[Math] Prove Symmetric Random Walk is a Martingale

filtrationsmartingalesrandom walkstochastic-processes

I'm given a probability space of ($\Omega$, $\mathcal{F}$, $\mathbb{P}$) and am asked to look into a symmetric random walk with its n-step defined as

$$
X_n = \Bigg\{
\begin{matrix}
1 & \text{with probability 1/2} \\
-1 & \text{with probability 1/2}
\end{matrix}
$$

where $X_n$ and $X_m$ are independent for all $n \ne m$. Setting
$0 = c_0 < c_1 < … <c_t$, let

$$
S_{s_i} = \sum_{n=1}^{c_i}X_n, \text{ i = 1, 2, …, t}
$$

where $S_0 = 0$.

As the title states, I want to show that this is a martingale. I know that a stochastic process $X=\{X_n\}$ is a martingale on filtered probability space ($\Omega$, $\mathcal{F}$, $\mathbb{F}=\{\mathcal{F}_n\}$,$\mathbb{P}$) if

  1. $X$ is adapted
  2. $\mathbb{E}|X_n|<\infty$ for all n
  3. $\mathbb{E}[X_{n+1}|\mathcal{F}_n]=X_n$ for all n

I've seen on multiple sites, including in a post here, that in order to prove the symmetric random walk is a martingale only the third bullet is proven. On most of the sites I've been to they simply say 1 and 2 are obvious or inherent, while on the post I linked above it states that only 3 must be shown when we

Assume that all variables are integrable and that the filtration we are working with is the natural filtration

My question is if this is an assumption that can be reasonably made here and if it is then how these things automatically lead to 1 and 2 being definite. Apologies is this is something obvious since I am just starting out and thank you for your time.

Best Answer

For the first point, notice that your probability space doesn't start with an inherent filtration. In this case, the natural filtration to work with is the one generated by the process $S_i$. That is $\mathcal{F}_i = \sigma(S_j : j \leq i)$. This is the smallest $\sigma$-algebra such that $S_j$ is $\mathcal{F}_i$-measurable for each $j \leq i$ and in particular it is part of the definition of $\mathcal{F}_i$ that $S_i$ is an adapted process.

The second point is obvious since, for each $i$, $S_i$ is a bounded random variable. We have $$|S_i| = |\sum^{c_i} X_n| \leq \sum^{c_i} |X_n| = c_i$$ and so $\mathbb{E}[|S_i|] \leq c_i < \infty$.

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