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I read from this link and thought that kernel density functions are used for solving the unrealistic normal distributions or specification errors. But when I read the description of kernel density in MATLAB, the estimate is based on a normal kernel function.

Then what is the point? I mean it still assumes normality and the problem is unsolved. Why use kernel density instead of using normal distribution when it is still based on assumption?

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    $\begingroup$ One way to look at it is to see that a kernel density estimate based on a Gaussian kernel is a mixture of normal distributions; a Gaussian mixture may be skew, or heavy-tailed, or light tailed, or multimodal. As such it doesn't assume the original distribution is of any particular form, only that it might reasonably approximated by a mixture. $\endgroup$ – Glen_b Aug 28 '13 at 23:20
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You seem to be confusing two things (the language of this is confusing, so it's understandable).

A Gaussian kernel does not assume a distribution is normal, e.g.:

set.seed(291021)

x <- c(rnorm(1000), runif(1000,0,1), rnorm(1000,10, 5)) 
       #Some weird combo
plot(density(x, kernel = "gaussian"))

whereas assuming that x is normal would yield:

mean(x)
sd(x)    
xnormed <- rnorm(3000, 3.54, 5.52)
lines(density(xnormed), col = 'red')
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  • $\begingroup$ The reference in the question deals with copula fitting. I guess the answer is missing the application of the copula to get the density estimation on the unit cube. $\endgroup$ – JohnRos Jun 29 '13 at 12:56
  • $\begingroup$ although Gaussian Kernel does not assume that x is normal like normaldist, the graphs are somewhat similar (both Gaussian kernel and normal dist). I wonder what make Gaussian kernel differ from or superior to normaldist. does it provide more reliable parameter when the real data are skewed? $\endgroup$ – user26979 Jun 29 '13 at 13:40
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    $\begingroup$ A "gaussian kernel" and the probability density function (PDF) for a Normal distribution are mathematically the same. $\endgroup$ – whuber Jul 1 '13 at 22:10

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