I am looking for a distribution where the probability density decreases quickly after some point away from the mean, or in my own words a "plateau-shaped distribution".
Something in between the Gaussian and the uniform.
8 Answers
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You may be looking for distribution known under the names of generalized normal (version 1), Subbotin distribution, or exponential power distribution. It is parametrized by location $\mu$, scale $\sigma$ and shape $\beta$ with pdf
$$ \frac{\beta}{2\sigma\Gamma(1/\beta)} \exp\left[-\left(\frac{|x-\mu|}{\sigma}\right)^{\beta}\right] $$
as you can notice, for $\beta=1$ it resembles and converges to Laplace distribution, with $\beta=2$ it converges to normal, and when $\beta = \infty$ to uniform distribution.
If you are looking for software that has it implemented, you can check normalp library for R (Mineo and Ruggieri, 2005). What is nice about this package is that, among other things, it implements regression with generalized normally distributed errors, i.e. minimizing $L_p$ norm.
Mineo, A. M., & Ruggieri, M. (2005). A software tool for the exponential power distribution: The normalp package. Journal of Statistical Software, 12(4), 1-24.
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@StrongBad's comment is a really good suggestion. The sum of a uniform RV and gaussian RV can give you exactly what you're looking for if you pick the parameters right. And it actually has a reasonably nice closed form solution.
The pdf of this variable is given by the expression:
$$\dfrac{1}{4a}\left[\mathrm{erf}\left(\dfrac{x+a}{\sigma\sqrt{2}}\right)-\mathrm{erf}\left(\dfrac{x-a}{\sigma\sqrt{2}}\right) \right]$$
$a$ is the "radius" of the zero-mean uniform RV. $\sigma$ is the standard deviation of the zero-mean gaussian RV.
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3$\begingroup$ Reference: Bhattacharjee, G. P., Pandit, S. N. N., and Mohan, R. 1963. Dimensional chains involving rectangular and normal error-distributions. Technometrics, 5, 404–406. $\endgroup$– Tim ♦Mar 26, 2016 at 11:14
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There's an infinite number of "plateau-shaped" distributions.
Were you after something more specific than "in between the Gaussian and the uniform"? That's somewhat vague.
Here's one easy one: you could always stick a half-normal at each end of a uniform:
You can control the "width" of the uniform relative to the scale of the normal so you can have wider or narrower plateaus, giving a whole class of distributions, which include the Gaussian and the uniform as limiting cases.
The density is:
$\frac{h}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2\sigma^2}(x-\mu+w/2)^2} \mathbb{I}_{x\leq \mu-w/2} \\ + \:\frac{h}{\sqrt{2\pi}\sigma}\quad\mathbb{I}_{\mu-w/2< x\leq \mu+w/2} \\ + \frac{h}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2\sigma^2}(x-\mu-w/2)^2} \mathbb{I}_{x > \mu+w/2} $
where $h = \frac{1}{1 + w/(\sqrt{2\pi}\sigma)}$
As $\sigma \to 0$ for fixed $w$, we approach the uniform on $(\mu-w/2,\mu+w/2)$ and as $w \to 0$ for fixed $\sigma$ we approach $N(\mu,\sigma^2)$.
Here are some examples (with $\mu=0$ in each case):
We might perhaps call this density a "Gaussian-tailed uniform".
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2$\begingroup$ Ach! I love attending formal balls wearing a Gausian-tailed uniform! ;) $\endgroup$– AlexisJun 6, 2018 at 21:05
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See my "Devil's tower" distribution in here [1]:
$f(x) = 0.3334$, for $|x| < 0.9399$;
$f(x) = 0.2945/x^2$, for $0.9399 \leq |x| < 2.3242$; and
$f(x) = 0$, for $2.3242 \leq |x|$.
The "slip-dress"distribution is even more interesting.
It is easy to construct distributions having whatever shape you want.
[1]: Westfall, P.H. (2014)
"Kurtosis as Peakedness, 1905 – 2014. R.I.P."
Am. Stat. 68(3): 191–195. doi:10.1080/00031305.2014.917055
public access pdf: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4321753/pdf/nihms-599845.pdf
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$\begingroup$ Hi Peter -- I took the liberty of giving the function and inserting an image as well as giving a full reference. (If memory serves I think Kendall and Stuart giving the details of a similar debunking in their classic text. If I remember correctly - it has been a long while - I believe they also discuss that it's not heavy-tailedness) $\endgroup$– Glen_bJul 28, 2016 at 1:02
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$\begingroup$ Thanks, Glen_b. I never said kurtosis measured what the tail-index numbers measure. Rather, my article proves kurtosis is, for a very broad class of distributions, nearly equal to E(Z^4 * I(|Z| > 1)). Thus, kurtosis clearly tells you nothing about the 'peak,' which is typically found in the range {Z: |Z| <1}. Rather, it is determined mostly by the tails. Call it E(Z^4 * I(|Z| > 1)) if the term "heavy-tailedness" has another meaning. $\endgroup$ Aug 31, 2016 at 19:40
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$\begingroup$ Also, @Glen_b which tail-index are you referring to? There are infinitely many. Tail crossings don't define "tailedness" properly. According to some tail crossing definitions of tail heaviness, N(0,1) is more "heavy-tailed" than .9999*U(-1,1) + .0001*U(-1000,1000), although the latter is obviously more heavy tailed, despite having finite tails. And, BTW, the latter has extremely high kurtosis, unlike N(0,1). $\endgroup$ Dec 19, 2017 at 1:06
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$\begingroup$ I can't find me saying "tail index" anywhere in my comment; I am not quite sure what you're referring to there when you say "which tail-index are you referring to". If you mean the bit about heavy-tailedness the best thing to do is check what Kendall and Stuart actually say; I believe there they actually compare the asymptotic ratio of densities for symmetric standardized variables, but it might have been survivor functions perhaps; the point was theirs, not mine $\endgroup$– Glen_bDec 19, 2017 at 5:49
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1$\begingroup$ Strange. Well, in any event, Kendall and Stuart got it wrong. Kurtosis is obviously a measure of tail weight as my theorems prove. $\endgroup$ Dec 22, 2017 at 0:05
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Lots of nice answers. The solution proffered here has 2 features: (i) that it has a particularly simple functional form, and (ii) that the resulting distribution necessarily produces a plateau-shaped pdf (not just as a special case). I'm not sure if this already has a name in the literature, but absent same, let us call it a Plateau distribution with pdf $f(x)$:
$$f(x) = k \frac{1}{1 + x^{2 a}} \quad \quad \text{for } x \in \mathbb{R}$$
where:
- parameter $a$ is a positive integer, and
- $k$ is a constant of integration: $k = \frac{a}{\pi} \sin \left(\frac{\pi}{2 a}\right)$
Here is a plot of the pdf, for different values of parameter $a$:
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As parameter $a$ becomes large, the density tends towards a Uniform(-1,1) distribution. The following plot also compares to a standard Normal (gray dashed):
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Another one (EDIT: I simplified it now. EDIT2: I simplified it even further, though now the picture doesn't really reflect this exact equation):
$$f(x) = \frac{1}{3 \cdot \alpha} \cdot \log{\left( \frac{\cosh{\left(\alpha \cdot a\right)}+ \cosh{\left(\alpha \cdot x\right)}} {\cosh{\left(\alpha \cdot b\right)}+ \cosh{\left(\alpha \cdot x\right)}} \right)} $$
Clunky, I know, but here I took advantage of the fact that $\log(\cosh(x))$ approaches a line as $x$ increases.
Basically you have control over how smooth is the transition ($alpha$). If $a = 2$ and $b = 1$ I guarantee it's a valid probability density (sums to 1). If you choose other values then you'll have to renormalize it.
Here is some sample code in R:
f = function(x, a, b, alpha){
y = log((cosh(2*alpha*pi*a)+cosh(2*alpha*pi*x))/(cosh(2*alpha*pi*b)+cosh(2*alpha*pi*x)))
y = y/pi/alpha/6
return(y)
}
f is our distribution. Let's plot it for a sequence of x
plot(0, type = "n", xlim = c(-5,5), ylim = c(0,0.4))
x = seq(-100,100,length.out = 10001L)
for(i in 1:10){
y = f(x = x, a = 2, b = 1, alpha = seq(0.1,2, length.out = 10L)[i]); print(paste("integral =", round(sum(0.02*y), 3L)))
lines(x, y, type = "l", col = rainbow(10, alpha = 0.5)[i], lwd = 4)
}
legend("topright", paste("alpha =", round(seq(0.1,2, length.out = 10L), 3L)), col = rainbow(10), lwd = 4)
Console output:
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = NaN" #I suspect underflow, inspecting the plots don't show divergence at all
#[1] "integral = NaN"
#[1] "integral = NaN"
And plot:
You could change a and b, approximately the start and end of the slope respectively, but then further normalization would be needed, and I didn't calculate it (that's why I'm using a = 2 and b = 1 in the plot).
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If you are looking for something very simple, with a central plateau and the sides of a triangle distribution, you can for instance combine N triangle distributions, N depending on the desired ratio between the plateau and the descent. Why triangles, because their sampling functions already exist in most languages. You randomly sort from one of them.
In R that would give:
library(triangle)
rplateau = function(n=1){
replicate(n, switch(sample(1:3, 1), rtriangle(1, 0, 2), rtriangle(1, 1, 3), rtriangle(1, 2, 4)))
}
hist(rplateau(1E5), breaks=200)
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Here's a pretty one: the product of two logistic functions.
(1/B) * 1/(1+exp(A*(x-B))) * 1/(1+exp(-A*(x+B)))
This has the benefit of not being piecewise.
B adjusts the width and A adjusts the steepness of the drop off. Shown below are B=1:6 with A=2. Note: I haven't taken the time to figure out how to properly normalize this.
