# Cross Validation for Kernel Density Estimation

In a post publihed in July, I mentioned the so called the Goldilocks principle, in the context of kermel density estimation, and bandwidth selection. The bandwith should not be too small (the variance would be too large) and it should not be too large (the bias would be too large). Another standard method to select the bandwith, as mentioned this afternoon in class is the cross-validation technique (described in Chiu (1991)). Here, we would like to minimize

$\mathbb{E}\left[\int [\widehat{f}_h(x)-f(x)]^2dx\right]$

The integral can be writen

$\int \widehat{f}_h(x)^2dx-2\int \widehat{f}_h(x)f(x)dx+\int f(x)^2dx$

Since the third component is constant, we have to minimize the expected value of the sum of the first two.

The idea is to approximate it as

$J(h)=\int \widehat{f}_h(x)^2dx-\frac{2}{n}\sum_{i=1}^n \widehat{f}_{(-i)}(X_i)$

which can easily be computed. Consider here some sample, with 50 observations, from a Gaussian distribution,

> set.seed(1)
> X=rnorm(50)

From Silverman’s rule of thumb (which should be appropriate here since the sample has a Gaussian sample) the optimal bandwidth is

> 1.06*sd(X)*length(X)^(-1/5)
[1] 0.4030127

Using the cross-validation technique mentioned above, compute

> J=function(h){
+ fhat=Vectorize(function(x) density(X,from=x,to=x,n=1,bw=h)$y) + fhati=Vectorize(function(i) density(X[-i],from=X[i],to=X[i],n=1,bw=h)$y)
+ F=fhati(1:length(X))
+ return(integrate(function(x) fhat(x)^2,-Inf,Inf)$value-2*mean(F)) + } > vx=seq(.1,1,by=.01) > vy=Vectorize(J)(vx) > df=data.frame(vx,vy) > library(ggplot2) > qplot(vx,vy,geom="line",data=df) The function has the following shape and the optimal value is > optimize(J,interval=c(.1,1))$minimum
[1] 0.4687553

$objective [1] -0.3355477 Note that, indeed, it is close to Siverman’s optimal bandwidth. ## 2 thoughts on “Cross Validation for Kernel Density Estimation” 1. italo says: this is insane code. 2. mgl says: Thank you for your great post. When I ran your code it resulted in set.seed(20160705) n <- 100 X <- rnorm(n) optimize(J,interval=c(.1,1)) minimum [1] 0.5030485$objective
[1] -0.2610623

Is it plausible bandwidth?

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