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Math Help - Method of moments

  1. #1
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    Exclamation Method of moments

    X is a discrete RV with P(X=1) = p , P(X=2)=1-p; three independent observations of X are made x1=1, x2=2,x3=2.
    a)find the method of moments estimate of p


    In order to find the method of moments estimate of p, I know I need to find the 1st moment and the 2nd moment.
    What I did is.... E(p) = 1/3 p +2/3 (1-p) ...am I on the right track? when I continue to find the 2nd moment E(p square)...it turns out weird... the final answer for this question is 1/3.

    d) if p has a prior distribution that is uniform on [0,1], what is its posterior density?
    Actually, I have difficulty in finding the posterior density. If anyone could do it as an exmaple for me. That would be perfect. Thanks!
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  2. #2
    MHF Contributor matheagle's Avatar
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    First of all, p is a probability.
    You want the expected value of X.
    For MOM you set the expected value equal to the sample mean.

     (1)(p)+(2)(1-p)={1+2+2\over 3}

    yup, I get  \hat p={1\over 3}
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  3. #3
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    Quote Originally Posted by cm917 View Post
    X is a discrete RV with P(X=1) = p , P(X=2)=1-p; three independent observations of X are made x1=1, x2=2,x3=2.
    a)find the method of moments estimate of p


    In order to find the method of moments estimate of p, I know I need to find the 1st moment and the 2nd moment.
    What I did is.... E(p) = 1/3 p +2/3 (1-p) ...am I on the right track? when I continue to find the 2nd moment E(p square)...it turns out weird... the final answer for this question is 1/3.

    d) if p has a prior distribution that is uniform on [0,1], what is its posterior density?
    Actually, I have difficulty in finding the posterior density. If anyone could do it as an exmaple for me. That would be perfect. Thanks!
    f(p | data) \propto f(p) \cdot f(data | p) where f(p) is the prior distribution of p, f(p | data) is the posterior distribution and f(dtat | p) is the likelihood function.

    Therefore f(p | data) \propto 1 \cdot p(1 - p)^2.

    Therefore f(p | data) = k p(1 - p)^2 where k is a normalising constant whose value is easily found to be 12.

    Therefore f(p | data) = 12 p(1 - p)^2.

    Note that E(p) = \frac{2}{5}.
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    Quote Originally Posted by matheagle View Post
    First of all, p is a probability.
    You want the expected value of X.
    For MOM you set the expected value equal to the sample mean.

     (1)(p)+(2)(1-p)={1+2+2\over 3}

    yup, I get  \hat p={1\over 3}
    But are we supposed to find E(x)?? I think we should find the E(p).... need more explaination..THANKS!
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  5. #5
    MHF Contributor matheagle's Avatar
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    E(X)= (1)(p)+(2)(1-p)

    \bar X={1+2+2\over 3}

    The idea is to set these equal.
    As I said yesterday, p is a number, NOT a random variable in the first part.

    E(3)=3, same with p, E(p)=p.
    YOU don't want E(p).
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    f(p | data) \propto f(p) \cdot f(data | p) where f(p) is the prior distribution of p, f(p | data) is the posterior distribution and f(dtat | p) is the likelihood function.

    Therefore f(p | data) \propto 1 \cdot p(1 - p)^2.

    Therefore f(p | data) = k p(1 - p)^2 where k is a normalising constant whose value is easily found to be 12.

    Therefore f(p | data) = 12 p(1 - p)^2.

    Note that E(p) = \frac{2}{5}.
    Thanks for your reply. But I don't understand how do you get the constant value 12. and how to get the E(P) = 2/5. since p is uniform distribution [0,1], the E(P) = a+b/2 ...?? is it?

    Thanks.
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  7. #7
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by cm917 View Post
    Thanks for your reply. But I don't understand how do you get the constant value 12. and how to get the E(P) = 2/5. since p is uniform distribution [0,1], the E(P) = a+b/2 ...?? is it?

    Thanks.

    The mean of a beta density is alpha over (alpha +beta).... 2/(2+3)
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  8. #8
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    Quote Originally Posted by cm917 View Post
    Thanks for your reply. But I don't understand how do you get the constant value 12. and how to get the E(P) = 2/5. since p is uniform distribution [0,1], the E(P) = a+b/2 ...?? is it?

    Thanks.
    Since f(p | data) is a pdf, \int_{0}^{1} f(p | data) \, dp = k \int_0^1 p(1 - p)^2 \, dp = 1. Therefore ....

    E(p) = \int_0^1 p f(p | data) \, dp = ....
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