|\eta_{n}|=n$.

$\mathbf{1})$ how did $|\eta_{1}|+\ldots+|\eta_{n}|$ become equal to $n$?

For third part, by using the general properties of conditional expectation, I got that

$E(\xi_{n+1}^{2}|\mathcal{F_{n}})=\xi_{n}^2 +2\xi_{n}E(\eta_{n+1}) +E(\eta_{n+1}^{2})$

But in the book they gave that $E(\xi_{n+1}^{2}|\mathcal{F_{n}})=1+\xi_{n}^2$. Which was then used to show that $E(\xi_{n+1}^2-(n+1))|\mathcal{F_{n}})=\xi_{n^2}-n$.

$\mathbf{2)}$ how did $2\xi_{n}E(\eta_{n+1}) +E(\eta_{n+1}^{2})$ becomes equal to 1?

I know that there is a similar question in this same website. But it was given i.i.d$N(0,1)$ in that case. That question is How to show $M_n = X_n^2-n$ is a martingale?

In the same question, one of the answers mentioned that

Because ${\{Y_n\}}$ is iid standard normal, $\mathbb{E} [Y_{n+1}^2 \space | \space \mathcal{F}_n \space]$ and $\mathbb{E} [Y_{n+1} \space | \space \mathcal{F}_n \space]$ are 1 and 0 respectively.

But normal distribution was not given in the question I'm currently working on. I also saw that i.i.d need not be normally distributed.

I'm just getting started in this area, so my knowledge is pretty basic.

  • $\begingroup$ Try to find $\mathbb E[\eta_{n}]=0$ and $\mathbb E[\eta_{n}^2]$ when $\mathbb P\{\eta_{n}=1\}=\mathbb P\{\eta_{n}=-1\}=\frac{1}{2}$ $\endgroup$
    – Henry
    Jun 15, 2023 at 0:42

1 Answer 1

  1. $\eta_n$ is $\pm1$, so $|\eta_n|=1$, so $|\eta_1|+\cdots+|\eta_n|=n$

  2. $\eta_n$ is $\pm1$, so $\eta_n^2=1$, so $$E[\eta_n^2|{\cal F}_n]=E[\eta_n^2]=E[1]=1$$. And $2\xi_nE[\eta_{n+1}]=0$ because $E[\eta_{n+1}|{\cal F_n}]=E[\eta_{n+1}]=0$

  • $\begingroup$ Thank you! Now I see that I should have tried to understand the question more properly $\endgroup$
    – A Y
    Jun 15, 2023 at 0:48

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