If I have a discrete-time non-martingale random process $X_n$ such that $X_n$ is non-negative and $E[X_n | X_0] \leq \delta$, $\delta > 0$, for all $n \geq 1$, then by using Markov's inequality, for $a > 0$, can I say the following:
$$P\left(\sup_{n \geq 1} X_n \geq a | X_0\right) \leq \frac{\delta}{a} \ \ \ \ \ \ (1)$$
My reasoning is this:
From Markov's inequality, if we consider a stopped process $X^{*}_n$ with some stopping time $T=\inf{\{T \leq M : M \in \mathbb{N}, X_T \geq a\}}$, then we have
$$P\left(\max_{1 \leq n \leq M} X_n \geq a | X_0\right) = P(X_T \geq a | X_0) \leq \frac{E[X_T | X_0]}{a} \leq \frac{\delta}{a}$$
Now, by taking the limit $M \rightarrow \infty$, I can replace the $\max{}$ operator by $\sup{}$ to obtain (1). However, I'm not sure that I can just take the limiting operator when the process is not a (super/sub)martingale. So any leads in this regard would really help me. Particularly, is taking the limit operator right for me to do, and why?
Best Answer
No. I gave a counter-example in my first comment, here is another one: Take $\{X_n\}_{n=0}^{\infty}$ i.i.d. exponentially distributed with mean 1, so that $$E[X_n|X_0]=1 \quad \forall n \in \{1, 2, 3,...\}$$ Clearly $\sup_{n\geq 1} X_n = \infty$ with prob 1 and so for any $a$: $$ P\left[ \sup_{n\geq 1} X_n \geq a \vert X_0\right] = P\left[ \sup_{n\geq 1} X_n \geq a\right]= 1$$
Details: Fix $\delta = 1, a=42$. Fix $M$ as a large positive integer. Define $K$ as the first positive integer $n$ such that $X_n\geq 42$ (define $K=\infty$ if there is no such positive integer). Define $T=\min\{M, K\}$.
We cannot claim $E[X_T|X_0]\leq 1$. Indeed:
\begin{align} E[X_T|X_0] &= E[X_T] \\ &= E[X_T|X_T\geq 42]P[X_T\geq 42] + E[X_T|X_T<42]P[X_T<42]\\ &\geq E[X_T|X_T\geq 42]P[X_T\geq 42] \\ &= (42+1)P[X_T\geq 42] \end{align} Since $\lim_{M\rightarrow\infty} P[X_T\geq 42] = 1$ we get $$ \lim_{M\rightarrow\infty} E[X_T|X_0] \geq 43$$