# Show that $\xi _{n}^2-n$ is a martingale

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.

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

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 $$E[\eta_{n+1}|{\cal F_n}]=E[\eta_{n+1}]=0$$