Another proof thread.

[Man of Knowledge]

See if you can prove this statement:

For a given random variable X, if the Laplace transformation of X does not exist then the Moments of X do not exist.

[premjan]

Does this have something to do with the series expansion for e^x? Wouldn't the implication go the other way?

[lpetrich]

The Laplace transform F(s) of a function f(t) is defined as:

F(s) = \int_0^\infty e^{-st} f(t) \,dt

where t is real and s may be complex. The moment integrals Mn are

M_n = \int_{-\infty}^\infty t^n f(t) \, dt

Or should the moment integral be defined here as

M_n = \int_0^\infty t^n f(t) \, dt

?

I would like a clarification, since the theorem statement
(Laplace transform is always infinite) -> (moments are all infinite)

is equivalent to
(some moments are finite) -> (Laplace transform is sometimes finite)

It is possible for me to think of a function where some of the moments are finite when the integration is over all the real numbers but where none of the moments are finite when the integration is over only the nonnegative real numbers. A function like
f(t) = s(t)e^{t^2}
where s(t) is +1 for t > 0, -1 for t < 0, and 0 for t = 0.

[premjan]

Some moments are finite will not get you laplace tranform finite, rather all the moments have to be finite.
http://upload.wikimedia.org/math/4/2/e/42eb1bb21ac88df4ccab203eac1c4175.png
I suppose the OP means that Laplace transform does not exist -> at least one of the moments do not exist.