Derivation question

  • Thread starter roadworx
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  • #1
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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?
 

Answers and Replies

  • #2
HallsofIvy
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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?
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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