Thread: 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)p²(1-p)³ + 10g(3)p³(1-p)²+5g(4)p⁴(1-p)+g(5)p⁵ = 0
I did try expanding out all the terms and getting the expression in terms of p, p², 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?

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!

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.