"0" isn't a member of the "multiplicative *group* of GF(p)", so there is no off-by-one problem. I guess since x->g^x is 1-1, it maps uniform distributions to uniform distributions. But I guess this reasoning only works for finite (countable??) sets. At 03:12 PM 10/14/2016, Allan Wechsler wrote:
I did misunderstand.
However, I think we have an off-by-one problem: g^k will never, in fact, take on the value 0.
It will be equally distributed over the nonzero values, of which there are only (p-1).
On Fri, Oct 14, 2016 at 6:02 PM, Dan Asimov <dasimov@earthlink.net> wrote:
But 2 is not a generator of [0,6] = Z/7. For a generator g of Z/p, the powers g^k will be all distinct for 0 <= k <= p-1; hence {g^k | 0 <= k < p} = Z/p.
So g^X is in fact uniform on Z/p when X is.
--Dan
On Oct 14, 2016, at 2:25 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Let p=7, g=2. For X in [0,6], g^X takes on the values 1, 2, 4, 1, 2, 4, 1. This isn't uniform: it takes the value 1 thrice, the values 2 and 4 twice, and 3, 5, and 6 never.
Am I misunderstanding your question?
On Fri, Oct 14, 2016 at 4:08 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Is this known or conjectured with a conjecture name?
If X is a uniform random distribution over [0,p-1], p prime, is g^X a uniform random distribution, for a generator g ?