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

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JICHAO ZHANG
JICHAO ZHANG 2023 年 4 月 15 日
コメント済み: JICHAO ZHANG 2023 年 4 月 16 日
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
  2 件のコメント
Cris LaPierre
Cris LaPierre 2023 年 4 月 15 日
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 2023 年 4 月 16 日
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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回答 (1 件)

Image Analyst
Image Analyst 2023 年 4 月 15 日
Are you 100% sure you need symbolic variables? Why can't you do it numerically with regular double variables? Try it.
  1 件のコメント
JICHAO ZHANG
JICHAO ZHANG 2023 年 4 月 16 日
i would like to double integral.
do you mean that i cancle the trapz commention to sum value?

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