2

Consider a non-negative stochastic process $X_t$ defined on probability space $(\Omega, \mathcal{F},\mathbb{P})$. Assume we have

\begin{align} \underset{t \rightarrow \infty}{\lim}\mathbb{E}[X_t]=0 \end{align}

Does this imply $X_t \overset{a.s.}{\to} 0$?

Fatou's lemma implies that if $X_t$ converges almost surely, it must converge to zero. (Connections between almost sure convergence and convergence in mean) Moreover, a zero mean non-negative random variable is almost surely zero. (A nonnegative random variable has zero expectation if and only if it is zero almost surely) However, I am not sure how to handle this question.

amWhy
  • 209,954
fes
  • 1,649
  • 2
    with the assumption of non-negativity, your question reduces to (a special case of) "Does $L^1$ convergence imply convergence a.s.?" See here for a counter-example: https://math.stackexchange.com/questions/138043/does-convergence-in-lp-implies-convergence-almost-everywhere – Rhys Steele Jun 05 '17 at 08:22
  • @nobody Do these counterexamples extend to my case? Looks a bit like that. – fes Jun 05 '17 at 08:29
  • Yes. (up to a trivial adaptation if you want a continuous time process) – Rhys Steele Jun 05 '17 at 08:31

1 Answers1

6

Consider a sequence of independent r.v.s. $\{X_t\}_{t\in \mathbb{N}}$ s.t. $$ \mathsf{P}(X_t=\sqrt{t})=t^{-1} \quad\text{and}\quad \mathsf{P}(X_t=0)=1-t^{-1}. $$ Then $$ \mathsf{E}X_t=\frac{1}{\sqrt{t}}\to 0 \quad\text{as }t\to\infty. $$ However, $\mathsf{P}(X_t\ge 1 \text{ i.o.})=1$.