Let $S_n=X_1+\dots+X_n$, where $X_1,X_2,\dots$ are independent and $\mathbb{E}(X_m)=0$, $\mathbb{E}(X_m^2)=\sigma_m^2\in(0,\infty)$ for all $m\geq1$. Let $\mathcal{F}_n=\sigma(X_1,\dots,X_n)$.
It is easy to show that for $c>0$, $(S_n+c)^2$ is a submartingale in $(\mathcal{F}_n)$.
I am tasked to use the submartingale above and Doob's inequality to show that for all $x>0$ we have $$\mathbb{P}\left(\max_{1\leq m\leq n}S_m\geq x\right)\leq\frac{\text{Var}(S_n)}{\text{Var}(S_n)+x^2}.$$ For $c>0$, It is obvious that $$\left\{\max_{1\leq m\leq n}S_m\geq x\right\}\subseteq\left\{\max_{1\leq m\leq n}(S_m+c)^2\geq(x+c)^2\right\}.$$ Then we can use Doob's inequality to get \begin{aligned}\mathbb{P}\left(\max_{1\leq m\leq n}S_m\geq x\right)&\leq\mathbb{P}\left(\max_{1\leq m\leq n}(S_m+c)^2\geq(x+c)^2\right)\\&\leq\frac{\mathbb{E}\{(S_n+c)^2\}}{(x+c)^2}=\frac{\text{Var}(S_n)+c^2}{(x+c)^2}\\&\leq\frac{\text{Var}(S_n)+c^2}{x^2+c^2}\end{aligned} I have checked, and there is no value of $c>0$ such that the RHS of this attained inequality equals the RHS of the inequality I am aiming for, so I must be missing something to make my inequalities stricter. Advice would be greatly appreciated, thanks!
Best Answer
In the line before you estimate $(x+c)^2\ge x^2+c^2$, it looks like if you set $c = \mathrm{Var}(S_n)/x$, you get the desired result.