# Chebyshev Inequality for Martingales

Suppose $\{X_n\}_{n \geq 1}$ is a square-integrable martingale with $E(X_1)=0$. Then for $c>0$:

$$P\left(\max_{i=1, \ldots, n} X_i \geq c\right) \leq \frac{\textrm{Var}(X_n)}{\textrm{Var}(X_n) + c^2}.$$

I imagine Doob's martingale inequality will come into play, but the details elude me.

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Got something from the answer below? –  Did Aug 11 '12 at 9:22

One can start from Doob's martingale inequality, which states that for every submartingale $(Y_n)_{n\geqslant0}$ and every $y\gt0$, $$\mathrm P\left(\max\limits_{0\leqslant k\leqslant n}Y_k\geqslant y\right)\leqslant\frac{\mathrm E(Y_n^+)}y\leqslant\frac{\mathrm E(|Y_n|)}y.$$ Applying this to $Y_n=(X_n+z)^2$ for some $z\gt0$ and to $y=(x+z)^2$ for some $x\gt0$, one gets $$\mathrm P\left(\max\limits_{0\leqslant k\leqslant n}X_k\geqslant x\right)\leqslant\mathrm P\left(\max\limits_{0\leqslant k\leqslant n}Y_k\geqslant y\right)\leqslant C_n(z),$$ with $$C_n(z)=\frac{\mathrm E(|Y_n|)}{y}=\frac{\mathrm E(X_n^2)+z^2}{(x+z)^2}.$$ Finally, for $z=\dfrac{\mathrm E(X_n^2)}{x}$, $C_n(z)=\dfrac{\mathrm E(X_n^2)}{\mathrm E(X_n^2)+x^2}$ hence the proof is complete.