Markov for Martingales

August 24, 2022

\[\newcommand{E}{\mathbb{E}} \renewcommand{\Pr}{\mathbb{P}} \newcommand{\D}{\mathcal{D}} \newcommand{\F}{\mathcal{F}} \newcommand{\gn}{\;|\;}\]Possibly one of the most useful inequalities in all of probability is Markov’s inequality, which states that for a non-negative random variable \(X\),

\[\Pr(X \geq \alpha )\leq \frac{\E[X]}{\alpha},\]for all \(\alpha>0\). The intuition is simple if looked at as an optimization problem. Suppose we fix the probability \(p_\alpha = \Pr(X\geq \alpha)\) and ask the question: How small can we make \(\E[X]\)? That is, we want to solve

\[\begin{align*} \min_{\D} \quad & \E_{X\sim D}[X] \\ \text{s.t}\quad & p_\alpha = \Pr(X\geq \alpha) \\ & X\geq 0. \end{align*}\]This is easy. If we could, we’d put all of \(X\)’s mass at 0, which would minimize \(\E[X]\) subject to \(X\geq 0\). But we have to move at least some of the mass to \(\alpha\) or beyond. So we’ll put mass \(p\) at \(\alpha\), and mass \(1-p\) at 0. It should be clear that any other alteration would simply increase \(\E[X]\). With these choices, we have \(\E[X] = \alpha p_\alpha\), and since this minimized \(\E[X]\), we have \(\E[X]\geq \alpha p_\alpha\), which is the desired result.

Admittedly, despite it’s usefulness, you can only make Markov’s inequality so interesting. However, it’s natural to wonder whether Markov-like inequalities hold for more than just non-negative random variables. For instance, do certain random *processes* exhibit this kind of behavior? The answer is yes.

Recall that a supermartingale \((X_t)\) adapted to the filtration \((\F_t)\) on the filtered space \((\Omega,\F,(\F_t),\Pr)\) obeys

\[\E[X_t\gn \F_\tau] \le X_\tau, \quad \text{for all }\tau\leq t.\]Ville’s inequality (mentioned in the intro post on game-theoretic probability) generalizes Markov’s inequality to supermartingales. In particular, it states that

\[\Pr(\exists t\geq 0: X_t \geq \alpha \gn \F_0) \leq \frac{X_0}{\alpha}.\]In other words, the probability – given the information at the beginning of the process – that the value of \(X_t\) *at any time* exceeds \(\alpha\), is bounded in Markov-like fashion. For this reason, it’s sometimes called an *infinite-horizon* extension of Markov’s inequality.

To prove it, we’ll show it holds for all finite times \(k\), and then take the limit as \(k\to\infty\). Define the stopping time \(T\) to be the first time at which \(X_t\) is at least \(\alpha\). Formally, \(T=\inf\{t:X_t\geq \alpha\}\). (Recall that a stopping time is simply a random \(T\) where, for all \(t\), one can determine if \(T=t\) given \(\F_t\).) Then, for some arbitrary \(k\), define the truncated stopping time \(\tau = \min(T,k)\). Clearly, both \(T\) and \(\tau\) are well-defined stopping times.

Recall Doob’s optional stopping theorem, which states that \(X_\tau\) is measurable (almost surely) since \(\tau\) is bounded (by construction). Thus, Markov’s inequality applies to \(X_\tau\):

\[\Pr(T< k)\leq \Pr(X_\tau\geq \alpha \gn \F_0) \leq \frac{\E[X_\tau \gn \F_0]}{\alpha} \leq \frac{X_0}{\alpha},\]where the first inequality follows since if \(T< k\), then \(\tau=T\) and \(X_\tau =X_T\geq \alpha\) by definition of \(T\). The final inequality holds because \((X_t)\) is a supermartingale. Since this holds for all \(k\), taking limits gives

\[\Pr(T<\infty) = \lim_{k\to\infty} \Pr(T<k) \leq \frac{X_0}{\alpha}.\]Noting that \(\Pr(T<\infty) = \Pr(\exists t\geq 0: X_t\geq \alpha)\) completes the proof.

Notice that there was nothing special about \(\F_0\). The same proof yields

\[\Pr(\exists t\geq T: X_t \geq \alpha \gn \F_T) \leq \frac{X_T}{\alpha},\]for any \(T\).

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