# Exercise - Martingales - Correct choice?

I have a final question on martingales. The statement of the problem is as follows:

Let $S_n=X_1+...+X_n$ with $S_0=0$, where $(X_n)_{n\geq1}$ are i.i.d. random variables with $\mathbb{E}[X_n]=0$ and $\mathbb{E}[X_n^2]=\sigma^2<\infty$. Let $M_n:=\max_{0\leq k \leq n}S_k$. Show that $$\mathbb{P}\left\{\frac{M_n}{\sqrt{n}}\geq a\right\}\leq \frac{\mathbb{E}[S_n^2]}{na^2}=\frac{\sigma^2}{a^2} \quad \text{ and } \quad \mathbb{E}\left[\left(\frac{M_n}{\sqrt{n}}\right)^2\right]\leq\frac{4}{n}\mathbb{E}[S_n^2]=4\sigma^2$$

My problem is that I have "troubles with the n's", I do not know which I have to fix and which not... So here is my reasoning, I hope somebody can tell me if it was correct or not, and if not, could you please correct me?

So the first inequality I have to show is for all $n$. So i fix $n \in \mathbb{N}$, and I consider the sequence $\left(\frac{S_k}{\sqrt{n}}\right)_k$ which is a martingale. Hence $\left(\left(\frac{S_k}{\sqrt{n}}\right)^2\right)_k$ is a submartingale. Since $\frac{M_n^2}{n}\leq \max_{0\leq k \leq n}\left(\frac{S_k}{\sqrt{n}}\right)^2$, I obtain for each $a^2>0$,

$$a^2\mathbb{P}\left\{\frac{M_n^2}{n}\geq a^2\right\}\leq a^2\mathbb{P}\left\{\max_{0\leq k \leq n}\left(\frac{S_k}{\sqrt{n}}\right)^2\geq a^2\right\}\leq\mathbb{E}\left[\left(\frac{S_n}{\sqrt{n}}\right)^2\right],$$ where the second inequality is obtained by Doob's maximal inequalities. Then by simple computations I obtain the result. Is this correct or not? If so, can I continue to prove the second inequality (a result of Doob's maximal inequalities) by using the same sequence $\left(\frac{S_k}{\sqrt{n}}\right)_k$?

Probably you mean $a^2 \cdot \mathbb{P} \left( \max \ldots \geq a^2 \right)$ instead of $a^2 \cdot \mathbb{P} \left( \max \ldots \geq a \right)$ ... –  saz Jan 27 '13 at 21:08
So far, it's fine (you should note that $$\frac{M_n^2}{n} = \max_{0 \leq k \leq n} \left(\frac{S_k}{\sqrt{n}} \right)^2$$ - so the first inequality in your proof is even an equality.) But what do you mean by "use the same sequence"? You already know that $\left( \frac{S_k^2}{\sqrt{n}} \right)_k$ is a submartingale - so where is the problem? –  saz Jan 27 '13 at 21:23
Ok, when you say this is okay, the rest works fine, I had only troubles with fixing $n$. And yes for the second part I also use $\left( \frac{S_k^2}{\sqrt{n}} \right)_k$ and simply apply the theorem. Thank you! –  Mathoman Jan 27 '13 at 21:27
Well, if you are not sure about fixing $n$, you could also use the following argumentation instead: $$\mathbb{P} \left( \frac{M_n^2}{n} \geq a^2 \right) = \mathbb{P} \left( M_n^2 \geq n \cdot a^2 \right) = \mathbb{P} \bigg( \sup_{0 \leq k \leq n} S_n^2 \geq \underbrace{n \cdot a^2}_{c} \bigg) \leq \frac{1}{c} \mathbb{E}(S_n^2)$$ so you simply have a constant (depending on $n$) when applying Doob's maximal inequality. –  saz Jan 27 '13 at 21:30