For independent centered random variables and $S_m=\sum_{j\leq m}X_j$,
$$
\mathbb P\left(\max_{m\leq n}S_m\geq x\right)
\leq\frac{\operatorname{Var}(S_n)}
{\operatorname{Var}(S_n)+x^2}.
$$
Apply the <Doob maximal inequality for a nonnegative submartingale> to $(S_m+c)^2$ and optimize at $c=\operatorname{Var}(S_n)/x$.

One-sided maximal inequality for independent centered sums

Put $V=\operatorname{Var}(S_n)=\sum_{m=1}^nv_m$. If $\max_{m\leq n}S_m\geq x$, then
$$
\max_{m\leq n}(S_m+c)^2\geq(x+c)^2.
$$
The <Doob maximal inequality for a nonnegative submartingale> therefore gives
$$
\mathbb P\left(\max_{m\leq n}S_m\geq x\right)
\leq\frac{\mathbb E[(S_n+c)^2]}{(x+c)^2}
=\frac{V+c^2}{(x+c)^2}.
$$
The right side is minimized at $c=V/x$, and substitution gives
$$
\boxed{\mathbb P\left(\max_{1\leq m\leq n}S_m\geq x\right)
\leq\frac{\operatorname{Var}(S_n)}
{\operatorname{Var}(S_n)+x^2}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 2 / b / Solution

For $\theta>0$, the <Exponential martingale for Brownian motion> and the <Doob maximal inequality for a nonnegative submartingale> give
$$
\mathbb P\left(\sup_{t\leq s}B_t\geq a\right)
\leq\exp\left(-\theta a+\frac12\theta^2s\right).
$$
Taking $\theta=a/s$ gives $e^{-a^2/(2s)}$. Apply the same argument to $-B$ and use the <union bound>:
$$
\boxed{\mathbb P\left(\sup_{t\leq s}|B_t|\geq a\right)
\leq2e^{-a^2/(2s)}.}
$$

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 1 / 1 / 3 / Solution

{c}

If $(X_m)_{m\leq n}$ is a nonnegative <submartingale>, then for every $a>0$,
$$
a\,\mathbb P\left(\max_{m\leq n}X_m\geq a\right)
\leq\mathbb E[X_n].
$$
Stop at the first crossing of $a$ and use the submartingale property on that event.

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