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Math Help - Definite integrals for Bayesian statistics

  1. #1
    bsk
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    Definite integrals for Bayesian statistics

    Hi,

    As far as I know this has an analytical solution but I'm not sure how to go about doing it. Just to clarify, I'm not looking for a numerical method.

    Sorry, don't know how to use maths symbols here:

    I need to find x and y, such that the integral of exp(-ax^2 - b), evaluated from -infinity to x, equals 0.025, and the same integral evaluated from y to infinity equals 0.0975.

    a and b are known constants.

    (The question is just to find the 95% confidence interval from a posterior distribution).

    Can anyone help?

    Thanks!
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  2. #2
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    Quote Originally Posted by bsk View Post
    Hi,

    As far as I know this has an analytical solution but I'm not sure how to go about doing it. Just to clarify, I'm not looking for a numerical method.

    Sorry, don't know how to use maths symbols here:

    I need to find x and y, such that the integral of exp(-ax^2 - b), evaluated from -infinity to x, equals 0.025, and the same integral evaluated from y to infinity equals 0.0975.

    a and b are known constants.

    (The question is just to find the 95% confidence interval from a posterior distribution).



    Can anyone help?
    Thanks!
    up to some factors which you can pick out for instance take the e^{-b} out of the integrand and note that e^{-ax^2} can be handled with the proper choice of \sigma, where \sigma is the square root of the variance of a normally distributed r.v., use the table for the normal distribution for answer.
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  3. #3
    bsk
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    Thanks, I think I've got it.
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  4. #4
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    just be sure u scale things properly in the end, for there is the \frac1{|\sigma|\sqrt{2\pi}} to consider, i.e., if X= \mu+\sigma{Z} is a normal r.v., then p(x), the probability density function of X is given by,
    p(x) = \frac1{|\sigma|\sqrt{2\pi}}exp(-\frac{(x-\mu)^2}{2\sigma^2})

    And thus P(a\leq{X}\leq{b}) = \int_a^bp(x)dx
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