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

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 ?

• Well, what have you tried for the other one condition?
– user9464
Dec 20, 2016 at 20:29
• The other one condition I understand it: "If I have information until time s, the best guess for time t is the value in time s". But mathematically I struggle to "play" with it. Dec 20, 2016 at 20:31
• Do you know what $E(X_t|F_s)=X_s$ means mathematically? Also, you might want to give a definition for $F_s$.
– user9464
Dec 20, 2016 at 20:33
• What do you mean by mathematically ? I understand that is the expected value of the r.v. X at time t given the filtration, the information we collected on the values of X up to time s Dec 20, 2016 at 20:35
• We learn that the expectation of x is the integral of x * f(x) dx but I don't see the utility here Dec 20, 2016 at 20:40

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

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.

• Thanks for the hint $Z_{n+1} = X_{n+1}*Z_n$. It makes sense to me now. You take out $Z_n$ out of the expected value, following $E(aX)=a*E(X)$, but why can we do that, if $Z_n$ is itself a product of $X_n$ ? Dec 20, 2016 at 20:57
• Thanks for the edit. I knew about the independence but not the first one. But does it mean I could have written $X_{n+1}*E(Z_n|F_n)$ and it would have been mathematically correct ? Dec 20, 2016 at 21:11
• I don't understand your question. Would you rephrase it?
– user9464
Dec 20, 2016 at 21:14
• The reason for being able to take out $Z_n$ in $E(X_{n+1}Z_n|F_n)$ is similar to $E(XY|X)=XE(Y|X)$.
– user9464
Dec 20, 2016 at 21:16
• Close. Though one should have $Z_n$ being $F_n$-measurable instead of $Z_n=F_n$ since $Z_n$ is a random variable both $F_n$ is the "information" given by $X_1,\cdots,X_n$.
– user9464
Dec 20, 2016 at 23:13

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.