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Hi im trying to understand conditional expectation and conditional probability based on sigma algebras. Therefore an answer in that flavour would be most useful.

So in a physical sense I can see what it means to condition on an event $A$ (i.e we know if this event happens or not). Then what does it mean to condition on a collection of events i.e a sigma algebra? and what is the meaning of asking for

1) $E[X|X]$

2) to go further, $E[X|f(X)]$

p.s I know conditioning on a r.v is just conditioning of the sigma algebra generated by that r.v. So I guess im just asking what it means physically to condition on a collection of events (and why when you do so you get something random back, unless those events were just {$A,A^{c},\Omega$,empty set}

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2 Answers 2

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By definition $Y=\mathbb E [X|X]$ is a random variable that is $\sigma(X)$-measurable and such that $\mathbb E[YU] = \mathbb E [X U]$ for all random variable $U$ that is $\sigma(X)$-measurable. Observe that $X$ is both $\sigma(X)$-measurable and satisfy obviously $\mathbb E [XU] = \mathbb E[X U]$ for any $U$ that is $\sigma(X)$-measurable. This means that $\mathbb E[X|X]=X$.

You can actually apply the same argument to get that for any random variable $X$ and any Borel measurable function $f$, $\mathbb E[f(X)|X]=f(X)$.

Regarding intuition, conditioning on a random variable is conditioning on the sigma algebra generated by this random variable which means that you condition on the possible knowledge that the outcome of this random variable may give you. Conditional expectation on a sigma algebra represent the partial expectation dependent on the possible knowledge that the sigma algebra may give you, this means that it will associate to any event in the sigma algebra a value that would be the expectation we get when we condition on the fact that this event actually happened.

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  • $\begingroup$ your last sentence - 'that would be the expectation' expectation of what? and also 'condition on the fact that this event actually happened' condition on the sigma algebra generated by this event right? $\endgroup$
    – Monty
    Commented Oct 20, 2018 at 21:10
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For a "physical" meaning it is useful to look at conditional expectations as projections. Consider a square-integrable random variable $X$ living on $(\Omega,\mathcal{F},\mathsf{P})$. The defining property of the conditional expectation of $X$ w.r.t. a $\sigma$-field $\mathcal{G}\subset\mathcal{F}$ ($\tilde{X}=\mathsf{E}[X\mid\mathcal{G}]$) can be writen as $\mathsf{E}[(X-\tilde{X})Z]=0$ for all bounded, $\mathcal{G}$-measurable $Z$. This means that $X-\tilde{X}$ is orthogonal to $L^2(\Omega,\mathcal{G},\mathsf{P})$, that is $\tilde{X}$ is the projection of $X$ onto $L^2(\Omega,\mathcal{G},\mathsf{P})$.

Intuitively, $\mathsf{E}[X\mid \mathcal{G}]$ acts like a smoothing operator. When $X$ is $\mathcal{G}$-measurable, it returns $X$. As $\mathcal{G}$ becomes coarser, one gets a "smoother" version of $X$ until $\mathcal{G}$ becomes a trivial $\sigma$-field in which case $\mathsf{E}[X\mid\mathcal{G}]$ is just a constant.

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