- #1
riemann01
- 2
- 0
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!
[tex]
LLF = -n\ln(pi)+n\ln(s)-\sum(\ln(s^2+(x-m)^2)),
[/tex]
First Derivatives:
[tex]
\frac {dL} {dm} = 2\sum(x-m) / \sum(s^2+(x-m)^2)
[/tex]
[tex]
\frac {dL} {ds} = n/s - 2\sum(s) / \sum(s^2+(x-m)^2)
[/tex]
Second Derivatives:
[tex]
\frac {d^2L} {dm^2} = (-2n(\sum(s^2+(x-m)^2)))+4\sum(x-m)^2)/(\sum(s^2+(x-m)^2))^2
[/tex]
[tex]
\frac {d^2L} {ds^2} =-n/s^2-2\sum(-s^2+(x-m)^2)/(\sum(s^2+(x-m)^2))^2
[/tex]
[tex]
\frac {d^2L} {dmds} =-4\sum(s(x-m))/(\sum(s^2+(x-m)^2))^2
[/tex]
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!
[tex]
LLF = -n\ln(pi)+n\ln(s)-\sum(\ln(s^2+(x-m)^2)),
[/tex]
First Derivatives:
[tex]
\frac {dL} {dm} = 2\sum(x-m) / \sum(s^2+(x-m)^2)
[/tex]
[tex]
\frac {dL} {ds} = n/s - 2\sum(s) / \sum(s^2+(x-m)^2)
[/tex]
Second Derivatives:
[tex]
\frac {d^2L} {dm^2} = (-2n(\sum(s^2+(x-m)^2)))+4\sum(x-m)^2)/(\sum(s^2+(x-m)^2))^2
[/tex]
[tex]
\frac {d^2L} {ds^2} =-n/s^2-2\sum(-s^2+(x-m)^2)/(\sum(s^2+(x-m)^2))^2
[/tex]
[tex]
\frac {d^2L} {dmds} =-4\sum(s(x-m))/(\sum(s^2+(x-m)^2))^2
[/tex]