Note that $a$ has a mean of 0.
My approach:
$$X_t=X_{t-1}+a_t$$ $$E[X_{t+1}\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}+2a\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]+E[2a\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]+0$$ $$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]$$ $$=X_{t-1}$$ Am I doing something wrong here? shouldn't the end product be $X_t$?
\begin{align} E[X_{t+1} \mid X_1, \ldots, X_t] &= E[X_t + a_{t+1} \mid X_1, \ldots, X_t] \\ &= X_t + E[a_{t+1} \mid X_1, \ldots, X_t] \\ &= X_t \end{align}
Let $\{X_t\}_{t\geq 1}$ be a sequence of independent random variables such that $\Pr\{X_t=1\}=\Pr\{X_t=-1\}=1/2$. Define $\mathscr{F_t}=\sigma(X_1,\dots,X_t)$ and $M_t=X_1+\dots+X_t$. We have (equalities between conditional expectations holding almost surely) $$ \mathbb{E}[M_{t+1}\mid\mathscr{F_t}] = \mathbb{E}[X_{t+1}+M_t\mid\mathscr{F_t}] = \mathbb{E}[X_{t+1}\mid\mathscr{F_t}] + \mathbb{E}[M_t\mid\mathscr{F_t}]. $$ But $X_{t+1}$ is independent of $\mathscr{F_t}$, therefore, $\mathbb{E}[X_{t+1}\mid\mathscr{F_t}]=\mathbb{E}[X_{t+1}]=0$. Also, $M_t$ is $\mathscr{F}_t$-measurable, hence, $\mathbb{E}[M_t\mid\mathscr{F_t}]=M_t$. It follows that $\mathbb{E}[M_{t+1}\mid\mathscr{F_t}]=M_t$ and $\{(M_t,\mathscr{F}_t)\}_{t\geq 1}$ is a martingale.
