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Math Help - Beta Distribution Proof

  1. #1
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    Post Beta Distribution Proof

    The probability p of an event is supposed to have a prior distribution
    Beta(a,B) so that its prior density is

    f(p) = [B(α, β)]^-1 p^(α-1) (1-p)^(β-1)


    Show that for any constant d,
    E[p(1-p)(p-d)2] is minimised wwhen d = (α+1) / (α+β+2)

    _____

    So far iv got

    E[p(1-p)(p-d)2] = ∫ p(1-p)(p-d)2 f(p) dp

    = ∫ [B(α, β)]^-1 p^α (1-p)^β (p-d)^2 dp

    = [B(α, β)]^-1 ∫ p^α (1-p)^β (p-d)^2 dp


    Any help will be appreciated
    Thanks a lot
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  2. #2
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    Quote Originally Posted by gethin08 View Post
    The probability p of an event is supposed to have a prior distribution
    Beta(a,B) so that its prior density is

    f(p) = [B(α, β)]^-1 p^(α-1) (1-p)^(β-1)


    Show that for any constant d,
    E[p(1-p)(p-d)2] is minimised wwhen d = (α+1) / (α+β+2)

    _____

    So far iv got

    E[p(1-p)(p-d)2] = ∫ p(1-p)(p-d)2 f(p) dp

    = ∫ [B(α, β)]^-1 p^α (1-p)^β (p-d)^2 dp

    = [B(α, β)]^-1 ∫ p^α (1-p)^β (p-d)^2 dp


    Any help will be appreciated
    Thanks a lot
    Expand (p -d)^2 and re-arrange the integral into:

    \frac{1}{B(\alpha, \beta)} \left[ \int_0^1 p^{\alpha + 2} (1 - p)^{\beta} \, dp - 2 d \int_0^1 p^{\alpha + 1} (1 - p)^{\beta} \, dp + d^2 \int_0^1 p^{\alpha} (1 - p)^{\beta} \, dp\right].

    Evaluate each integral using the formula here: Beta Integral -- from Wolfram MathWorld

    Simplify the resulting expression using the formula here: Beta Function -- from Wolfram MathWorld

    You now have a simple quadratic function in d. Find the value of d that minimises this quadratic.
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  3. #3
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    Thanks alot
    Been stuck on that one for a few days
    Much appreciated
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  4. #4
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    Post

    Sorry I thought I got this but im not getting the right answer...

    If I expand

    [B(α, β)]^-1 . ∫ p^(α+2) (1-p)^β dp -2d . ∫ p^(α+1) (1-p)^β dp +d^2 . ∫ p^α (1-p)^β dp

    I get

    [B(α, β)]^-1 . [B(α+3, β+1) - 2d] . [B(α+2, β+1) + d^2] . B(α+1, β+1)

    I can substitue for the gamma function and the factorial function but they dont seem to cancel and lead to d = (α+1) / (α+β+2)

    Can you explain to me where to go from here

    Thanks alot
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  5. #5
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    Quote Originally Posted by gethin08 View Post
    Sorry I thought I got this but im not getting the right answer...

    If I expand

    [B(α, β)]^-1 . ∫ p^(α+2) (1-p)^β dp -2d . ∫ p^(α+1) (1-p)^β dp +d^2 . ∫ p^α (1-p)^β dp

    I get

    [B(α, β)]^-1 . [B(α+3, β+1) - 2d] . [B(α+2, β+1) + d^2] . B(α+1, β+1)

    I can substitue for the gamma function and the factorial function but they dont seem to cancel and lead to d = (α+1) / (α+β+2)

    Can you explain to me where to go from here

    Thanks alot
    Please go back and look at the expression I posted. At a glance, it does not look to me like you've done the basic calculation properly.
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  6. #6
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    aw rite im with you
    i read you expression the wrong way
    got the right answer now though
    thanks again
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