Suppose I have a nonnegative stochastic process $X_t$. Furthermore, suppose the following is true:
$$\limsup_t \frac{1}{t}E\left[ \log X(t)\right] \leq a < 0$$ for some constant $a \in \mathbb{R}$. Due to the nonegativity of $X(t)$, does it follow that:
$$\lim_{t \rightarrow \infty} E[X(t)] = 0$$
I would think it should since if $E[X(t)]$ converges to something else (say $c \in \mathbb{R})$, then $\limsup_t \frac{1}{t}E[\log X(t)]$ converges to $0$ which violates the assumption of it being strictly less than $0$. Or am I thinking of this wrong? I would think there is a way to argue that $E[X(t)] \rightarrow 0$. The reason why is because a paper I'm reading has the following for a nonnegative stochastic process $Y_t$:
$$\limsup_t \frac{1}{t}\log Y_t \Longrightarrow lim_t Y_t = 0 \textit{a.s.}$$
But now I have expectations instead.
Let $P\{X_t=e^{-4t}\}=P\{X_t=e^{2t}\}=0.5$. Then
$$\frac{1}{t}\mathbb{E}[\ln X_t]=\frac{1}{t}(-t)=-1$$
But $$\mathbb{E}X_t=\frac{1}{2}\left(e^{-4t}+e^{2t}\right)\rightarrow \infty\text{ as }t\rightarrow\infty$$