Assume $X_0=0$. The process evolves with the following rules.
- $X_{t+1} = X_{t} +1 $ if $X_t<10$;
- If $X_t\ge 10$, $$ X_{t+1} = \begin{cases} X_t + 1 &\text{with probability } 1/2\\ X_t -3 &\text{with probability } 1/2 \end{cases} $$
Can we show that $E X_t$ is uniformly bounded by some constant?
It is a pretty standard exercise though if you see such stuff for the first time, it may be not obvious. Note that for sufficiently large $k$ ($>13$, or something like that), we have the equality $$ P(X_{t+1}\ge k)=\frac 12[P(X_t\ge k-1)+P(X_t\ge k+3)] $$ Now let $a>0$ be the positive root of $e^{a}+e^{-3a}=2$. Then we can easily show by induction that $P(X_t\ge k)\le Ce^{-ak}$ for all $k$ for some $C>0$. Indeed, for small $k$, it holds because the LHS is at most $1$ and the $RHS$ is at least $1$ if $C$ is large enough, and for large $k$ it follows from the induction assumption and the recurrence above (the base is obvious). Thus, we have an exponential tail and, in particular, all moments finite.
All we really use here is the constant drift towards the origin near infinity.