A strange thing my professor told me about expectation

[AxiomOfChoice]

Suppose you know $E[X] = 0$ for a given (continuous) random variable. Does that mean $E[|X|] < \infty$? This is what my professor told me today, though it doesn't really make much sense...

 

[mathman]

It is a subtle point. If E(|X|) is infinite, then according to some definitions, E(X) does not exist.
On the other hand if the distribution is Cauchy { density 1/[π(1+x²)]}, then E(X)=0 by symmetry even though E(|X|) is infinite.

 

[AxiomOfChoice]

It is a subtle point. If E(|X|) is infinite, then according to some definitions, E(X) does not exist.
On the other hand if the distribution is Cauchy { density 1/[π(1+x²)]}, then E(X)=0 by symmetry even though E(|X|) is infinite.

Ok, thanks. Looking back in our textbook, the definition of "expectation" begins: "Suppose $\int_\Omega |X(\omega)| dP(\omega) <\infty$. Then we define the expectation..." So I guess that's the way it is!

 

[micromass]

It is a subtle point. If E(|X|) is infinite, then according to some definitions, E(X) does not exist.
On the other hand if the distribution is Cauchy { density 1/[π(1+x²)]}, then E(X)=0 by symmetry even though E(|X|) is infinite.

If X is Cauchy distributed, then E[X] doesn't exist. So saying E[X]=0 is wrong there. It's as wrong as saying $\int_{-\infty}^{+\infty}{xdx}=0$. The integral simply does not exist.