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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?

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2 Answers 2

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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.

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  • $\begingroup$ From where does $X_{n+1} = Y_{n+1} + X_n$ come from? Sorry if this is silly $\endgroup$
    – Nipster
    Commented Apr 10, 2021 at 12:57
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    $\begingroup$ By definition, $X_{n+1}=Y_1+Y_2+\dots+Y_n+Y_{n+1}=X_n+Y_{n+1}$. $\endgroup$ Commented Apr 10, 2021 at 13:00
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$$E(X_t|\mathcal F_s)=E(\sum\limits_{i=1}^{t}Y_i|Y_1,Y_2,..,Y_s)$$ $$=\sum\limits_{i=1}^{s}Y_i+E(\sum\limits_{i=s+1}^{t}Y_i|Y_1,Y_2,..,Y_s)$$ $$=X_s+E(\sum\limits_{i=s+1}^{t}Y_i)=X_x+0=X_s.$$

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. $\mathcal F_s$.

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