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}

  • 1
    $\begingroup$ Yes, it is correct. $\endgroup$
    – saz
    Commented Dec 29, 2013 at 14:40

1 Answer 1


To elaborate on Saz's comment slightly:

Yes the expectation of a supermartingale (resp. sub) is decreasing (resp. increasing), for precisely the reason you use.

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!

  • $\begingroup$ Why does $S_n\to -\infty$ when $X_n\ge0$ for all $n$ (hence $S_n\ge0$)? $\endgroup$
    – Phil-ZXX
    Commented Dec 29, 2013 at 15:47
  • $\begingroup$ @Tom that is a typo $X_n=-1$ with probability $1-1/n^2$. $\endgroup$
    – Lost1
    Commented Dec 29, 2013 at 15:53

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