Find maximum of a function.

조회 수: 2 (최근 30일)
ektor
ektor 2015년 8월 17일
댓글: JMP Phillips 2015년 8월 20일
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
  댓글 수: 4
Walter Roberson
Walter Roberson 2015년 8월 19일
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 2015년 8월 19일
편집: ektor 2015년 8월 19일
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.

댓글을 달려면 로그인하십시오.

채택된 답변

JMP Phillips
JMP Phillips 2015년 8월 19일
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.
  댓글 수: 2
ektor
ektor 2015년 8월 19일
Hi JMP, Thanks. I made a small modification, N=10 now becomes N=1. Does this make sense now?
JMP Phillips
JMP Phillips 2015년 8월 20일
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?

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Numbers and Precision에 대해 자세히 알아보기

태그

Community Treasure Hunt

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

Start Hunting!

Translated by