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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.

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If $X$ is integrable and independent of $Y$ then $E[X\mid Y]=E(X)$. –  Davide Giraudo Dec 16 '12 at 21:57
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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? –  Mejjem Dec 16 '12 at 22:09

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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 $1_D$?

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Thanks helped alot! –  Mejjem Dec 16 '12 at 23:17

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