# Expectation of (super/sub)-martingales

A stochastic process $X=(X_n)_{n\in\mathbb N}$ on the filtered probability space $(\Omega,\mathcal F,(\mathcal F_n)_{n\in\mathbb N},\mathbb P)$ that is a martingale has the property that $$\mathbb E[X_m\mid \mathcal F_n]=X_n \quad\quad\quad\forall m\ge n$$ ($\le$ for super-martingales and $\ge$ for sub-martingales). Now I was wondering if this implies that $$\mathbb E[X_m]=\mathbb E[X_n] \quad\quad\quad\forall m\ge n$$ (as before, $\le$ for super and $\ge$ for sub) as it would make sense intuitively.

Hence my question: are the following steps correct?

Let $m\ge n$ and set the relation $\sim$ to be $=$, $\le$ or $\ge$ if $X$ is a martingale, super-martingale or sub-martingale respectively. Then \begin{array} \mathbb E[X_m\mid \mathcal F_n]\sim X_n \quad\Rightarrow\quad \mathbb E[\mathbb E[X_m\mid \mathcal F_n]]\sim\mathbb E[X_n] \quad\Rightarrow\quad \mathbb E[X_m]\sim\mathbb E[X_n] \end{array}

• Yes, it is correct.
– saz
Dec 29, 2013 at 14:40

However, it is very important to notice this does not necessary tells us how the martingale behave. Here is an example of a SUB-martingale, which diverges to $-\infty$.
Let $X_n = n^3$ with probability $1/n^2$, $X_n= -1$ for $n\geq 1$ with probability $1-1/n^2$ and $X_n$ are all independent
Notice $E(X_n)>0$. $S_n = \sum_{i=1}^n X_n$ is submartingale. (let $S_0=0$), but by Borel-Cantelli lemma $X_n =-1$ eventually, so $S_n\rightarrow -\infty$ with probability 1, even though $E(S_n)\rightarrow\infty$. The moral of the story is that expectation can mislead!
• Why does $S_n\to -\infty$ when $X_n\ge0$ for all $n$ (hence $S_n\ge0$)? Dec 29, 2013 at 15:47
• @Tom that is a typo $X_n=-1$ with probability $1-1/n^2$. Dec 29, 2013 at 15:53