For a supermartingale $\{X_k\}$, applying Doob's maximal inequality yields the tail probability bound $$P\left( \left| X_k \right| \geq N \right) \leq P\left( \sup_{0\leq t \leq k} \left| X_t \right| \geq N \right) \leq \frac{EX_0 + 2EX_k^-}{N},$$ where $X_k^-$ denotes the negative part of $X_k$. I wonder whether there is a similar inequality for the "tail expectation", i.e., $$E\left( \left| X_k \right| \mathbb{1}_{\{\left| X_k \right| \geq N\}} \right) \sim O(1/N),$$ given $\{X_k\}$ is bounded below. The guess is based on the intuition that the supermartingale decreases over time in the expectation sense, and the "mass" beyond the threshold $N$ should be small for a large $N$.
1 Answer
Here is a partial answer if $X$ is furthermore $L^2$ using Cauchy-Schwartz and Doob's maximal inequality we have :
$$E\left( \left| X_k \right| \mathbb{1}_{\{\left| X_k \right| \geq N\}} \right)\leq E\left( \left| X_k^2 \right|\right)^{1/2}. E\left( \mathbb{1}_{\{\left| X_k \right| \geq N\}} \right)^{1/2}\leq ||X_k||_2.\left(\frac{EX_0 + 2EX_k^-}{N}\right)^{1/2}$$
so we have : $$E\left( \left| X_k \right| \mathbb{1}_{\{\left| X_k \right| \geq N\}} \right) \sim O(1/N^{1/2})$$