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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
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| ``Re: Why is the expected value not zero if theta = 0?''
by stevecheng on 2005-08-16 16:48:29 |
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| > By "absolutely convergent," I assume that you mean that the integral converges for *any* subinterval of r ... and indeed, the antiderivative of rf(r)dr is ln(1 + r^2) / (2 * pi), which diverges on the interval [0, inf), for example.
That's one way to phrase the definition;
the easier (and more general) formulation
is just E |R| < \infty.
In your example, r/1+r^2 ~ 1/r asymptotically
so E |R| = \infty.
> Is there such a thing as a distribution for which the mean does exist but the variance (or second moment) does not? Or are those two always a "package deal?"
Certainly not, although the counterexamples are probably
all fairly artificial.
For example, take the density
f(t) = |t|^{-2.5}, |t| >= 1
f(t) = 0, |t| < 1
multiplied by the appropriate normalizing factor.
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