Find maximum of a function.

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 Commenti

John D'Errico
John D'Errico il 17 Ago 2015
Well, first, you would need to define what gidraw is.
ektor
ektor il 18 Ago 2015
Modificato: ektor il 18 Ago 2015
thanks John. gidraw is a scalar. For example gidraw=0.5 (see above).
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 il 19 Ago 2015
Modificato: ektor il 19 Ago 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.

Accedi per commentare.

 Risposta accettata

JMP Phillips
JMP Phillips il 19 Ago 2015

0 voti

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 Commenti

ektor
ektor il 19 Ago 2015
Hi JMP, Thanks. I made a small modification, N=10 now becomes N=1. Does this make sense now?
JMP Phillips
JMP Phillips il 20 Ago 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?

Accedi per commentare.

Più risposte (0)

Categorie

Tag

Community Treasure Hunt

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

Start Hunting!

Translated by