How to find argmax for the function below?

2020 1 29
1 个回答
更新时间：2020 1 30
11 次查看（30 天）

0 个投票

Below are my codes. The LLF function has two unknowns, PD and rho, I make them into x variable that has x(1) and x(2) in the function. I tried to use fminsearch(-LLF) to find the argmax but it did not work. I used the fzero instead, but it gives me an error: Error using fzero (line 246)
FZERO cannot continue because user-supplied function_handle ==>...........
Index exceeds the number of array elements (1).
% x(1) = PD, x(2) = rho
r1 = 0.051;
r2 = 0.27;
r3 = 0.037;
r4 = 0.116;
r5 = 0.222;
r6 = 0.121;
r7 = 0.026;
r8 = 0.025;
r9 = 0.02;
r10 = 0.14;
LLF = @(x) log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r1))).*normpdf((sqrt(1-x(2)).*norminv(r1)-norminv(x(1)))./sqrt(x(2)))) +...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r2))).*normpdf((sqrt(1-x(2)).*norminv(r2)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r3))).*normpdf((sqrt(1-x(2)).*norminv(r3)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r4))).*normpdf((sqrt(1-x(2)).*norminv(r4)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r5))).*normpdf((sqrt(1-x(2)).*norminv(r5)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r6))).*normpdf((sqrt(1-x(2)).*norminv(r6)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r7))).*normpdf((sqrt(1-x(2)).*norminv(r7)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r8))).*normpdf((sqrt(1-x(2)).*norminv(r8)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r9))).*normpdf((sqrt(1-x(2)).*norminv(r9)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r10))).*normpdf((sqrt(1-x(2)).*norminv(r10)-norminv(x(1)))./sqrt(x(2))));
x0 = [0,0];
x = fzero(LLF,x0);

3 个评论
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John D'Errico
John D'Errico 2020-1-29
Um, fzero is not a minimizer. It is a root finder, that only works in ONE variable, NOT two. You cannot use fzero.
You have two variables, so you cannot use a tool like fminbnd.
You CAN use fminsearch. It is a minimization tool. So all you need to do is negate the function. Why it did not work is a mystery, since you have not accurately told us what you did.
Yijia Qiao
Yijia Qiao 2020-1-29
So I changed it to negative function. It gives me an exiting message as follows: Exiting: Maximum number of function evaluations has been exceeded
- increase MaxFunEvals option.
Current function value: Inf
% x(1) = PD, x(2) = rho
r1 = 0.051;
r2 = 0.27;
r3 = 0.037;
r4 = 0.116;
r5 = 0.222;
r6 = 0.121;
r7 = 0.026;
r8 = 0.025;
r9 = 0.02;
r10 = 0.14;
LLF = @(x) -log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r1))).*normpdf((sqrt(1-x(2)).*norminv(r1)-norminv(x(1)))./sqrt(x(2)))) -...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r2))).*normpdf((sqrt(1-x(2)).*norminv(r2)-norminv(x(1)))./sqrt(x(2)))) - ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r3))).*normpdf((sqrt(1-x(2)).*norminv(r3)-norminv(x(1)))./sqrt(x(2)))) - ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r4))).*normpdf((sqrt(1-x(2)).*norminv(r4)-norminv(x(1)))./sqrt(x(2)))) - ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r5))).*normpdf((sqrt(1-x(2)).*norminv(r5)-norminv(x(1)))./sqrt(x(2)))) - ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r6))).*normpdf((sqrt(1-x(2)).*norminv(r6)-norminv(x(1)))./sqrt(x(2)))) - ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r7))).*normpdf((sqrt(1-x(2)).*norminv(r7)-norminv(x(1)))./sqrt(x(2)))) - ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r8))).*normpdf((sqrt(1-x(2)).*norminv(r8)-norminv(x(1)))./sqrt(x(2)))) - ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r9))).*normpdf((sqrt(1-x(2)).*norminv(r9)-norminv(x(1)))./sqrt(x(2)))) - ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r10))).*normpdf((sqrt(1-x(2)).*norminv(r10)-norminv(x(1)))./sqrt(x(2))));
x0 = [0,0];
x = fminsearch(LLF,x0);
Walter Roberson
Walter Roberson 2020-1-30
You need to add options like I showed.

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回答（1 个）

Walter Roberson
Walter Roberson 2020-1-29

0 个投票

r1 = 0.051;
r2 = 0.27;
r3 = 0.037;
r4 = 0.116;
r5 = 0.222;
r6 = 0.121;
r7 = 0.026;
r8 = 0.025;
r9 = 0.02;
r10 = 0.14;
LLF = @(x) log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r1))).*normpdf((sqrt(1-x(2)).*norminv(r1)-norminv(x(1)))./sqrt(x(2)))) +...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r2))).*normpdf((sqrt(1-x(2)).*norminv(r2)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r3))).*normpdf((sqrt(1-x(2)).*norminv(r3)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r4))).*normpdf((sqrt(1-x(2)).*norminv(r4)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r5))).*normpdf((sqrt(1-x(2)).*norminv(r5)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r6))).*normpdf((sqrt(1-x(2)).*norminv(r6)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r7))).*normpdf((sqrt(1-x(2)).*norminv(r7)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r8))).*normpdf((sqrt(1-x(2)).*norminv(r8)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r9))).*normpdf((sqrt(1-x(2)).*norminv(r9)-norminv(x(1)))./sqrt(x(2)))) + ...
log(sqrt(1-x(2))./(sqrt(x(2)).*normpdf(norminv(r10))).*normpdf((sqrt(1-x(2)).*norminv(r10)-norminv(x(1)))./sqrt(x(2))));
x0 = [0,0];
LLFmax = @(x) -LLF(x);
options = optimset('MaxFunEvals', 1e6, 'MaxIter', 1e6);
[bestx, fval] = fminsearch(LLFmax, x0, options)
when it eventually stops complaining it ran out of iterations, it will have fval of Inf and the first element of the output will be 0. Tis reflects that if you set x(1) to be 0 then you get out complex infinities, or NaN. The maximum is not well defined unless you add constraints.

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