# If $\{X_n\}$ is a martingale, then $E[X_n-X_{n-1}]=0$

Apparently this should be quite simple, but I have been trying for a while and can't seem to get this. Let $\{X_n\}$ be a martingale, then we have: $$E[X_n-X_{n-1}]=0$$ According to some notes I found this is a simple consequence of the two facts:

1. $E[X_{n+1}-X_n|\mathcal{F}_n]=0$, where $\mathcal{F_n}$ is the corresponding filtration.
2. $E[E[X|Y]]=E[X]$ for $X,Y$ random variables.

I understand both of these, but can't see how to put them together. My main confusion is how to relate them as 1. deals with condition expectation values with respect to sigma algebras, while 2. deals with expectations with respect to random variables.

• "My main confusion is how to relate them as 1. deals with condition expectation values with respect to sigma algebras, while 2. deals with expectations with respect to random variables." These are the same since $E(X|Y)$ is defined as $E(X|\sigma(Y))$ where $\sigma(Y)$ denotes the sigma-algebra generated by $Y$.
– Did
Commented Oct 11, 2014 at 10:13
• Thanks Did, that was helpful. Commented Oct 11, 2014 at 10:43

Note that the equality

$$\mathbb{E}(\mathbb{E}(X \mid Y)) = X$$

$$\mathbb{E}(\mathbb{E}(X \mid Y))= \mathbb{E}X.$$
In fact, this holds not only for $\sigma$-algebras generated by a random variable, but for any $\sigma$-algebra, i.e.
$$\mathbb{E}(\mathbb{E}(X \mid \mathcal{F})) = \mathbb{E}X$$
for any $\sigma$-algebra $\mathcal{F}$. Consequently, we find
$$\mathbb{E}(X_n-X_{n-1}) = \mathbb{E}( \mathbb{E}(X_n-X_{n-1} \mid \mathcal{F}_{n-1})) = 0.$$