If we have a martingale $M_i$ and some constant $c$, is $cM_i$ also a martingale? I have checked the conditions of a martingale: (1) $\mathbb{E}[|cM_i|] < \infty$ and (2) $\mathbb{E}[cM_{i+1} | cM_i, \dots, cM_1] = \mathbb{E}[cM_{i}]$ (not very sure about this step because of the $c$'s in the conditional) so I am inclined to say it is. I just want some confirmation in case I missed something in my argument and also an explanation of the steps to get the result in (2) if it indeed is.
Also related but I was not able to prove or disprove, would $M_i + c$ be a martingale? Thanks!
Yes. Technically, the process $(M_i)$ is a martingale with respect to some filtration $(\mathcal{F}_i)$, i.e. martingale property is that $\mathbb{E}[M_{i+1} \mid \mathcal{F}_i] = M_i$. Usually the filtration is the canonical filtration, i.e. $\mathcal{F}_i = \sigma(M_j : j \leq i$) is the sigma-algebra (or "information") generated by the sequence so far. So in your case, the canonical filtration of $(c M_i)$ is the same as the original $(M_i)$ -- intuitively, the information generated by the two processes are the same. So for your point (2) to check, you can argue along the lines of $\mathbb{E}[c M_{i+1} \mid c M_1, \dots, c M_i] = c \mathbb{E}[ M_{i+1} \mid c M_1, \dots, c M_i ] = c \mathbb{E}[ M_{i+1} \mid M_1, \dots, M_i ] = c M_i$, for example.