Sufficient conditions for divergence of continuous-time stochastic process Let $X_t$ be a real-valued stochastic process (if it helps, we can assume that it is a component of a multivariate diffusion or jump-diffusion process).
I'm looking for sufficient conditions under which $X_t$ diverges in probability, i.e.
$$\forall a>0: \quad P(|X_t|<a)\stackrel{t\rightarrow\infty}{\rightarrow} 0.$$
It is relatively easy to come up with necessary conditions. 
For example, if $X_t$ diverges in probability, then its moments of even order diverge.
However, the divergence of even-ordered moments is not sufficient.
Indeed, a simple counterexample is $X_t=t X_0$, where
$$X_0=\begin{cases}
0& \text{with probability 1/2}, \\
1& \text{with probability 1/4}, \\
-1& \text{with probability 1/4}.
\end{cases}$$
For this process, for $k\in\mathbb{N}$ we have $\mathbb{E}[X_t^{2n}]=t^{2n}/2\rightarrow\infty$ as $t\rightarrow\infty$ but at the same time we have $\forall a>0$ that $P(|X_t|<a)\stackrel{t\rightarrow\infty}{\rightarrow} 1/2$.
Are there any standard theorems that give sufficient conditions?
 A: It appears you are looking for sufficient conditions in terms of expectations of functions of $X_t$. If so, the following may be offered: 
\begin{equation}
\text{$|X_t|$ diverges to $\infty$ in probability iff $E\frac1{1+|X_t|}\to0$;} \tag{*} 
\end{equation}
here everywhere the convergence is as $t\to\infty$. 
Indeed, $|X_t|$ diverges to $\infty$ in probability iff $Y_t:=1/|X_t|$ converges to $0$ in probability, that is, iff $P(Y_t>b)\to0$ for each real $b>0$. Let $f(y):=\frac y{1+y}$. 
Then $I\{y>b\}\le f(y)/f(b)$ for $b>0$ and $y\ge0$, where $I\{\cdot\}$ is the indicator function. So, $P(Y_t>b)\le Ef(Y_t)/f(b)=E\frac1{1+|X_t|}/f(b)\to0$ if $E\frac1{1+|X_t|}\to0$, which proves the "if" part of (*). (In fact, here we use the Markov inequality.) 
Using inequality $f(y)\le I\{y>b\}+f(b)$ for $b>0$ and $y\ge0$, we see that 
\begin{equation}
E\frac1{1+|X_t|}= Ef(Y_t)\le P(Y_t>b) +f(b)\to f(b)
\end{equation}
for each real $b>0$ if $Y_t$ converges to $0$ in probability. So, the "only if" part of (*) follows as well, by letting $b\downarrow0$. (This part also follows by the Lebesgue dominated convergence theorem.) 
It should also be clear that, in (*), one can replace $E\frac1{1+|X_t|}$ by $E\frac1{1+X_t^2}$ or $E\exp(-X_t^2/2)$ or, more generally, by $Eg(|X_t|)$, where $g$ is (say) any continuous function which is strictly decreasing to $0$ on $[0,\infty)$. 
In particular, this allows one to express the condition that $|X_t|$ diverges to $\infty$ in terms of the characteristic functions $c_t(u)=Ee^{iuX_t}$ of $X_t$ -- say, by writing $e^{-x^2/2}=\int_{-\infty}^\infty e^{iux}\frac1{\sqrt{2\pi}}e^{-u^2/2}\,du$ and $\frac2{1+x^2}=\int_{-\infty}^\infty e^{iux}e^{-|u|}\,du$ for real $x$, whence 
$E\exp(-X_t^2/2)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty c_t(u)e^{-u^2/2}\,du$ and 
$E\frac1{1+X_t^2}=\frac12\,\int_{-\infty}^\infty c_t(u)e^{-|u|}\,du$, and so, 
\begin{equation}
\text{$|X_t|\to\infty$ in probability iff $\int_{-\infty}^\infty c_t(u)e^{-u^2/2}\,du\to0$ iff $\int_{-\infty}^\infty c_t(u)e^{-|u|}\,du\to0$.}  
\end{equation}
