I need to find a symmetric low-kurtosis distribution class, which includes the uniform, the triangular and the normal Gaussian distribution. The Irwin-Hall distribution (sum of standard uniform) offers this characteristic, but is not treating non-integer orders $N$. However, if you e.g. simply independently sum up e.g. 2 standard uniform $[0,1]$ and one 3rd with a smaller range like $[0,0.25]$ you will indeed nicely obtain a more general and smoothly extended version of Irwin-Hall for any arbitrary order (like $N=2.25$ in this case). However, I wonder if it is possible to find a practical closed formula for the CDF?

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    $\begingroup$ "Smoothly extended" raises some thorny questions. In the thread at stats.stackexchange.com/questions/41467, the poster observes that the smoothness of the Irwin-Hall distribution changes abruptly from one (integral) value of $n$ to the next. This already suggests we should not expect there to be any mathematically "nice" closed form that is parameterized by real values of $n$. Moreover, there is no such closed formula even for the Irwin-Hall distribution itself. $\endgroup$
    – whuber
    Oct 29 '13 at 14:28
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    $\begingroup$ Hi, I made detailed sampling experiments and look to histograms of such generalized Irwin-Hall distribution. Indeed, introducing non-integer N helps to avoid jumps in the behavior! Also e.g. the kurtosis increases smoothly with real-valued N. If this would be not the case it would be indeed not nice. I think it should be possible to extend the Irwin-Hall summing formula for CDF in some way. $\endgroup$
    – user32038
    Oct 29 '13 at 14:46
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    $\begingroup$ The summation is not a "closed formula" in the usual sense of the word because the number of terms increases without bound as $n$ varies. This is an important distinction, because true closed formulas do exist: the characteristic function of the Irwin-Hall distribution, $\left((\exp(it) - 1)/(it)\right)^n,$ exists for non-integral $n$ and so its inverse Fourier Transform answers your question--that is, if you consider that a closed practical formula! $\endgroup$
    – whuber
    Oct 29 '13 at 15:09
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    $\begingroup$ Hi whuber! I need a suitable implementation e.g. in Pascal. For moderate N (like 20) the well-known Irwin-Hall CDF formula is no problem at all, but I do not want to spend too much computing time, e.g. for an integration, (inverse) Fourier transform or whatever. Of course, the Fourier transform approach is elegant, but not so accurate, because I am highly interested in CDF(x) for "large" x, so the tail area matters for me! $\endgroup$
    – user32038
    Oct 29 '13 at 19:44
  • $\begingroup$ Hi, could you sketch our in more detail how you would go from Fourier transform of PDF to the plain CDF? (although I generally believe this method has oscillation problems in the tails, e.g. giving negative PDF if we try to evaluate the integral...). $\endgroup$
    – user32038
    Nov 4 '13 at 7:32

Well, this isn't really a full answer, will come back later to complete ...

Brian Ripley's book Stochastic simulation have the closed pdf formula as exercise 3.1 page 92 and is given below: $$ f(x) = \sum_{r=0}^{\lfloor x \rfloor} (-1)^r \binom{n}{r} (x-r)^{(n-1)}/ (n-1)! $$ An R implementation of this is below:

makeIH  <-  function(n) Vectorize( function(x) {
                            if (x < 0) return(0.0)
                            if (x > n) return(0.0)
                            X  <-  floor(x)
                            r <- seq(from=0,  to=X)
                            s <-  (-1)^r * choose(n, r)*(x-r)^(n-1)/factorial(n-1)
                            } )

which is used this way:

fun3  <-  makeIH(3)
 abline(v=1, col="red")
 abline(v=2, col="red")

and gives this plot:

enter image description here

The unsmoothness at the integer values $x=1, x=2$ can be seen, at least with a good eyesight ....

(I will come back to complete this later)

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    $\begingroup$ +1 I'd be interested in seeing what the completion of this is, if you have time. ... As an aside, while I know that there's a lack of smoothness here (in the sense of there being discontinuities in the second derivative) I can't say that I really perceive it as not-smooth (my eye tends to say "no kinks, it looks smooth"). I assume that if I was riding a roller coaster along that curve, I'd be sure to feel the change though. $\endgroup$
    – Glen_b
    Mar 13 '17 at 21:31
  • $\begingroup$ For the moment out of time ... later $\endgroup$ Mar 13 '17 at 21:38
  • $\begingroup$ Too see the lack of smoothness probably we will need to choose the plotting points to include the integers ... $\endgroup$ Nov 13 '18 at 13:30

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