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Math Help - Posterior distribution

  1. #1
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    Posterior distribution

    Question:
    Let the unknown probability that a basketball player makes a shot successfully be Θ. Suppose your prior on Θ is uniform on [0,1] and that she then makes two shots in a row. Assume that the outcomes of the two shots are independent.
    What is the posterior density of Θ?
    What would you estimate the probability that she makes a third shot to be?

    My attempt:
    If I let X be the event that she made two shots then I have the posterior as
    f(Θ|X) = f(X,Θ)/f(X)
    Given that making a shot is binomial,
    f(X,Θ) = (n C x)*Θ^x*(1-Θ)^n-x
    where (n C x) is the binomial coefficient, n choose x.
    Then f(Θ|X) = f(X,Θ)/f(X)
    If I integrate out the marginal distribution from f(X|Θ) over the interval [0,1], I get
    f(X) = 1/(n+1)

    Hence,
    f(Θ|X) = [(n+1)*n!/x!(n-x)!]*Θ^x*(1-Θ)^n-x
    = [Γ(n+2)/Γ(x+1)*Γ(n-x+1)]*Θ^x*(1-Θ)^n-x
    which is beta(x-1,n-x+1)

    Does this look correct?

    Then for the second question, would I just take the expected value of the beta with these parameters? I.e. if y ~ beta(1,1), E[y] = 0.5?
    Here, I got the parameters from the fact that we now need 1 success in 1 try.

    Thanks in advance for any input you may have.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by speedBoots View Post
    Question:
    Let the unknown probability that a basketball player makes a shot successfully be Θ. Suppose your prior on Θ is uniform on [0,1] and that she then makes two shots in a row. Assume that the outcomes of the two shots are independent.
    What is the posterior density of Θ?
    What would you estimate the probability that she makes a third shot to be?

    My attempt:
    If I let X be the event that she made two shots then I have the posterior as
    f(Θ|X) = f(X,Θ)/f(X)
    p(\theta|x)=\frac{p(x|\theta)p(\theta)}{p(x)}

    where:

    p(x|\theta)=\frac{2!}{x!(2-x)!}\theta^x (1-\theta)^{2-x} ,\ x\in\{0,1,2\}

     <br />
p(\theta)=1 ,\ \theta \in [0,1]<br />

    and:

     <br />
p(x)=\int_0^1 p(x|\theta)p(\theta)\; d\theta<br />

    where we are interested in the case x=2 .

    Then the probability of making the third shot will be:

    p_3=p(\theta|2)

    CB
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