Find maximum of a function.

4 vues (au cours des 30 derniers jours)
ektor
ektor le 17 Août 2015
Commenté : JMP Phillips le 20 Août 2015
Dear all,
I have the following code that aims at maximixing LL_all with respect to "gidraw" which is scalar.
%the data
n=1;
Xfi=[ -1.4+randn(n,1) 0.5+0.6*randn(n,1)];
beta=[3 2 ]';
yi=randn(n) ;
gidraw=0.5;
lam=Xfi*beta + gidraw;
int=yi.*lam-exp(lam);
int=sum(int);
%the function
LL_all = -(log(normpdf(gidraw,2,sqrt(0.5)))+int);
So, my question is what is the maximum of LL_all? That is, how can I find the maximum of this function? Also can I calculate the Hessian matrix(which is a scalar more precisely) of this function?
Thanks in advance
  4 commentaires
Walter Roberson
Walter Roberson le 19 Août 2015
I thought I understood what you wanted, but lam is an n x 1 vector and yi is an n x n array, so yi .* lam is an error. yi*lam would be well defined and would be n x 1 which would be fine for adding exp(lam) which would be n x 1 as well. The sum() in the next line would then be creating a scalar. Is that what is desired?
ektor
ektor le 19 Août 2015
Modifié(e) : ektor le 19 Août 2015
Hi Walter, I made a small modification in the above code to make things easier. n=10 now becomes n=1. Yes, this is what is desired. Basically I want to find the posterior mode of LL_all.

Connectez-vous pour commenter.

Réponse acceptée

JMP Phillips
JMP Phillips le 19 Août 2015
If gidraw is a scalar, your problem doesn't make sense, as LL_all is just a number. If gidraw is vector, the maximum of -[log(normpdf) + int] is +Inf.
  2 commentaires
ektor
ektor le 19 Août 2015
Hi JMP, Thanks. I made a small modification, N=10 now becomes N=1. Does this make sense now?
JMP Phillips
JMP Phillips le 20 Août 2015
still confused sorry, your code just returns a number. Don't you need to specify a range of inputs for the distribution, so that you can find the maximum within the range? In your equation for LL_all, which is the vector input, is it Int?

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Computational Geometry dans Help Center et File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by