I have been trying to solve the following problem. Let $Z_1, Z_2, \ldots$ be an iid collection of random variables such that $P(Z_i=-1) = 3/4$ and $P(Z_i = 3) = 1/4$. Let $X_0 = 5$ and $X_n = 5 + \sum_{i=1}^n Z_i$ for $n \geq 1$. Finally let $T = \inf\{ n \geq 1: X_n = 0 \ \text{or} \ Z_n>0\}$. Find $E(X_T)$.
Now, it seems to me a reasonable strategy here would be to employ the optional stopping theorem and one of its conditions is a bounded stopping time $T$. In this particular case, I believe that $T \leq 5$ since
$$T = \inf\{ n \geq 1: X_n = 0 \ \text{or} \ Z_n>0\} \leq \inf\{ n \geq 1: X_n = 0\} \leq 5, $$
as if $Z_1 = Z_2 = \ldots = Z_5 = -1$ then $X_5 = 0$. Is my reasoning correct here or I do I need a different argument? Thank you in advance.
Your reasoning is not correct because the inequality $$ T_0:=\inf\left\{n\geqslant 1: X_n=0\right\}=\inf\left\{n\geqslant 1:\sum_{k=1}^n Z_k=-5\right\}\leqslant 5 $$ is not true, as by example $Z_1=3$ and $Z_k=-1$ for $k\in\{2,\ldots,9\}$ gives $T_0=9$. That is: the infimum above represent the first time that a random sequence $\{X_n\}_{n\in \mathbb N}$ took the zero value.
What the expression "$T$ is bounded" means is that $\Pr [T\leqslant n]=1$, for some $n\in \mathbb{N}$. In this case is easy to see that this is true for the original stopping time defined by the condition $X_n=0$ or $Z_n>0$ as this stopping time is bounded because the condition $Z_n<0$ can be true just finitely many times before $X_n=0$ (however $T_0$ is not bounded, as can be seen easily with the example above).