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I was wondering how the ordinary expectation value $E(X)$ is related to $E(X|\mathcal{F})$ where $\mathcal{F} \subset \mathcal{E}$ where the latter is supposed to be the sigma algebra on our probability space.

My first thought was that $E(X|\mathcal{E}) = E(X),$ but this is clearly wrong, as $X$ is $\mathcal{E}$ measurable and thus $E(X|\mathcal{E})= X.$

Then I noticed that by the total law of expectation $E(E(X|\mathcal{F}))=E(X)$ we have something like a tower property for the standard expectation value, but in the sense that $E(.)$ wins over any $E(.|\mathcal{F}).$ Spoken in terms of tower properties, this would mean that if $E(X)$ can be represented as a conditional expectation, it must be a maximally small sigma algebra. So my guess is $E(X|\{\emptyset, \Omega\})=E(X),$ is this true?

At first glance, it seems to fulfill all the properties of the conditional expectation, so my guess is yes, but I would like to have your confirmation.

Danielokw
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2 Answers2

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Yes, this is true. By the definition of the conditional expectation, $\operatorname E(X\mid\{\emptyset,\Omega\})$ is a random variable $Y$ that is $\{\emptyset,\Omega\}$-measurable and $\operatorname E[XI_A]=\operatorname E[YI_A]$ for every $A\in\{\emptyset,\Omega\}$. If $A=\Omega$, then $I_A=1$ and $\operatorname EX=\operatorname EY$, if $A=\emptyset$, then $I_\emptyset=0$. Hence, $Y$ is a random variable that is equal to $\operatorname EX$ almost surely. We can think about $\{\emptyset,\Omega\}$ as a $\sigma$-algebra that contains no information and the best way in some sense to predict the value of a random variable $X$ is $\operatorname EX$ if we have no additional information (see this answer).

Community
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Cm7F7Bb
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Yes, you are right. As constants are $\{\emptyset,\Omega\}$-measurable, the constant $E[X]$ is, moreover we have $$ \int_\Omega E[X]\, dP = \int_\Omega X\, dP $$ and likewise for the empty set, so $E[X]$ is the conditional expectation with respect to the trivial $\sigma$-algebra $\{\emptyset, \Omega\}$

martini
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