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 have a problem that can be resolved if i show that $$E(\varepsilon_k\mid\sigma(\varepsilon_1,\ldots,\varepsilon_{k-1}))=E(\varepsilon_k)$$ where $\varepsilon_1,\ldots,\varepsilon_k$ $\sim \mathcal{N}(0,1)$ and i know they are independent.

I dont know where to even start. Any proof or help would be great.

share|cite|improve this question
If $X$ is integrable and independent of $Y$ then $E[X\mid Y]=E(X)$. – Davide Giraudo Dec 16 '12 at 21:57
use the definition of conditional expectation. – Nate Eldredge Dec 16 '12 at 21:57
Davide Giraudo: Is that not what i want to show? @NateEldredge: The Definition of conditional expectation is the real-valued random variable satisfying that $$ \int_D X dP = \int_D E(X| \mathbb{D}) dP $$ for every $D \in \mathbb{D}$. Can you give me a Hint on how to use that? – Martin Dec 16 '12 at 22:09
up vote 1 down vote accepted

By the definition of conditional expectation, it suffices to show $\int_D \varepsilon_k \,dP = \int_D E[\varepsilon_k]\,dP$ for every $D \in \sigma(\varepsilon_1, \dots, \varepsilon_{k-1})$. In other words, to show that $E[1_D \varepsilon_k] = E[1_D] E[\varepsilon_k]$. But what do you know about the random variables $\varepsilon_k$ and $1_D$?

share|cite|improve this answer
Thanks helped alot! – Martin Dec 16 '12 at 23:17

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.