我做了一个用MLE方法模拟估计参数的算法，使用函数fmincon 然后将似然函数写出解析形式带入来估计参数，但是结果都为NaN，哪位大侠帮忙解答一下？？谢谢！程序如下
b=20;A=[0,0,0,0,0,0];
x0 = [-100;-100;1;-1;-1;-1];    % Starting guess at the solution
[x1,fval] = fmincon(@funy2,x0,A,b);
disp(x1);


function f = funy2(x)
%
global Y X;
%
randn('seed', 0)            
%
n=200;

beta=[2 4 3 3] ;%
X=ones(n,4);
Y=ones(n,1);


for i=1:n
X(i,1) = 1 + 2.*randn(1);%x1t
X(i,2) = 1 + 1*randn(1);  %x2t
X(i,3)=X(i,1)*beta(1)+beta(2);%cd
X(i,4)=X(i,2)*beta(3)+beta(4);%cs
Y(i,1) =min(X(i,3),X(i,4));    %qt
end

m=200;

function3=0;
for j=1:m

    fc1=(1/x(1))*exp(-(1/(2*(x(1)^2)))*((Y(j)-x(4)-X(j,1)*x(3)).^2));
    fc2=(1/x(2))*exp(-(1/(2*(x(2)^2)))*((Y(j)-x(6)-X(j,2)*x(5)).^2));


    fa = @(h) (1/x(1))*exp(-(1/(2*(x(1)^2)))*((h-x(4)-X(j,1)*x(3)).^2));
    Fa = quad(fa,Y(j),inf);


    fb = @(g) (1/x(2))*exp(-(1/(2*(x(2)^2)))*((g-x(6)-X(j,2)*x(5)).^2));
    Fb = quad(fb,Y(j),inf);

    function1=fc1*Fb+fc2*Fa;
    function2=log(function1);
    function3=function3+function2;
end

      f=-function3;

end