Derivation of Exponential Equation

[roadworx]

hi,

I have:

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

I am trying to work out why this simplifies to:

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)

Any ideas?

 

[HallsofIvy]

hi,

I have:

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

I am trying to work out why this simplifies to:

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)

Any ideas?

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

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

Now, that denominator: is that
2^{\mu^2}
or
2^{2\mu}
or just 2\mu^2

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