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In Non-parametric statistics one requirement for the kernel is : $$ K(-u)=K(u) $$ for all values of $u$. This requirement ensures that the average of the corresponding distribution is equal to that of the sample used. Can anyone prove or show me how this criteria comes from even nature of kernel?

Reference:https://en.wikipedia.org/wiki/Kernel_(statistics)

It looks like that: $$ E(X)= \int_{-\infty}^{\infty}x K(x)dx =0 $$ This is in definition section in non-parametric statistics. Thanks.

MUH
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  • It follows directly from the results at http://stats.stackexchange.com/questions/46843 . – whuber Apr 01 '16 at 19:18
  • @whuber: Can you explain how the follows from your link? from this we can say that average of distribution is zero. How will average of sample=0? Also, this does not seem to have any utility then. – MUH Apr 01 '16 at 20:16
  • The average of the kernel is added to the average of the empirical distribution (the data). Thus, the mean will be preserved if and only if the mean of the kernel is zero: that is its utility. – whuber Apr 01 '16 at 20:59
  • Why do we need to add the two averages? You have some data from which you are trying to estimated density function, so you want to obtain the mean and other parameters etc. Now, this property will help us to get the mean easily as according to statement, average of the corresponding distribution is equal to that of the sample used. Why do we need to add two averages? can you explain it a bit? Forgive me for my ignorance. – MUH Apr 01 '16 at 21:08
  • If the average of the kernel is nonzero, that kernel is effectively creating a *systematic* shift of the density estimate. Do you really want to built such a bias into your procedures, especially when it's so easy to fix? – whuber Apr 01 '16 at 21:09
  • Ok, now I understand this. Actually, I didn't know what do we do with this kernel to obtain the density. So, it's pretty much its application that answers this question. This is what should answer this question. https://en.wikipedia.org/wiki/Kernel_density_estimation. Thanks @whuber a lot. – MUH Apr 01 '16 at 21:18

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