How to check these sequences generated by i.i.d random variables are martingales? Let $\{Y_n\}_{n\geq 1}$ be a sequence of independent, identically distributed random variables.
$P(Y_i=1)=P(Y_i=-1)=\frac12$
Set $S_0=0$ and $S_n=Y_1+...+Y_n$ if $n\geq 1$
I want to check if the following sequences are martingales.
   $$M_n^{(1)}=\frac{e^{\theta S_n}}{(\cosh{\theta})^n}$$
$$M_n^{(2)}=\displaystyle\sum_{k=1}^n sign(S_{k-1})Y_k, n\geq 1,M_0^{(2)}=0$$
$$M_n^{(3)}=S_n^2 -n$$
I have no idea how to answer these questions. I think to answer these questions, knowledge of moment generating functions and its  natural logarithm   Cumulant is necessary.
If any member knows the correct answers to these questions may reply with correct answers.
 A: For $M_n^{(1)}$ we must first check that the process is integrable. Note that $\theta\mapsto |e^\theta+e^{-\theta}|$ is an even function which is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$ and hence attains its minimum $2$ at $\theta=0$. Now,
\begin{align}
\mathbb E[|M_n^{(1)}|] &= \mathbb E\left[\left|\frac{2e^{\theta S_n}}{(e^\theta+e^{-\theta})^n} \right|\right]\\ &\leqslant \sup_{\theta\in\mathbb R}\frac2{|e^\theta+e^{-\theta}|^n} \mathbb E[e^{\theta S_n}]\\
&\leqslant 2^{-n} e^{n|\theta|}\\
&<\infty.
\end{align}
As for the martingale property, for each positive integer $n$ we have
\begin{align}
\mathbb E[M_{n+1}^{(1)}] &= \mathbb E\left[\frac{2e^{\theta S_{n+1}}}{e^\theta + e^{-\theta}}\mid\mathcal F_n \right]\\ &= \frac 2{e^\theta+e^{-\theta}} \mathbb E[e^{\theta S_n}e^{\theta Y_{n+1}}\mid\mathcal F_n]\\
&=\frac{2e^{\theta S_n}}{e^\theta + e^{-\theta}}\mathbb E[e^{\theta Y_{n+1}}]\\
&=\frac{2e^{\theta S_n}}{e^\theta + e^{-\theta}}\\
&= M_n^{(1)}.
\end{align}
For $M_n^{(2)}$ integrability follows readily from the bound
$$
\mathbb E[|M_n^{(2)}] = \mathbb E\left[\left|\sum_{k=1}^n\mathrm{sgn}(S_{k-1})Y_k \right|\right] \leqslant \sum_{k=1}^n 1\cdot1 = n.
$$
As for the martingale property,
\begin{align}
\mathbb E[M_{n+1}^{(2)}\mid\mathcal F_n] &= \mathbb E\left[\sum_{k=1}^{n+1}\mathrm{Sgn}(S_{k-1}Y_k\mid\mathcal F_n \right]\\
&=\mathbb E[\mathrm{Sgn}(S_n)Y_{n+1}\mid\mathcal F_n] + \mathbb E\left[\sum_{k=1}^n\mathrm{Sgn}(S_{k-1})Y_k\mid\mathcal F_n \right]\\
&= \mathrm{Sgn}(S_n)\mathbb E[Y_{n+1}] +\mathbb E[M_n^{(2)}\mid\mathcal F_n]\\
&= 0 + M_n^{(2)}\\
&= M_n^{(2)}.
\end{align}
For $M_n^{(3)}$, integrability follows from
$$
\mathbb E[|M_n^{(3)}|] = \mathbb E[|S_n^2-n|] \leqslant \mathbb E[S_n^2] + n = 2n<\infty.
$$
As for the martingale condition,
\begin{align}
\mathbb E[M_{n+1}^{(3)}\mid\mathcal F_n] &= \mathbb E[S_{n+1}^2-(n+1)\mid\mathcal F_n]\\
&=\mathbb E[(S_n+Y_{n+1})^2 \mid\mathcal F_n] - (n+1)\\
&= \mathbb E[S_n^2 + 2S_nY_{n+1} + Y_{n+1}^2\mid\mathcal F_n] - (n+1)\\
&= S_n^2 + 2S_n\mathbb E[Y_{n+1}] + \mathbb E[Y_{n+1}^2] -(n+1)\\
&= S_n^2 -n\\
&= M_n^{(3)}.
\end{align}
A: $M_n^{(1)}$ is a martingale. It satisfies the conditional expectation:
$E\left(M_{n+1}^{(1)} | S_n \right)= M_n^{(1)}$
