Unable to perform assignment because value of type 'sym' is not convertible to 'double'.

조회 수: 14 (최근 30일)
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
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.
JICHAO ZHANG
JICHAO ZHANG 2023년 4월 16일
Error using trapz (line 66)
Point spacing must be a scalar specifying uniform spacing or a vector of x-coordinates for each data point.
Error in timerCALLfinite (line 46)
U=trapz(0:0.001:1,trapz(0:0.001:1,infi.*inverlap1(LAP,alpha,R,ztau,br),2));

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답변 (1개)

Image Analyst
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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