Re: Random choice of numbers

quasi <quasi@xxxxxxxx> writes:

As a followup, here are some questions relating to the generalized
form ...

For n in N, n > 2, let p(n) be the probability that if n numbers are
chosen at random from the interval (0,1), the (n-1)'th power of the
last number chosen is greater than the product of the first (n-1)
numbers.

Based on results in this thread, we have p(3) = 5/9.

(1) Prove or disprove: p(n) is rational, for all n.

As geo noted, we can consider X_j = exp(-T_j) for independent exponential
random variables T_j with rate 1, and
p(n) = P{(n-1) T_n < sum_{j=1}^{n-1} T_j}.
Now sum_{j=1}^{n-1} T_j has a Gamma distribution with shape parameter n-1 and
scale parameter 1, so

p(n) = int_0^infty x^(n-2) exp(-x)/(n-2)! (1 - exp(-x/(n-1))) dx
= 1 - ((n-1)/n)^(n-1)

Assuming (1) is true, write p(n) = a(n)/b(n), as a fraction in lowest
terms, and consider the following two problems ...

(2) Prove or disprove: b(n) = n^(n-1), for all n.

See above. Note gcd(n-1,n) = 1.

(3) Prove or disprove: a(n) is always composite, unless n = 3,4,6,8.

a(n) = n^(n-1) - (n-1)^(n-1). This is also prime for n=168.

Of course if n-1 is divisible by k, a(n) is divisible by n^k - (n-1)^k.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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