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.

  • 2
    $\begingroup$ 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: math.stackexchange.com/questions/138043/… $\endgroup$ – Rhys Steele Jun 5 '17 at 8:22
  • $\begingroup$ @nobody Do these counterexamples extend to my case? Looks a bit like that. $\endgroup$ – fesman Jun 5 '17 at 8:29
  • $\begingroup$ Yes. (up to a trivial adaptation if you want a continuous time process) $\endgroup$ – Rhys Steele Jun 5 '17 at 8:31

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, $\mathsf{P}(X_t\ge 1 \text{ i.o.})=1$.

  • $\begingroup$ What does i.o. mean? $\endgroup$ – fesman Jun 5 '17 at 8:54
  • $\begingroup$ "infinitely often". $\endgroup$ – d.k.o. Jun 5 '17 at 8:55
  • $\begingroup$ Why $\mathbb P (X_t \geq 1 \text{ i.o.}) = 1$? $\endgroup$ – Falrach Oct 16 '18 at 9:29
  • 2
    $\begingroup$ Ah, Borel-Cantelli. Nevermind. $\endgroup$ – Falrach Oct 16 '18 at 9:32

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