Can somebody help me with determining whether $Z_{n}=\log(2n+S_{n})$ is a martingale, supermartingale or submartingale with $S_{n}=\sum_{i=1}^{n}X_{i}$ and the are i.i.d. random variables with $P(X_i = 1) = P(X_i = -1) = \frac{1}{2}$.
What I have so far is that $E[log(2n+S_{n})|F_{n-1}]\leq \log E[2n+S_{n}|\mathcal{F}_{n-1}]$ by Jensen but now I am stuck in working to $Z_{n-1}$.
For each $n$ we have $2n+S_n>0$ a.s. so \begin{align} \mathbb E[|Z_n|] &= \mathbb E[\log(2n+S_n)]\\ &\leqslant \log(\mathbb E[2n+S_n])\\ &= \log(2n + \mathbb E[S_n])\\ &= \log2 + \log n<\infty. \end{align} Further, \begin{align} Z_{n+1} &= \log(2(n+1) + S_{n+1})\\ &= \log(2 + X_{n+1} + 2n + S_n)\\ &= \log(2 + X_{n+1} + e^{Z_n}), \end{align} so \begin{align} \mathbb E[Z_{n+1}\mid\mathcal F_n] &= \mathbb E[\log(2 + X_{n+1} + e^{Z_n})\mid\mathcal F_n]\\ &= \mathbb E[\log(2 + X_{n+1} + e^{Z_n})\mid\mathcal F_n, X_{n+1}=-1]\mathbb P(X_{n+1}=-1) + \mathbb E[\log(2 + X_{n+1} + e^{Z_n})\mid\mathcal F_n, X_{n+1}=1]\mathbb P(X_{n+1}=1)\\ &= \frac12\mathbb E[\log(1+e^{Z_n})\mid\mathcal F_n] + \frac12\mathbb E[\log(3+e^{Z_n})\mid\mathcal F_n]\\ &= \frac12\mathbb \log(1+e^{Z_n}) + \frac12\log(3+e^{Z_n})\\ &\geqslant \frac12Z_n + \frac12Z_n\\ &=Z_n, \end{align} so that $Z_n$ is a submartingale.