0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

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

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.