Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm a little bewildered on how to get the following proved...

Suppose we make the following assumptions:

Let $f_n$ be a measurable function on $\mathbb{R}^n$. Let $Z_1,Z_2,\cdots$ be independant random variables and $\mathcal{F}_n = \sigma(Z_1,\cdots, Z_n)$. Let $X_n = f_n(Z_1,\cdots, Z_n)$ and assume that $\mathbb{E}|X_n|<\infty$ and $\mathbb{E}f_n(z_1,\cdots, z_{n-1}, Z_n) = f_{n-1}(z_1,\cdots, z_{n-1})$ for all $n$.

Apparantly then $X_n$ defines a martingale...I just feel like i'm missing some basic understanding about martingales to see why $\mathbb{E}[X_{n+1}|\mathcal{F}_n] = X_n$

holds in this case. Would someone be so nice to enlighten m

share|cite|improve this question
It follows from the assumption on $\Bbb Ef_{n+1}(z_1,..,z_n,Z_{n+1})$. – Berci Oct 23 '12 at 15:42
up vote 3 down vote accepted

$\mathbb{E}f_n(z_1,\cdots, z_{n-1}, Z_n) = f_{n-1}(z_1,\cdots, z_{n-1})$ holds for all $(z_1,\cdots, z_{n-1}) \in \mathbb{R}^{n-1}$,so for all $\omega \in \Omega$,

$\mathbb{E}[X_n|\mathcal{F}_{n-1}](\omega) = \mathbb{E}f_n(Z_1(\omega),\cdots, Z_{n-1}(\omega), Z_n) = f_{n-1}(Z_1(\omega),\cdots, Z_{n-1}(\omega)) = X_{n-1}(\omega)$

which means $\mathbb{E}[X_n|\mathcal{F}_n] = X_{n-1} $ a.s.

share|cite|improve this answer
Thanks, this is very enlighting indeed. – DinkyDoe Oct 23 '12 at 16:32
Je heet DinkyDoe, tuurlijk... – BallzofFury Oct 23 '12 at 20:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.