Expected value inequality for $X_n^T$

Question

Let $X_n$ be a bounded random process and let $T$ be a stopping time that is finite almost surely.

Suppose we have the inequality $E(X_n) \geq c$, where $c$ is a constant.

Does it follow that $E(X_n^T) \geq c$ as well?

Here $X_n^T = X_{T \wedge n}$, where $T \wedge n$ is the minimum of $T$ and $n$.

If this is correct, an argument or even a sketch of an argument would be appreciated. If this is incorrect, then a counterexample would be great; also, under what conditions do we have such an inequality for $E(X_n^T)$?

Answer 1

This is not correct. Let $\mathbb{P}(X_1 = 2c) = \mathbb{P}(X_1 = 0) = \frac 12$ and $X_2 = 0$ if $X_1 = 2c$ and $X_2 = 2c$ if $X_1 = 0$. Then if $T = \inf\{X_n = 0\}$ we have $T \le 2 < \infty$, but $\mathbb{E}[X_2^T] = 0$.

I'm not sure what the tightest conditions to get this inequality are, but $X$ being a supermartingale is certainly sufficient.

Answer 2

As long as $X_n$ is a supermartingale, meaning $E[X_{n+1}|X_n]\le X_n$ for all $n\in \mathbb N$, then we can conclude that $E[X_n ^T]\ge E[X_n]$. This follows from the general fact that if $X_n$ is a supermartingale and $S$ is a stopping time which is bounded above by $b$ almost surely, then $E[X_S]\ge E[X_b]$. This is proven in Probability: Theory and Examples by Durrett, p. $212$, theorem $5.4.1$.

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Here is a counter-example, Let $X_n$ be an iid sequence of die rolls, $X_n\sim \operatorname{Unif}\{1,\dots,6\}$.

Let $T=\inf\{n:X_n=1$}.

Then $E[X_n]=3.5$, but $E[X_n^T]\approx 1$ when $n$ is large. As $n$ increases, it becomes overwhelmingly likely that a one is rolled at least once in the first $n$ rolls, which case $E[X_n^T]=1$.