# Expectation: Binomial Distribution

• Nov 9th 2012, 02:28 AM
Mick
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)py(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?
• Nov 9th 2012, 06:22 AM
HallsofIvy
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!
• Nov 9th 2012, 12:43 PM
Mick
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.