Proof that $E[X_t] \leq E[G_t]$ for $t \geq 0$ implies $E[X_\tau] \leq E[G_\tau]$ for bounded stopping times $\tau$?

Question

Let $(X_t)_{t \geq 0}$ and $(G_t)_{t \geq 0}$ be non-negative stochastic processes with right continuous sample paths on some probability space. Assume that $\mathbb{E}[X_t] \leq \mathbb{E}[G_t]$ holds for all $t \geq 0$. Is there an elegant argument which shows that $\mathbb{E}[X_{\tau}] \leq \mathbb{E}[G_{\tau}]$ holds for all bounded stopping times $\tau$?

Answer 1

It looks like this statement is not true in general. Consider the following discrete-time processes. Let $(X_k)_{0 \leq k \leq n}$ be a sequence of i.i.d. random variables distributed as the shifted Bernoulli $$ P(X_0 = 1) = P(X_0 = 2) = \frac 1 2 , $$ and define $(G_k)_{0 \leq k \leq n}$ by $G_k = 4X_k - 4$, which takes either of values 0 or 4. Then $E(X_k) = 3/2$ and $E(G_k) = 2$. Let $\tau$ be the first time $k$ such that $X_k$ is equal to 1, setting $\tau = n$ if none of the random variables $X_k$ is equal to 1. The idea is that if $n$ is large enough, there is very likely to be some time at which $X_k = 1$, in which case we would have $X_\tau = 1 > 0 = G_\tau$. Indeed, $$ E(G_\tau - X_\tau) = 2 P(\tau = n) - 1 P(\tau < n) = 2 \cdot \Big( \frac 1 2 \Big)^{n+1} - \left( 1 - \Big( \frac 1 2 \Big)^{n+1} \right) ,$$ and this is strictly negative even for $n = 1$.

In light of this example, we can define continuous-time processes by extending $(X_t)_{0 \leq t \leq n}$ to non-integer times in a right-continuous fashion, defining $X_t$ to be $X_{\lfloor t \rfloor}$ above. Defining $G_t$ in the same way and $\tau$ exactly as before gives the same result for $E(G_\tau - X_\tau)$.