Derivatives of Cauchy Distribution

[riemann01]

Hi guys,

I would like to ask you where you spot the mistake in the derivatives of the loglikelihood function of the cauchy distribution, as I am breaking my head :( I apply this to a Newton optimization procedure and got correct m, but wrong scale parameter s. Thanks!

<br /> LLF = -n\ln(pi)+n\ln(s)-\sum(\ln(s^2+(x-m)^2)),<br />

First Derivatives:
<br /> \frac {dL} {dm} = 2\sum(x-m) / \sum(s^2+(x-m)^2)<br />
<br /> \frac {dL} {ds} = n/s - 2\sum(s) / \sum(s^2+(x-m)^2)<br />

Second Derivatives:
<br /> \frac {d^2L} {dm^2} = (-2n(\sum(s^2+(x-m)^2)))+4\sum(x-m)^2)/(\sum(s^2+(x-m)^2))^2<br />
<br /> \frac {d^2L} {ds^2} =-n/s^2-2\sum(-s^2+(x-m)^2)/(\sum(s^2+(x-m)^2))^2<br />
<br /> \frac {d^2L} {dmds} =-4\sum(s(x-m))/(\sum(s^2+(x-m)^2))^2<br />

 

[statdad]

I would point out that if you have a term like this:

<br /> \sum_{i=1}^n {\left(\ln (10 + x_i^2)\right)}<br />

then the first derivative is

<br /> \sum_{i=1}^n {\frac{2x_i}{10+x_i^2}<br />

and not

<br /> \frac{\sum_{i=1}^n {2x_i}}{\sum_{i=1}^n (10+x_i^2)}<br />

 

[riemann01]

Thanks for the reply statdad, you are correct but x is known and is a data set, what we are looking is m and s, for instance:

<br /> \sum \frac{1}{10+(x-m)^2}, -\sum \frac{-2(x-m)}{(10+(x-m)^2)^2} <br />

The first - comes from the formula of fraction differentiation and the second minus from differentiating the (x-m).

 

[statdad]

I realize full well what this is about: my point came from an apparent type in your first post. You essentially wrote

<br /> \frac{dL}{dm} = \frac{2\sum{(x-m)}}{\sum{(10+(x-m)^2)^2}}<br />

I made a poor choice in using x_i in my example - I was merely trying to hint that you can't distribute the sum to numerator and denominator. My point was this: if you are starting with the first derivatives as written, the fact that they (seem to be, in the typing of your first post) incorrect could be the cause of your subsequent problem.