求E1_mean的程序没问题，function定义也没问题，问题出在fsolve。fsolve是为了求每个E1_mean元素对应的x(1),x(2),也就是V和sigmaV，E1_mean包含10000个数据，怎样才能一次性求出对应的10000对V和sigmaV? matlab为R2013b。程序如下：
global E1_mean
beta01=zeros(1,10000);beta01(1,: )=0.00000901;beta1=0.93095390;beta2=0.06165110;r=zeros(1,10000);r(1,: )=0.00196230； theta=zeros(1,10000);theta(1,: )=-0.00032120 ;path1=zeros(245,10000); path1(1,: )=0.001218392 ^2;R1=zeros(245,10000);
for i=2:245
path1(i,: )=beta01+beta1*path1(i-1,: )+beta2*path1(i-1,: ).*(randn(1,10000)-theta).^2 ;
R1(i,: )=r+(path1(i,: ).^0.5).*randn(1,10000);
end
R1_cumsum=cumsum(R1,1);
E1=zeros(244,10000);
for i=1:244
E1(i,: )=180970333.4*exp(R1_cumsum(i+1,: ));
end
E1_mean=mean(E1,1);
function F = blsprice(x)
global E1_mean j
% x(1) : V; 银行总资产
% x(2) : sigmav; 资产波动率
rf = 3.6/100; %无风险利率
B = 2055510000000.00 /(10^12) ; %股权的执行价格 B/（10^12）
sigmaE = 0.545240875 ;
T = 1; % 时间周期
t = 0; % 时间
for j=1:10000
d1 = (log(x(1)/(0.97*B))+rf*(T-t)+x(2)^2*(T-t)/2) / (x(2)*sqrt(T-t));
d2 = d1-x(2)*sqrt(T-t);
F(1) = x(1)* normcdf(d1) - 0.97*B* normcdf(d2) -(E1_mean(1,j))/10^12;
F(2) = x(1)*normcdf(d1)*x(2) - sigmaE*(E1_mean(1,j))/10^12;
end
global j
V1=zeros(1,10000);
sigmaV1=zeros(1,10000);
for j=1:10000
x0 = [1.2;1];
options=optimset('Display','off','MaxFunEvals',10000,'MaxIter',10000);
[x,fval] = fsolve(@blsprice,x0,options);
V1(1,j)=x(1);
sigmaV1(1,j)=x(2);
end
程序运行后显示Error using erfc
Input must be real and full.
Error in normcdf>localnormcdf (line 124)
p(todo) = 0.5 * erfc(-z ./ sqrt(2));
Error in normcdf (line 46)
[varargout{1:max(1,nargout)}] =localnormcdf(uflag,x,varargin{:});
Error in blsprice (line 20)
F(1) = x(1)* normcdf(d1) - 0.97*B*normcdf(d2) -(E1_mean(1,j))/10^12;
Error in trustnleqn (line 200)
F= feval(funfcn{3},reshape(xTrial,sizes.xRows,sizes.xCols),varargin{:});
Error in fsolve (line 366)
[x,FVAL,JACOB,EXITFLAG,OUTPUT,msgData]=...
Error in solve_blsprice (line 9)
[x,fval] = fsolve(@blsprice,x0,options);
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