1
$\begingroup$

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.

$\endgroup$
  • 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
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.