# What is the expectation of $X$ given $X$

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}

• Commented Oct 20, 2018 at 16:12

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 $$f$$, $$\mathbb E[f(X)|X]=f(X)$$.
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.