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I'm struggling with the concept of conditional expectation, when the sigma algebra on which it is conditioned isn't generated by a partition.

If $(\Omega,\mathcal{F},P)$ is a probability field such that $\mathcal{F}$ is generated by a partition $\Lambda_n$.

Then we know that:

$E[X|\mathcal{F}]$ = $E_1$$[X|\mathcal{F}]$ $I(\omega \in \Lambda_1)$ +$E_2$$[X|\mathcal{F}]$ $I(\omega \in \Lambda_2)$ + $E_3$$[X|\mathcal{F}]$ $I(\omega \in \Lambda_3)$ + .....

Where $E_i[.]$ is the expectation calculated as per the conditional probability $P(.|\Lambda_i)$

Hence when $\omega$ is in $\Lambda_i$ the conditional expectation gives the expectation of random variable X given that the observed event is $\Lambda_i$ and hence use the modified conditional probability rather than the original one. However this interpretation is only valid as long as the conditioning sigma algebra is generated by a partition. Is there a similar interpretation for a general case?

i.e what will it physically represent?

Any help will be greatly appreciated!

Thanks!

Best,
Adwait

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It looks to me that you are assuming that $\mathscr{F}$ is generated by a partition, not $\mathscr{A}$. If $\mathscr{A}$ where generated by a partition and $X$ were $\mathscr{A}$ measurable, then $X$ would be constant on each of the parts $\Lambda_k$. –  Liviu Nicolaescu Apr 29 '13 at 9:26
    
Sorry for the typo! Replaced $\mathcal{A}$ with $\mathcal{F}$ –  Adwaitvedant Mathkar Apr 29 '13 at 9:32
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I'm not sure, if this is the right place to ask this question as it isn't really research level (see FAQ), please ask at math.stackexchange.com instead. [Did you read en.wikipedia.org/wiki/Conditional_expectation?] –  j.p. Apr 29 '13 at 9:45
    
Yup have read the wikipedia article. But it doesn't give an intuitive explanation for general sigma algebras. Have posted it here since I didn't find any (satisfactory) explanation in a graduate level text book. –  Adwaitvedant Mathkar Apr 29 '13 at 9:59
    
Apparently simultaneously crossposted at MSE. –  Did Apr 29 '13 at 12:33
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1 Answer

The best intuition that I have for conditional expectation is that it's a projection. Also, try thinking about the conditional expectation as a Radon Nikodym derivative.

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