If a random process $\{Y_t\}_{t\ge0}$ whose moment of first order is uniformly bounded, does $Y_t/t$ converges to $0$ a.s.?

51 Views Asked by Bumbble Comm At 29 Apr 2026 - 11:08

I have problem in reading the paper in the proof Theorem 3.6 (in Page 14):

I am puzzled here: it says $X(t)/t$ converges a.s. to $v$, and $X(t)-R(t)$ is tight, then concluding that $R(t)/t$ converges a.s. to $v$.

I think the author want to say that he concluded $(X(t)-R(t))/t \rightarrow 0$ a.s. by the tightness, but I failed.

There is one counterexample for uniformly bounded first order moment process $Y_n$ whose time is discrete and $Y_n/n$ doesn't converges to $0$ a.s.: $Y_n$ are mutually independent, and $P(Y_n=n)=1/n, P(Y_n=0)=1-1/n$.

Therefore, I think I need some property of the trajectory to have it, but the proof of the paper is really not clear to me.
However, it is easy to see uniformly finite first moment obtains the convergence to $0$ in $L^1$.

It might cost you some time to read the paper, and it is very grateful for your help!