Unable to perform assignment because value of type 'sym' is not convertible to 'double'.
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       이전 댓글 표시
    
function [U]=timerpathCALLfinite(s0,v0,sigma,kappa,K,B)%varibles
s0=100;
v0=0.001;
K=100;
T=2;
steptime=256;
dt=T/256;
B=0.001;
r=0.01;
sigma=0.25;% not larger then 0.1
kappa=0.1;
rho=0.5;
v=zeros(T*steptime);%volatility price
v(1)=v0;%initial price;
zpath(1)=(2*sqrt(v0))/sigma;
deta=(4*kappa)/(sigma^2); % Bessel process parameter
nu=deta/2-1;  % bessel model index
sum(1)=0;
p(1)=0;
for j=1:steptime*T% search stopping time
    v(j+1)=v0*exp((kappa-0.5*sigma^2)*(j*dt)-sigma*sqrt(j*dt)); %volatility
    sum=sum+dt*((v(j)+v(j+1))/2);% cumulation volitility
    if sum>=B
        tau=j+1;
        br=tau*dt;
        break;
    end
end
if sum>=B % at stopping time exercise
    syms ztau R 
    %%insert expectation
    d=sqrt((1-rho^2)*B);
    vaps=-rho*(2*kappa/(sigma^2)-0.5).*(0.5*(1./ztau+1/zpath(1)).*B)+(r.*br)-(B/2)+rho.*(ztau-zpath(1));
    d1=((log(s0/K)+vaps+(1-rho^2)*B))./d;
    d2=d1-d;
    infi=s0.*exp(vaps-r.*br).*normcdf(d1)-K.*exp(-r.*br).*normcdf(d2);  %inner expectation
    %%intervsion laplace transform to density
    alpha=18.4/(2*B);  % laplace parameter
    sh=sinh(R.*sqrt(alpha./2)); % large value
    pe=besseli(sqrt(alpha),2.*sqrt(2.*alpha.*zpath(1).*ztau)./sh);%bessel function
    LAP=real(pe.*sqrt(2.*alpha).*(ztau.^(nu+1))./(sh.*8*(zpath(1)^(nu))).*exp(-((nu^2)*br*(sigma^2))/8-((zpath(1)+ztau).*sqrt(2*alpha).*coth(R.*sqrt(alpha/2)))));
    %%% DOUBLE integral outer expectation
    U=trapz(0:1,trapz(0:1,infi.*inverlap1(LAP,alpha,R,ztau,br),2));
    function[disthree]=inverlap1(LAP,alpha,R,ztau,br)
        v0=0.001;
        B=0.001;
        sigma=0.25;% not larger then 0.1
        kappa=0.1;
        v(1)=v0;%initial price;
        zpath(1)=(2*sqrt(v0))/sigma;
        deta=(4*kappa)/(sigma^2); % Bessel process parameter
        nu=deta/2-1;  % bessel model index
        H2=LAP/2;
        for t=1:1:15
            alphaP=complex(alpha,(t*pi)/B);
            sh1=sinh(R.*sqrt(alphaP/2));
            pe1=besseli(sqrt(alpha+(t*pi)/B),2.*sqrt(2.*alphaP.*zpath(1).*ztau)./sh1);
            H2=H2+((-1)^t).*real(pe1.*sqrt(2.*alphaP).*(ztau.^(nu+1))./(sh1.*8.*(zpath(1)^(nu))).*exp(-((nu^2).*br.*(sigma^2))./8-((zpath(1)+ztau).*sqrt(2.*alphaP).*coth(R.*sqrt(alphaP./2)))));
        end
        SU=zeros(12);
        SU(1)=H2;
        for I=1:12
            NT=15+I;
            alpha2=complex(alpha,(NT*pi)/B);
            sh2=sinh(R.*sqrt(alpha2/2));
            pe2=besseli(sqrt(alpha+(NT*pi)/B),2.*sqrt(2.*alpha2.*zpath(1).*ztau)./sh2);
            SU(I+1)=SU(I)+((-1)^NT).*real(pe2.*sqrt(2.*alpha2).*(ztau.^(nu+1))./(sh2.*8.*(zpath(1).^(nu))).*exp(-((nu^2).*br.*(sigma^2))./8-((zpath(1)+ztau).*sqrt(2.*alpha2).*coth(R.*sqrt(alpha2./2)))));
        end
        AVGSU=0;
        C=[1,11,55,165,330,462,462,330,165,55,11,1];
        for J=1:12
            AVGSU=AVGSU+C(J)*SU(J);
        end
        U=exp(18.4/2)/B;
        disthree=(U.*AVGSU)./2048;
    end
댓글 수: 2
  Cris LaPierre
    
      
 2023년 4월 15일
				Please share the full error message (all the red text).
Note that the code you have shared here cannot be run due to syntax errors.
답변 (1개)
  Image Analyst
      
      
 2023년 4월 15일
        Are you 100% sure you need symbolic variables?  Why can't you do it numerically with regular double variables?  Try it.
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