# Showing a stochastic process is a (discrete) martingale

Given a sequence of iid random variables $$(Y_i)_{i=1}^\infty$$ on a probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$ such that $$\mathbb{E}|Y_i| < \infty$$ and $$\mathbb{E}Y_i = 0$$, consider the discrete time process given by $$X_0 := 0, \quad X_n = \sum_{i=1}^n Y_i, \quad n \in \{1,2,...\}.$$ and also the filtration given by $$\mathcal{F_n} = \sigma(Y_1, \dots, Y_n)$$ and show that $$(X_n)_{n=1}^\infty$$ is a (discrete) martingale with respect to $$(\mathcal{F_n})_{n=1}^\infty$$.

So far, in answering this question I believe that I have proven the first two properties of a martingale:

• We have that, since $$X_n$$ is the sum of $$\mathcal{F}$$-measurable variables, we know that it is too $$\mathcal{F}$$-measurable for each $$n$$. Hence it is adapted.

• $$\mathbb{E}|X_n| \leq \mathbb{E}(\sum_{i=1}^n| Y_i|) = \sum_{i=1}^n\mathbb{E}| Y_i| < \infty$$.

However, I am not sure about how one could go about proving the final property of a martingale, that $$\mathbb{E}(X_t | \mathcal{F_s}) = X_s$$ for all $$s \leq t$$.

Might anyone have any ways of demonstrating such a proof?

Since we are in discrete time, it is enough to show that $$\mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}]=X_n$$ a.s. for all $$n\in\mathbb{N}$$. Notice that $$X_{n+1}=Y_{n+1}+X_n$$, so that $$\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]=\mathbb{E}[Y_{n+1}\mid\mathcal{F}_n]+\mathbb{E}[X_n\mid\mathcal{F}_n]=\mathbb{E}[Y_{n+1}]+X_n=X_n\text{ a.s.},$$ since $$Y_{n+1}$$ and $$\mathcal{F}_n$$ are independent, and $$X_n$$ is $$\mathcal{F}_n$$-measurable.
• From where does $X_{n+1} = Y_{n+1} + X_n$ come from? Sorry if this is silly Commented Apr 10, 2021 at 12:57
• By definition, $X_{n+1}=Y_1+Y_2+\dots+Y_n+Y_{n+1}=X_n+Y_{n+1}$. Commented Apr 10, 2021 at 13:00
$$E(X_t|\mathcal F_s)=E(\sum\limits_{i=1}^{t}Y_i|Y_1,Y_2,..,Y_s)$$ $$=\sum\limits_{i=1}^{s}Y_i+E(\sum\limits_{i=s+1}^{t}Y_i|Y_1,Y_2,..,Y_s)$$ $$=X_s+E(\sum\limits_{i=s+1}^{t}Y_i)=X_x+0=X_s.$$
In the second equality I have used the fact that $$\sum\limits_{i=1}^{s}Y_i$$ is already measurable w.r.t $$\mathcal F_s$$. In the next equality I have used the independence of $$Y_{s+1},Y_{s+1},...,Y_t$$ w.r.t. $$\mathcal F_s$$.