Why is the variance of the Parzen density estimator infinite?

[crbazevedo]

Hello everyone,

I'm new to this forum and I'm glad to have found such a high quality resource where we can have such valuable guidance and discussions.

I've read somewhere that the variance of [tex]p(x) = {\frac{1}{n}}\sum_{i=1}^{n}\delta(x-x_i) \forall x \in \Re[\tex],

in which [tex]D_n = \left\{x_1, \cdots, x_n\right\}[\tex] independent realizations of a continuous random variable [tex]X[\tex], and [tex]\delta[\tex] is the dirac delta function, is infinite, irrespective to [tex]n[\tex] and [tex]D_n[\tex].

My questions are straightforward:

1) Why is that (my guess that the integral involved in the variance calculation does not converge)?;

2) What does it mean to have infinite variance in practical terms?;

3) Any practical examples where a distribution with infinite variance would be useful?;

Please note that I don't have a strong background in statistics.

Any help will be much appreciated.

Cheers,
Carlos

[bpet]

I've read somewhere that the variance of [tex]p(x) = {\frac{1}{n}}\sum_{i=1}^{n}\delta(x-x_i) \forall x \in \Re[\tex],

in which [tex]D_n = \left\{x_1, \cdots, x_n\right\}[\tex] independent realizations of a continuous random variable [tex]X[\tex], and [tex]\delta[\tex] is the dirac delta function, is infinite, irrespective to [tex]n[\tex] and [tex]D_n[\tex].


Assuming you mean to find the variance of the random variable with density p(x), this would be infinite only if the variance of X is infinite (for example if X has the Cauchy or Pareto distribution).