Re: MGF vs characteristic function
- From: "Lou Thraki" <louthraki@xxxxxxxxx>
- Date: 18 Aug 2006 10:56:17 -0700
Eric B wrote:
Hi all,
Can anyone explain to me why for any pdf the characteristic function
always exists but the MGF (moment generating function) may not ?
For example - The Pareto distribution ...
Thanks
For any pdf P(x), the integral
integral( P(x), x=-inf..inf)
exists, and is equal to 1. You can then understand that
integral( P(x)*f(x), x=-inf..inf )
exist for any function for which |f(x)| <= c for all x
and some fixed positive c. This is for example the
case when
f(x) = exp( z*x )
and z is imaginary. In that case the integral is the
characteristic function with the imaginary part of
z as its argument.
When z has a positive real part, exp( z*x ) grows
indefinitely for positive x, so the integral may not
be defined if P(x) is non-zero all the way up to inf.
The same when z has a negative real part, and
P(x) is non-zero al the way down to -inf.
So you may say that the mgf is always defined on
the imaginary axis, but this is the characteristic
function with as its argument the imaginary part of
the argument of the mgf.
.
- References:
- MGF vs characteristic function
- From: Eric B
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