Let's take $X_1,X_2,...\sim i.i.d.$ with finite expected value. We have $S_n = \Sigma_{k=0}^{n}$ and filtration $\mathcal{F}_n$ that is generated by process $X_n$,( can we write that $ \forall_n ~ \sigma (X_n)\subset \mathcal{F}_n ) $.

We aim to show that $S_n-nEX_1$ is a martingale with respect to filtration $\mathcal{F}_n$.

It is obvious that $$ E|S_n-nEX_1| \leq E|S_n| + nE|X_1|\leq \infty $$ because expected value is finite.

And also that $S_n$ is measurable wrt $\mathcal{F}_n$ because filtration is generated by $X_n$ and $nEX_1$ is measurable wrt $\mathcal{F}_n$ because it is constant.

There is some troubles with third condition.

$$E( S_n-nEX_1 | \mathcal{F}_{n-1} ) = E(S_n | \mathcal{F}_{n-1}) - E(nEX_1) $$ Firstly we use independence and we get $$ = S_{n-1}+EX_n - nE(EX_1) $$ I have no idea what to do now to deal with the problem.


1 Answer 1


You are almost done. $S_{n-1}+EX_n - nE(EX_1)=S_{n-1}+EX_1 - nE(EX_1)=S_{n-1}-(n-1)EX_1$.

  • $\begingroup$ It looks good but i have one question why we can write $EX_n=EX_1$? $\endgroup$
    – nodis6
    Commented Nov 29, 2022 at 12:01
  • $\begingroup$ i.i.d. means independent and identically distributed. $X_n$ has the same distribution, hence the same mean as $X_1$ @nodis6 $\endgroup$ Commented Nov 29, 2022 at 12:06
  • $\begingroup$ All ritght it is clear to me thank yoou. $\endgroup$
    – nodis6
    Commented Nov 29, 2022 at 12:09

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