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Statistics:minimizing an interval for a standard normal distribution

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Generalize: For arbitrary 0 < p < 1, show that the method giving a and b produces the minimum length interval.

    Hint: It might be helpful to use local extrema for the inverse function of the distribution function.


    2. Relevant equations
    The method is is talking about is locating the z scores using (1-p)/2 and [1-(1-p)/2]


    3. The attempt at a solution
    Let a be the area on the tail end of the distribution not included in p
    Let b be the other end so that

    a+b=1-p and b=1-p-a

    Then the points A and B are the end points of the interval containing p.

    B-A = (F^-1)(p+a)-(F^-1)(a)

    This is where I am stuck. I know f(y). So d/dy(F(y))=f(y) and then (f^-1)'(y)=1/(f'(f^-1)(y))

    I am not sure how to proceed.
     
  2. jcsd
  3. Nov 8, 2012 #2

    Ray Vickson

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    Homework Helper

    It is helpful to assume the normal distribution is standard (mean = 0, variance = 1). Do you see why you can do that without loss of generality? Now, if F(z) is the standard normal cdf, you want to minimize b-a, subject to the constraint F(b) - F(a) = p. Hint: Lagrange multipliers.

    RGV
     
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