Expectation: Binomial Distribution

Y~Binomial(5,p). I have to show that if there is a function g(y) such that E[g(Y)]=0 for every p in (0,1) then g(y)=0 for y=0,1,2,3,4,5.

So I start by denoting the expectation, E[g(Y)]= the sum from y=0 to 5 of g(y)(5Cy)p^{y}(1-p)^{5-y} and set it equal to zero.

This becomes: g(0)(1-p)^{5 }+5g(1)p(1-p)^{4} +10g(2)p^{2}(1-p)^{3 }+ 10g(3)p^{3}(1-p)^{2}+5g(4)p^{4}(1-p)+g(5)p^{5} = 0

I did try expanding out all the terms and getting the expression in terms of p, p^{2}, etc as you know that p is non-zero, but then it would create a big system of equations in terms of g(y). So what would I do next please?

Re: Expectation: Binomial Distribution

You don't say what it is you are trying to do, or what problem you are trying to solve, so I don't see how anyone can say what you should do next!

Re: Expectation: Binomial Distribution

A random variable Y follows the binomial distribution, n=5, probability p unspecified. Say there is a function g of the random variable, g(Y).

**What I have to show:** If E[g(Y)]= 0, then the function g(y)=0 for all values of y.