# Martingale transform question

I was reading my notes and I was having trouble understanding theorem 4.3 below. I understand essentially what it is saying, but to me its simplying stating something rather intuitive? That given $\{C_n\}$ and $\{X_n\}$ then $\{C \bullet X\}$ is also a martingale. But how does it relate to the fact that "you cant beat the system"?

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Suppose you decided to buy some stock. Right after the $(n-1)$-th step, you bought $C_n$ stocks with value $X_{n-1}$ per unit stock. Now at $n$-th step, you sells the stocks. Your total winning at this trade is then $C_n (X_n - X_{n-1})$ So repeating this process, your total winning up to $n$-th step is $(C \bullet X)_n$. That it is a martingale null at 0 particularly implies that then we always have $$\mathbb{E}[(C \bullet X)_n] = \mathbb[(C \bullet X)_0]= 0.$$ So in average, it is impossible for you to win against the system, unless you have infinite wealth. – Sangchul Lee Mar 17 '12 at 18:11
Thanks @sos440, the only thing I'm not clear with is, how exactly is it a martingale null at $0$? Also, if my winnings is $C_n(X_n - X_{n-1})$ is positive at every round, wouldnt I make money? Sorry if im getting confused, would appreciate it if you could elaborate further especially on the martingale null part. – Heijden Mar 17 '12 at 18:57
Just observe that $\mathbb{E}[(C \bullet X)_1 | \mathcal{F}_0] = \mathbb{E}[C_1 (X_1 - X_0) | \mathcal{F}_0] = C_1 (\mathbb{E}[X_1 | \mathcal{F}_0] - X_0) = 0$. So it is consistent that ths defining sum vanishes when $n = 0$. (The summation symbol $\sum_{a \in A}$ is defined to yield zero if $A$ is empty.) And of course, you can earn some money in some particular instances, but what it tells is that the expected gain and expected loass balance exactly, so no net expected winning. We use probability to forecast something, so we have to consider every possible case, not a particluar one. – Sangchul Lee Mar 17 '12 at 20:07
thanks @sos440 ! Most appreciated, I understand it now :) – Heijden Mar 18 '12 at 23:17