Suppose, $G$ is a finite group, $A$ is its finite symmetric set of generators. Let's define $\lVert g \rVert_A$ as the Cayley distance (in respect to $A$) from $g$ to $e$.
Now, suppose $\{X_n\}_{n=1}^\infty$ are i.i.d. random elements of $G$. Let's define a random walk on $G$ with step $X_i$ as the sequence of random elements $\{Z_n\}_{n = 0}^\infty$ defined by the recurrence:
$Z_n = \begin{cases} e & \quad n = 0 \\ Z_{n-1}X_n & \quad n > 0 \end{cases}$
It is known, that for the random walks on groups there is a following "Law of Large Numbers":
If $E \lVert X_i \rVert_A < \infty$ then $\exists \alpha \in \mathbb{R}$ such that $\frac{\lVert Z_n \rVert_A}{n} \overset{p}{\underset{n \to \infty}{\longrightarrow}} \alpha$
Here $\overset{p}{\underset{n \to \infty}{\longrightarrow}}$ stands for convergence in probability.
Note, that the $\alpha$ in this theorem (unlike the classical Law of Large Numbers) is generally not equal to $E \lVert X_i \rVert_A$. For example, $\alpha$ is always $0$ for finite $G$. However, the inequality $\alpha \leq E \lVert X_i \rVert_A$ always holds as a consequence of triangle inequality for Cayley metric.
My question is: Is the inverse statement also true? Namely:
Does $\exists \alpha \in \mathbb{R}$ such that $\frac{\lVert Z_n \rVert_A}{n} \overset{p}{\underset{n \to \infty}{\longrightarrow}} \alpha$ always imply $E \lVert X_i \rVert_A < \infty$?