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], <br /> <br /> 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 $$D_n[\tex]. <br /> <br /> My questions are straightforward: <br /> <br /> 1) Why is that (my guess that the integral involved in the variance calculation does not converge)?; <br /> <br /> 2) What does it mean to have infinite variance in practical terms?; <br /> <br /> 3) Any practical examples where a distribution with infinite variance would be useful?; <br /> <br /> Please note that I don't have a strong background in statistics.<br /> <br /> Any help will be much appreciated.<br /> <br /> Cheers,<br /> Carlos$$[/tex][/tex][/tex][/tex][/tex]

 

[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], <br /> <br /> 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 $$D_n[\tex].$$[/tex][/tex][/tex][/tex][/tex]

[tex][tex][tex][tex][tex]$$<br /> Assuming you mean to find the variance of <i>the random variable with density</i> p(x), this would be infinite only if the variance of X is infinite (for example if X has the Cauchy or Pareto distribution).$$[/tex][/tex][/tex][/tex][/tex]