Assume $(X_t)_{t \geq 0}$ is a real valued strong Markov process starting in $x \in \mathbb{R}$, and define $T_{R}(x):=\inf(s \geq 0 | X_s \notin (-R+x,R+x) \}$. Is the following statement true?
$ \mathbb{E}_x[T_R(x)] \asymp R^{\alpha} $ is equivalent to $ \mathbb{E}_x[(X_t-x)^2] \asymp t^{2/\alpha} $ for a $\alpha>0$.
Here $\asymp $ means that here exists $c_1,c_2,t_0>0$ such that $ c_2 t^{2/\alpha} \leq \mathbb{E}_x[(X_t-x)^2] \leq c_1 t^{2/\alpha}. $ for $t>t_0$.
For example for the Brownian Motion the statement is true for $\alpha=2$ because $\mathbb E_x[T_{r}(x)]=R^2$ and $\mathbb{E}_x[(B_t-x)^2]=t^{2/2}$.
Maybe this holds true for general stochastic processes?
$\textbf{EDIT}$
Set $R=t^{1/d_W}$ then one obtains \begin{align*} \mathbb{E}_x[T_R(x)] \asymp t \end{align*} So the expected time leaving the ball $(-t^{1/d_w}+x,-t^{1/d_w}+x)$ is linear in $t$. So in a heuristic way one would expect:
\begin{align*} \mathbb{E}_x[(X_t-x)^2] \asymp t^{2/d_w}. \end{align*}
Thanks in advance for any partial answer.