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!



  • $\begingroup$ 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 $\Lambda_k$. $\endgroup$ Apr 29, 2013 at 9:26
  • $\begingroup$ Sorry for the typo! Replaced $\mathcal{A}$ with $\mathcal{F}$ $\endgroup$ Apr 29, 2013 at 9:32
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    $\begingroup$ 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?] $\endgroup$
    – j.p.
    Apr 29, 2013 at 9:45
  • $\begingroup$ 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. $\endgroup$ Apr 29, 2013 at 9:59
  • $\begingroup$ Apparently simultaneously crossposted at MSE. $\endgroup$
    – Did
    Apr 29, 2013 at 12:33

1 Answer 1


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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