# Derivation question

1. May 3, 2009

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?

2. May 3, 2009

### HallsofIvy

Staff Emeritus
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 $2\mu^2$.