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Derivation question

  1. May 3, 2009 #1
    hi,

    I have:

    [tex]exp \left(-\frac{\lambda}{2^\mu^2} \sum_{i=1}^n \frac{(x_i-\mu)^2}{x_i}\right)[/tex]

    I am trying to work out why this simplifies to:

    [tex]exp\left(\frac{n \lambda}{\mu}\right) exp \left( -\frac{\lambda}{2^\mu^2} \sum_{i=1}^n x_i -\frac{ \lambda}{2} \sum_{i=1}^n \frac{1}{x_i} \right) [/tex]

    Any ideas?
     
  2. jcsd
  3. May 3, 2009 #2

    HallsofIvy

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    Staff Emeritus
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    Multiply out that square: [itex](x_i- \mu)^2= x_i^2- 2\mu x_i+ \mu^2[/itex]. Dividing by [itex]x_i[/itex] gives
    [tex]\frac{(x_i- \mu)^2}{x_i}= x_i- 2\mu+ \frac{mu^2}{x_i}[/tex]
    so that sum can be written as three different sums.

    That second sum is [itex]\sum_{i=1}^n \mu= n\mu[/itex] and that is the exponential
    [tex] exp\left(\frac{n\lambda}{\mu}\right)[/tex]

    Now, that denominator: is that
    [tex]2^{\mu^2}[/tex]
    or
    [tex]2^{2\mu}[/tex]
    or just [itex]2\mu^2[/itex]

    It looks like it is supposed to be 2 to a power but to get that result it must be [itex]2\mu^2[/itex].
     
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