Re: Uniform sum distribtuion

How beautiful! Could you please give a reference to where this discuss
is
proved!

Herman Rubin wrote:
> In article <1114137213.004407.142710@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> <liuwei99@xxxxxxxxxxxxxxx> wrote:
> >X uniform distributed on the interval [0,1]
> >Y uniform distributed on the interval [0,2]
> >Z uniform distributed on the interval [0,3]
>
> >What is the probability density function of X+Y+Z
> >on the interval[0,6]
>
> Here is a general method for doing problems like this,
> going back to the 18th century.
>
> The Laplace transform of the distribution of a uniform
> random variable on (u, v), u < v, is
>
> [exp(-ut) - exp(-vt)]/[(v-u)t],
>
> and the Laplace transform of the density x^k on
> (0, infinity) is 1/[k!*t^{k+1}].
>
> So the Laplace transform here is
>
> (1-exp(-t))*(1-exp(-2t))* (1-exp(-3t))/(6*t^3).
>
> The product of the exponential factors is
>
> 1 - u - u^2 + u^4 + u^5 - u^6,
>
> where u = exp(-t)).
>
> So the density is 1/3 times
>
> x_^2 - (x-1)_^2 - (x-2)_^2 + (x-4)_^2 + (x-5)_^2 - (x-6)_^2,
>
> where a_ is the maximum of a and 0. As we have symmetry,
> it is only necessary to compute the density up to 3, for
> which we get
>
> x^2/3 0 < x < 1
>
> (2x-1)/3 1 < x < 2
>
> (-x^2+6x-5)/3 2 < x < 4,
>
> and complete it by symmetry.
>
> --
> This address is for information only. I do not claim that these
views
> are those of the Statistics Department or of Purdue University.
> Herman Rubin, Department of Statistics, Purdue University
> hrubin@xxxxxxxxxxxxxxx Phone: (765)494-6054 FAX:
(765)494-0558

.

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