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Math Help - Expectation: Binomial Distribution

  1. #1
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    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?
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  2. #2
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    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!
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  3. #3
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    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.
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