# Is $S_n-nEX_1$ a martingale

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.

You are almost done. $$S_{n-1}+EX_n - nE(EX_1)=S_{n-1}+EX_1 - nE(EX_1)=S_{n-1}-(n-1)EX_1$$.
• It looks good but i have one question why we can write $EX_n=EX_1$? Commented Nov 29, 2022 at 12:01
• i.i.d. means independent and identically distributed. $X_n$ has the same distribution, hence the same mean as $X_1$ @nodis6 Commented Nov 29, 2022 at 12:06