John wrote:
>
> Please help me solve this problem.
> Let t and u are independent uniform distributions with range from 0 to 1,
> and a, b, and c are constants.
>
> I tried to compute the expectation of (t^a)*(u^b)*[(1-t-u)^c].
> Since t and u are independent, finding the expectation of t and u is
> finding the integral of (t^a)*(u^b)*[(1-t-u)^c] with respect to t and,
> then, u.
>
> I rewrote (t^a)*(u^b)*[(1-t-u)^c] to different forms but no luck.
> Somehow, I believe the result is either Hypergeometric function or Beta
> function.
I know this isn't quite what you asked, but it looks to me as if it
might be nicer integrated on 0<t, 0<u, t+u<1 - in which case I'd bet
sight unseen on the obvious three-parameter variation of the beta
function, Gamma(a) Gamma(b) Gamma(c) / Gamma(a+b+c).
This should presumably be called the "Tribeta" function and symbolized
with a neoGrecian letter like a B but with three lobes - for precedent,
consider the "digamma" function, derived from the gamma function and
symbolized with an F-like letter used in non-classical Greek, and the
further "trigamma" and "tetragamma" functions, found in some tables,
whose symbols, completely unknown to classicists, look like thirtysecond
and sixtyfourth note rests...
Presumably the resulting "Tribeta" distribution would be useful in the
Bayesian study of three-outcome events, modelling three-element mixtures
in geology, etc.
-Robert Dawson
.
.
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