When I was reading a book about filtering, I saw a practice question stating as: Prove $\text{Var} X \geq |X - E(X|Y)|^2$.
I thought there might be a typo, so instead of proving this, I started to prove:
$$\text{Var} X \geq E|X - E(X|Y)|^2.$$
However, then I realized that there's no information about whether $EX^2$ is finite or not. Does this inequality still hold without the assumption that $EX^2 < \infty$?
For example, what if $\text{Var} X$ and $E|X - E(X|Y)|^2$ are both infinity?
This book talks about infinity stuff a little bit in the previous chapter, so I do believe the author have considered it. This may be an easy question, but I'm still confused.
UPDATE:
Maybe I can be more specific...
When $EX^2$ is finite, then it's easy to show this inequality holds. When $EX^2$ is infinite, then we only need to worry about two cases:
- $VarX$ and $E|X - E(X|Y)|^2$ are both infinite, then can we even compare two infinite value? (This may not be very important...)
- $VarX$ is finite, while $E|X - E(X|Y)|^2$ is infinite.
We have to show that the second case is impossible. Then the proof is complete. So how to show that the second case is impossible?
We usually only define the variance of a random variable which has finite second moment, but if you wanted to extend the definition to those random variables with with only finite first moment then you would just get infinite variance. So in this case the inequality is trivial as both sides are equal to infinity.