# Doob-style second moment martingale inequality

Let $\{X_k\}_{k=0}^{\infty}$ be a martingale, supposing $X_0 = 0$ and $E[{X_n}^2] <\infty$. Prove that $$P\left(\max_{1\le k \le n} X_k \ge r \right) \le \frac{E[{X_n}^2]}{E[{X_n}^2] + r^2}$$ for all $r>0$.

I have tried to emulate the proof of the Doob maximal inequality as follows. Let $T = \min\{k\, : \, X_k \ge r\}$. Then by Jensen's inequality, $\{{X_k}^2\}$ is a submartingale, so that $E[{X_{T\wedge n}}^2] \le E[{X_n}^2]$ via optional stopping theorem. Since $${X_{T\wedge n}}^2 \ge \mathbb{1}_{T\le n} r^2 + \mathbb{1}_{T> n} {X_n}^2$$ it follows $$E[{X_{T\wedge n}}^2] \ge r^2 P(T\le n) + E[{X_n}^2 \mathbb{1}_{T>n}]$$ If this is the right approach, then we next will have to prove $E[{X_n}^2 \mathbb{1}_{T>n}] \ge P(T\le n) E[{X_n}^2]$ but I am not able to show this. I have tried Cauchy-Schwarz and Jensen's, both to no avail. So I expect this approach is too weak.

Another observation I made is the following: By Cauchy-Schwarz, $$E[X_{T\wedge n} \mathbb{1}_{T>n} ]^2 \le E[{X_{T\wedge n}}^2 \mathbb{1}_{T>n}] P(T>n) \le r^2 P(T>n)^2$$ since $X_{T\wedge n} \le r$ when $T>n$. But also $$E[X_{T\wedge n} \mathbb{1}_{T> n} ]^2 = E[X_{T\wedge n} \mathbb{1}_{T\le n} ]^2\ge E[r\cdot \mathbb{1}_{T\le n}]^2 = r^2 P(T\le n)^2$$ by using optional stopping theorem to see $E[X_{T\wedge n}] = E[X_0] = 0$. Thus $P(T\le n) \le P(T>n)$. But I don't know how to proceed from here. Any help/hints are much appreciated!

• You might be interested in the Cantelli Inequality – Gautam Shenoy Jan 22 '16 at 18:31

## 1 Answer

1. $\{X_k\}$ cannot be a nonnegative MTG with $X_0=0$. So, I assume it is just a MTG.

2. Using the original Doob's weak inequality, for some $z >-r$ (and noticing that $\mathbb{E}X_n=0$):

$$P\{\max_{1\le k\le n}{X_k}\ge r\}\le P\{\max_{1\le k\le n}{(X_k+z)^2}\ge (r+z)^2\}\le\frac{1}{(r+z)^2}\mathbb{E}(X_n+z)^2=\frac{1}{(r+z)^2}(\mathbb{E}X_n^2+z^2)$$

1. Optimizing the RHS over $z$ we get

$$\frac{d}{dz}\frac{1}{(r+z)^2}(\mathbb{E}X_n^2+z^2)=\frac{2z}{(r+z)^2}-\frac{2(\mathbb{E}X_n^2+z^2)}{(r+z)^3}=0 \Rightarrow$$

$$z^*=\frac{\mathbb{E}X_n^2}{r}$$

1. Finally,

$$P\{\max_{1\le k\le n}{X_k}\ge r\}\le \frac{1}{(r+\mathbb{E}X_n^2/r)^2}(\mathbb{E}X_n^2+(\mathbb{E}X_n^2/r)^2)=\frac{\mathbb{E}X_n^2}{\mathbb{E}X_n^2+r^2}$$