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.