SmoothHistogram with sharp boundary [duplicate]

Question

When using SmoothHistogram (to plot probability distribution function), I noticed that the edge of the figure is automatically smoothed towards zero probability. However, in some probability distributions (or some data), this is not the case. For example,

SmoothHistogram[RandomVariate[GammaDistribution[1, 2], 10000]]

Note that at nearly zero value of x-axis, the probability distribution drops. This is incorrect. To see that, here is the result from Histogram, which shows correctly the sharp boundary:

Histogram[RandomVariate[GammaDistribution[1, 2], 10000]]

Note that the Histogram plot is correct, as one can check using

Plot[PDF[GammaDistribution[1, 2], x], {x, 0, 20}, Filling -> Axis]

Is it possible to tell SmoothHistogram not trying to interpolate the end points with zero?

Answer 1

The nature of the kernel smoothing process will not allow you to have a discontinuity. If you use SmoothKernelDistribution, then you can set the shape of the kernel used for the smoothing (SmoothHistogram doesn't seem to have this option). For your problem, you could choose the kernel to be (for instance) a decaying exponential and then you would be smoothing only to the right.

data = RandomVariate[GammaDistribution[1, 2], 10000]; d2 = 
SmoothKernelDistribution[data, Automatic, If[# > 0, Exp[-#], 0] &];
Plot[PDF[d2, x], {x, -2, 6}]

Unfortunately, even using such a one-sided smoothing kernel does not do what you want. It fixes the problem of showing values to the left of zero (as happens with a Gaussian kernel) but will still not allow discontinuities.