Hiya, so I have this weird problem. Professor kinda threw it out of left field, and I need advice.

You don't necessarily have to solve it, but if you get me on the right track by about 3pm today (October 2) then I will give you 200fg. (Fg is not an issue, have plenty of things I can sell). Let me know!

I was puzzled at first.

Now I guess this was written by a statistician, rather than by a mathematician

There are some confusing parts in the problem :

"We have a coin ... an unbiased estimate of p". This is the most confusing part in my opinion.
We are not conducting some tests to find a constant p (a hidden value of a constant).
Actually p is a variable all along the problem, and we are asked to find a general scheme that can apply to every p (or at least "for some p").

"Is there some other function f(r) such that..." :
The question can be understood as : "How many choices do we have to find (n+1) constants a_0, a_1, a_2, ... , a_n such that :
SIGMA ( a_r * p^r * (1-p)^(n-r) ) = p
FOR EVERY 0<p<1 (or at least "for some p") ?

The minimal hypothesis here is : for at least (n+1) different values of p (a missing information).

Otherwise (if p were seen as a constant), you could simply plug in f(r) = p (for every r=0...n) :
E(t) = p and V(t) = 0 < p(1-p)/n
But this is not what is expected here, for sure !

Now to the solution :

A) The (n+1) polynomials g_r (r=0...n) are linearly independant.

Since they are (n+1) independant vectors in a (n+1) dimensional vector space (the set of degree < n+1 polynomials), they form a base of that vector space.

C) Since they form a base, every vector has a unique expression (decomposition) of the form : a_0*g_0 + a_1*g_1 + ... + a_n*g_n, where a_0, a_1... a_n are constants (real or complex numbers, depending on the field you're working on).

D) We already know a decomposition for the specific vector X (identity polynomial) :
X = SIGMA (r=0 to r=n ; (r/n) * Combinatory(r,n) * g_r(X) )

E) We are asked for different coefficients in the decomposition of the same vector X : X = SIGMA (r=0 to r=n ; f_r * Combinatory(r,n) * g_r(X) )
So it is safe to conclude that f_r = r/n for every r = 0 ... n.

Notice that the equality in the vector space of degree < n+1 polynomials comes from the matching of at least n+1 different values ("for some p" = for at least (n+1) different p's).

The part concerning V(t) is simply useless... Amazing ?

The linearity of (g_r), r=0...n can be proven by writing a linear combinaison equals to zero :

0 = a_0 * g_0 + a_1 * g_1 + ... + a_n * g_n

That is, for every x :

0 = a_0 * g_0(x) + ... + a_n * g_n (x)

Now plug in x=0 :
0 = a_0 * 1 + 0 + 0 + ... + 0
a_0 = 0

Plug in a_0 = 0 :

0 = a_1 * g_1 + ... + a_n * g_n
ie : 0 = a_1 * X * ( 1-X )^(n-1) + ... + a_n * X^n

You are now able to factorize by X :
0 = X * [ a_1 * (1-X)^(n-1) + a_2 * X * (1-X)^(n-2) + ... + a_n * X^(n-1) ]

The product of those 2 polynomials is zero : at least 1 must be zero (and it's not the first) :
0 = a_1 * (1-X)^(n-1) + a_2 * X * (1-X)^(n-2) + ... + a_n * X^(n-1)

And repeat : plug in x = 0, see that a_1 = 0, etc
You have proven that every a_0, a_1 ... a_n are all zero.

This post was edited by feanur on Oct 2 2013 06:08am

Very well explained solution

As someone who does stats, I don't see much ambiguity with the wording of the problem

P is a constant, but it is an unknown "true value" that we are trying to construct a best estimate with using r. We can't straight up use P as the estimate because we have no idea what it is; it'd defeat the purpose of the exercise. Maybe I'm misunderstanding your misunderstanding though...

And I'm not a statistician

My problem is to use p as the variable (or indeterminate) of the given polynomials g_r.
If p is a constant, then those polynomials act upon a set of 1 element (and not a usual field like the real numbers).
There's nothing to say in this case, and the vector-space theory won't help !

That's a good point. I'm not sure how that works either lol.

Edit:
Well, I guess when you're looking at constructing an estimate, the estimate must be unbiased for all viable P. Therefore, in the E[f(r)] = P condition, P is treated as a variable.

This post was edited by zackill4 on Oct 2 2013 06:52am

Great! Thanks a lot both of you. feanur, I need to sell a TF2 key or something before I can pay you the fg. Will get that done asap though (will be later today / tomorrow).