Product of i.i.d r.v. is Martingale?

Question

I face a problem where I need to show that the new r.v. that I create is a Martingale.

I set that $(X_i)_{i=0,1,\dots}$ is a r.v. with $E(X_i) = 1$ and that $Z_n = X_1\cdot X_2\cdot \ldots\cdot X_n$, then I need to show that $(Z_n)_{n=0,1,\dots}$ is a martingale.

I know that to say that it is a martingale, it has to fulfill those conditions:

• $E[|X_t|] < \infty$ for all t,
• $E[X_t|F_s] = X_s$ for all s < t.

Can I say that $E(Z_n) = 1^n$ and then say that it is a martingale ?

Answer 1

What you want to show is for each $n$, $$ E(Z_{n+1}|\mathcal{F}_n)=Z_n $$ where $\mathcal{F}_n=\sigma(X_1,\cdots,X_n)$.

Now writing $Z_{n+1}=X_{n+1}Z_n$, you get $$ E(Z_{n+1}\mid\mathcal{F}_n)=E(X_{n+1}Z_n\mid\mathcal{F}_n)=Z_nE(X_{n+1}|\mathcal{F}_n)=Z_nE(X_{n+1})=Z_n. $$

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One has $E(X_{n+1}Z_n\mid\mathcal{F}_n)=Z_nE(X_{n+1}|\mathcal{F}_n)$ because $Z_n$ is $\mathcal{F}_n$-measurable.

One has $E(X_{n+1}|\mathcal{F}_n)=E(X_{n+1})$ because $X_{n+1}$ and $\mathcal{F}_n$ are independent.

You can check Corollary 3.7.3 in Oloffson and Adersson's book you mentioned in the comment for these two arguments if you don't have the measure theory background.

Answer 2

Consider $$\mathbb{E}(Z_t|Z_s=z) = \mathbb{E}(z\cdot\Pi_{i=s+1}^tX_i) = z\cdot\Pi_{i=s+1}^t\mathbb{E}(X_i) = z.$$ This is the idea behind a martingale without getting bogged down in the measure theory. In essence, things aren't getting worse or better on average. Thus in expectation the future is simply the current realization.