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Math Help - Spot the distribution: A difference between two normal CDFs

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    Spot the distribution: A difference between two normal CDFs

    Does anyone know if this is a known distribution with known properties? It would assist me in writing a model if I knew I were working with a known family of distributions.

    My pdf is constructed out of a normal pdf conditional on knowing \mu is in some interval. For clarity: let f(x) be the normal pdf and  F(x) the normal cdf, with the usual parameters \mu, \sigma^2. The distribution family I am trying to find the name for is the conditional pdf given that \mu lies in some range:
    f(x \mid \mu \in [\underline \mu, \overline \mu]) = \frac{1}{\overline \mu - \underline \mu} \int_{\underline \mu}^{\overline \mu} f(x) d\mu
    There are a few ways to write this integral. I include all of them in case anyone has seen any of them before:

    • Use F(x \mid \mu , \sigma^2) to represent the normal CDF with parameters \mu, \sigma^2.

      Then, my conditional PDF can be written:  \frac{F(-x \mid -\overline \mu, \sigma^2) - F(-x \mid -\underline \mu, \sigma^2)}{\overline \mu - \underline \mu}


    •  \frac{1}{2} \frac{1}{\overline \mu - \underline \mu} \left(\text{Erf}\left[\frac{\overline \mu -x}{\sqrt{2} \sigma }\right]-\text{Erf}\left[\frac{\underline \mu -x}{\sqrt{2} \sigma }\right] \right) \ , where Erf is the error function.


    • A third way is to write the integral explicitly:  \frac{1}{\sqrt{\pi}} \frac{1}{\overline \mu - \underline \mu} \int_{\frac{\underline \mu -x}{\sqrt{2} \sigma }}^{\frac{\overline \mu -x}{\sqrt{2} \sigma }} e^{-t^2} \, dt

    The pdf seems to have a number of nice properties. For example, it is symmetric and the mean is  \frac{\overline \mu + \underline \mu}{2} . Its variance is  \sigma^2 + \frac{(\overline \mu-\underline \mu)^2}{12} . Its graph looks a little like a generalized normal, and indeed it seems to tend toward the uniform as  \overline \mu - \underline \mu \rightarrow \infty . I'm looking for guidance because I don't want to reinvent the wheel!

    Many, many thanks if you know this!
    Last edited by cgari; December 14th 2009 at 05:17 PM.
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