I am working on an exercise that asks me to prove the boundedness of the following stochastic process:
Let $X_i$ be iid r.v. s.t. $P(X=1)=P(X=-1)=1/2$
Let $S_n = X_1+\ldots+X_n, n\in \mathbb{N}$
Let $\alpha>0$ and let $Z_n = (\exp(S_n-\alpha n)), n\in \mathbb{N}$
How can I show that $|Z_{n+1}-Z_n|<=k, \forall n\in \mathbb{N}$?
I tried different approaches but I keep getting stuck. I do know that $S_n$ makes "jumps" of at most $1$ while $\alpha\cdot n$ makes jumps of size at most $\alpha$ but since the exponential is there I am not sure how this translates into a boundedness of the difference between the two exponentials. Passing to the logarithm doesn't help because of the minus sign in the middle (or at least I think). I also thought that $$\frac{\exp(S_{n+1})}{\exp(S_n)}\leq e \implies \exp(S_{n+1})\leq\exp(S_n+1).$$ There probably is some (in)equality that I don't know, could you please help me?
Notice that $|Z_{n+1} - Z_n| \in \{ |\exp(1-\alpha)-1|Z_n, |\exp(-1-\alpha)-1|Z_n\}$ by looking at the possible values of $X_{n+1}$. In particular, if $X_{n+1} = 1$ then $|Z_{n+1} - Z_n| = |\exp(1-\alpha)-1|Z_n$.
Suppose $\alpha < 1$ and there is a $k$ such that for every $n$, $|Z_{n+1} - Z_n| \leq k$ with probability $1$. Pick $n$ sufficiently large, so that $$\exp((1-\alpha)n) > \frac{k}{|\exp(1-\alpha)-1|}.$$ Then with probability $2^{-(n+1)}$, $X_k = 1$ for $k = 1, \dots, n+1$. In particular, with positive probability $S_{n+1} = n+1, X_{n+1} = 1$ and hence $$|Z_{n+1} - Z_n| = |\exp(1-\alpha)-1| \exp((1-\alpha)n) > k$$ so that the result you stated doesn't hold.
If $\alpha \geq 1$ then notice that $S_n \leq n$ implies that $Z_n = \exp(S_n - \alpha n) \leq \exp((1-\alpha)n) \leq 1$. Hence, in this case, $|Z_{n+1} - Z_n| \leq Z_{n+1} + Z_n \leq 2$.