Setting up properly the fminunc function
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Dear all,
I am trying to maximize this function
T=1000;
z=randn(T,1);
u=randn(T,1);
k1=0.01;
k2=0.01;
options=optimset('LargeScale','off','display','off','TolFun',0.0001,'TolX',0.0001,...
'GradObj','off', 'Hessian','off','DerivativeCheck','off');
for t=1:T
[x,fval,exitflag,output,G_sum,H]=fminunc('funct',u(t),options,...
z,k1, k2,t, u,T);
end
where
function LL= funct(x,z,k1, k2,t, u,T)
if t==1
u=[x; u(2:T) ];
elseif t==T
u=[u(1:T-1);x ];
else
u=[u(1:t-1);x;u(t+1:T) ];
end
e2=(z-u).^2;
kk= - 0.5*x^2/k2;
LL = -( -.5/k1*sum(e2(t:T))+kk) ;
end
However, I am not sure if the function is properly set up in terms of 'x'. As you can see 'x' changes position within 'u'.
I suspect that an alternative approach would be
function LL= funct(x,z,k1, k2,t, u,T)
u(t)=x;
e2=(z-u).^2;
kk= - 0.5*x^2/k2;
LL = -( -.5/k1*sum(e2(t:T))+kk );
end
Which of the two is more efficient? Is there any alternative solution?
Thanks in advance.
Accepted Answer
Walter Roberson
on 27 May 2019
T=1000;
z=randn(T,1);
u=randn(T,1);
k1=0.01;
k2=0.01;
options = optimset('LargeScale','off','display','off','TolFun',0.0001,'TolX',0.0001,...
'GradObj','off', 'Hessian','off','DerivativeCheck','off');
x = zeros(1, T); fval = zeros(1,T); exitflag = zeros(1,T);
output = cell(1,T); G_sum = cell(1,T); H = cell(1,T);
for t = 1:T
[x(t), fval(t), exitflag(t) ,output{t}, G_sum{t}, H{t}] = fminunc(@(x) funct(x, z, k1, k2, t, u, T), u(t), options);
end
function LL = funct(x, z, k1, k2, t, u, T)
u(t) = x;
e2 = (z-u).^2;
kk = - 0.5*x^2/k2;
LL = -( -.5/k1*sum(e2(t:T))+kk );
end
If you only include e2(t:T) in your sum, then it is not clear to me why you are calculating e2 over the whole vector? Why not, for example,
for t = 1:T
[x(t), fval(t), exitflag(t) ,output{t}, G_sum{t}, H{t}] = fminunc(@(x) funct(x, z, k1, k2, u(t:end)), u(t), options);
end
function LL = funct(x, z, k1, k2, urest)
urest(1) = x;
e2 = (z-urest).^2;
kk = - 0.5*x^2/k2;
LL = -( -.5/k1*sum(e2)+kk );
end
3 Comments
ektor
on 28 May 2019
Walter Roberson
on 28 May 2019
The first of those calculates for all t values, throwing away all of the intermediate results and keeping only the last result. It would not leave you with an x for t = 1, an x for t = 2, and so on, only with an x for t = T.
ektor
on 28 May 2019
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