[corrected: replace p by p-1] "X is a uniform random distribution over [0,p-1], p prime" should have been "X is a uniform random distribution over [0,p-2], p prime", although [0,p-1]=[0,p-2] (mod p-1), so they are the same set. It isn't an off-by-one error in the range set but a clumsy (but still accurate) definition of the domain set. At 04:55 PM 10/14/2016, Mike Speciner wrote:

Sure there is, unless the question is rephrased to say [0,p-1), or [1,p-1], or [0,p-2], ...

On 14-Oct-16 19:49, Henry Baker wrote:

...

"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 ?