# Does $X_t\geq 0$, $\underset{t \rightarrow \infty}{\lim}\mathbb{E}[X_t]=0$ imply $X_t \overset{a.s.}{\to} 0$?

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.

• 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/… – Rhys Steele Jun 5 '17 at 8:22
• @nobody Do these counterexamples extend to my case? Looks a bit like that. – fesman Jun 5 '17 at 8:29
• Yes. (up to a trivial adaptation if you want a continuous time process) – 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, $\mathsf{P}(X_t\ge 1 \text{ i.o.})=1$.
• Why $\mathbb P (X_t \geq 1 \text{ i.o.}) = 1$? – Falrach Oct 16 '18 at 9:29