matlab doesn't calibrate via lsqnonlin when initial values depart from the actual parameters

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Daniel 2013-7-6

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https://ww2.mathworks.cn/matlabcentral/answers/81282-matlab-doesn-t-calibrate-via-lsqnonlin-when-initial-values-depart-from-the-actual-parameters

Hi guys,

I have written a program that calibrates two parameters from the option prices. I have simulated 10 option prices for known parameters and now try to check my calibration function.

If I set initial values of the paramters close to the actual one (the difference is not more that 0.4) then it calibrates the right parameters in 4 seconds. When I set the initial values slightly different from the initial (e.g. I set it equal to 2 when actual is 1), or when I change the upper and lower bounds and run the calibration, matlab calibration doesn't converge - it is always busy and I have to kill the process. ( I even left it for 1 night, the result was still "busy")

Please help me, changing the accuracy and number of function evaluation doesn't help!

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Daniel 2013-7-6

编辑：Daniel 2013-7-6

在 MATLAB Online 中打开

%%%%%%%%%%Calibration part  %%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%initial parameters and lower\upper bounds %%
x0=[0.2,-1.5];
lb = [0.001,-2];
ub = [1.2,0];
%%%%%%%%%%reading market data from from xls file %%%%%%%%%%%%
mktdata=xlsread('marketdata');
K = mktdata(2:11,1);
T = mktdata(1,2);
mktprice = mktdata(2:11,2);
S=50;
%%%%%%%%%%%%%Calibration - Determining the parameters %%%%%%%%%
options=optimset('TolFun',1e-3);
x = lsqnonlin(@LSQDiff,x0,lb,ub,options);
toc;

Then follows the code for each function:

    function [ Diff ] = LSQDiff( x )
%%%%%%%%%%%%%%%%%%%Allocate memory %%%%%%%%%%%%%%%%%
Diff=zeros(length(K),length(T));
%%%%%%%%%%Calculate the matrix of differences %%%%%%%%%%%%%%%
    for i=1:length(T)
     Diff(:,i)= mktprice(:,i)-Putprice(S,K,T(i),x(1),x(2));
    end
reshape(Diff,length(T)*length(K),1);

next one

function [Put]=Putprice(S,K,T,c,beta)
%%%%%%%declare the parameters for previously calibrated a(t) and b(t)
a1=0;  a2=10;  b1=0;  b2=0.02;  r1=0;  r2=0.05;  q1=0;  q2=0;  syms u;
%%%%%%%%%%%%%%%%%specify r(t) and q(t)%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
r=r1*u+r2;
q=q1*u+q2;
rfunc=@(u)r1*u+r2;
qfunc=@(u)q1*u+q2;
%%%%%%%%%%%%%%%%%specify a(t) and b(t) functions %%%%%%%%%%%%%%%%%%%%%
a=a1*u+a2;
b=b1*u+b2;
alpha=r-q+b;
bfunc=@(u) b1*u+b2;
taufunc=@(u)a^2*exp(-2*abs(beta)*int(alpha,u,0,u));
rbfunc=@(u) r1*u+r2+b1*u+b2;
alphafunc=@(u) r1*u+r2+b1*u+b2-q1*u-q2;
res=taufunc(u);
resfunc=matlabFunction(res);
%%%%%%%%%%%%%%%%%compute integral for tau!%%%%%%%%%%%%%%%%%%%%%%%%%%%%
tau=integral(@(u)resfunc(u),0,T);
%%%%%%%%%%%%%%%%%%continue with the pricing %%%%%%%%%%%%%%%%%%%%%%%%%
nuplus=(c+0.5)/abs(beta);
delta=2*(nuplus+1);
x=1/abs(beta)*S^abs(beta);
k=1/abs(beta).*(K.^abs(beta)).*exp(-abs(beta).*integral(alphafunc,0,T));
%%%%%%%%%%%%%%Clean Price of the option %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Put_clean=exp(-integral(rbfunc,0,T)).*K.*(x^2/tau)^(1/(2*abs(beta))).*Phiminus(-0.5/abs(beta),k.^2/tau,delta,x^2/tau)...
    -exp(-integral(qfunc,0,T))*S.*Phiminus(0,k.^2/tau,delta,x^2/tau);
%%%%%%%%%%%%%Default part of the option %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  Def=1-exp(-integral(bfunc,0,T))*(x^2/tau)^(1/(2*abs(beta)))*momentum(-1/(2*abs(beta)),delta,x^2/tau);
  Put_def=K.*exp(-integral(rfunc,0,T)).*Def;
  Put=Put_clean+Put_def;
  end

Last one

function y=Phiminus(p,k,delta,alpha)
temp1=2^p*exp(-alpha/2);
tol=10^(-60);
%if I set here smaller tolarenace value, my calibrated parameters significantly differ from the real ones (0.6 insteadof 1). But matlab does it in 4 seconds, so it should be ok.
n=0;
x=1;
%f=zeros(10,length(k));
    while(x>tol)
    f(n+1,:)=temp1*(alpha/2)^n*gamma(p+delta/2+n).*...
gammainc(k/2,p+delta/2+n)./(gamma(n+1)*gamma(delta/2+n));
    x=f(n+1,1);
    n=n+1;
    end
z=sum(f);
y=z';
end

Hope it's readable and not too long. I would really appreciate any help.

Matt J 2013-7-6

编辑：Matt J 2013-7-6

OK. Well, given it's complexity, it might be easier to help if you write the function you're trying to minimize in mathematical form, rather than code form.

For now, I'll just remark that non-differentiable operations like abs(beta) are probably "illegal", since lsqnonlin uses smooth algorithms.

Daniel 2013-7-6

编辑：Daniel 2013-7-6

http://postimg.org/image/8bhnhe5h9/

this is the function that has two parameters beta and с (с is inside the delta)

r(u) and b(u) are constants

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Answer 1

Matt J 2013-7-6

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/81282-matlab-doesn-t-calibrate-via-lsqnonlin-when-initial-values-depart-from-the-actual-parameters#answer_91006

It sounds like you have a highly ill-conditioned problem which could be slowing convergence. The non-differentiability menationed in my Comment above might also be a factor.

Because it's only a 2-variable problem, you can probably investigate what's going on by making a surf() plot of the objective function (the squared residual) and see if it's shaped funny.

It might also help to terminate the algorithm early, using MaxIterations or similar, and seeing where it's getting stuck. Then, look for this location on the surf plot and see if the local shape of the surface is behaving oddly there.

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Matt J 2013-7-6

Did you locate which line?

Daniel 2013-7-6

编辑：Daniel 2013-7-7

在 MATLAB Online 中打开

it stops here

%%%%%%%%%%%%%Default part of the option %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Def=1-exp(-integral(bfunc,0,T))*...(x^2/tau)^(1/(2*abs(beta)))*momentum(-1/(2*abs(beta)),delta,x^2/tau);

The problem is in hypergeom function, for some values of the parameters it gives NaN

That's actually very bad for the calibration, because I don't know which upper and lower bounds and initial values I should set...

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