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Conditional exponential distribution and exponential evidence

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    This is a subset of a larger problem I'm working on, but once I get over this hang up I should be good to go. I have a set of measurements [itex]x_n[/itex] that are exponentially distributed

    [tex]p(x_n|t)=e^{-(x_n-t)} I_{[x_n \ge t]}[/tex]

    and I know that t is exponentially distributed as

    [tex]p(t)=e^{-t}I_{[t\ge0]}[/tex]


    2. Relevant equations
    marginal probability
    [tex]p(x)=\int p(x|t) p(t) dt[/tex]


    3. The attempt at a solution
    So the probability of N observations of x are
    [tex]p(\mathbf{x}|t)=e^{-s(x)} e^{Nt} I_{[\textrm{min}(x_n) \ge t]}[/tex]
    where
    [tex]s(x)=\sum_{n=1}^N x_n[/tex]

    Which means that
    [tex]p(\mathbf{x},t)=e^{-s(x)} e^{t(N-1)} I_{[\textrm{min}(x_n) \ge t]} I_{[t\ge0]}[/tex]

    If I want to find p(x) it should be
    [tex]p(\mathbf{x})=\int_0^{x_{min}} e^{-s(x)}e^{t(N-1)} I_{[\textrm{min}(x_n) \ge t]}I_{[t\ge0]} dt[/tex]
    [tex]p(\mathbf{x})=e^{-s(x)}\frac{1}{N-1}e^{t(N-1)}|^{t=x_{min}}_{t=0}I_{[\textrm{min}(x_n) \ge t]}I_{[t\ge0]}[/tex]
    [tex]p(\mathbf{x})=e^{-s(x)}\frac{1}{N-1}I_{[\textrm{min}(x_n) \ge t]}I_{[t\ge0]}(e^{x_{min}(N-1)}-1)[/tex]

    The issue is that this function isn't normalized. Are my limits wrong, or should I renormalize?
     
    Last edited: Apr 3, 2013
  2. jcsd
  3. Apr 4, 2013 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    The formula for ##p(\mathbf{x})## should not have ##t## in it.

    Anyway, why would you need to re-normalize? Your ##p(\mathbf{x})## integrates to 1 when integrated over ##\mathbb{R}_{+}^N##. If you don't believe it, try the simple cases of N = 2 and N = 3 first.
     
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