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