Help with a minimize problem

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Roberto López
Roberto López 2017-3-5
编辑： Walter Roberson 2017-3-6
f
unction b = two_var(v)
x = v(1);
y = v(2);
b = 242*log(1-x-y)+120*log(2*x-x^2-2*x*y)+79*log(2*y-y^2-2*x*y)+33*log(2*x*y);
end
v = [0.5,0.5]
a = fminsearch(@two_var,v);
I'm trying to minimize this function with the next code. The answer is x=0.247 and y=0.173
I leave the link where the problem comes from. Page 4
https://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-fall-2006/lecture-notes/lecture12.pdf

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采纳的回答

John D'Errico
John D'Errico 2017-3-5
编辑：John D'Errico 2017-3-5
Time to think about your function. ALWAYS do that. Plot it if possible.
fun = @(x,y) 242*log(1-x-y)+120*log(2*x-x^2-2*x*y)+79*log(2*y-y^2-2*x*y)+33*log(2*x*y);
fun(.5,.5)
ans =
-Inf
fun(-1,-1)
ans =
609.016175390547 + 625.176938064369i
fun(.247, .173)
ans =
-455.717889710442
So some values of x and y generate -inf. Some generate complex numbers. That is completely expected, since the log function does nasty stuff at 0 or negative values. Well, nasty in terms of what fminsearch will expect.
Basic rule: fminsearch expects a continuous, real valued function of the inputs.
If that presumption fails, then expect all hell to break loose in the eyes of fminsearch.
ezsurf(fun)
So, what do we see? First, the function is certainly unbounded from below, going to -inf along the line
x + y = 1
Next, it appears to have a MAXIMUM at the location you describe.
So honestly, I think you are confused. Are you trying to MAXIMIZE this function?
fminseach is a MINIMIZATION tool. You can make it maximize by negating the function as returned. It still minimizes, but the negative of your original function, so a maximum.
fun2 = @(xy) -fun(xy(1),xy(2));
[xymax,fmax] = fminsearch(fun2,[.2 .2])
xymax =
0.246457832567394 0.17315985403955
fmax =
455.717447610379
  1 个评论
Roberto López
Roberto López 2017-3-6
Thank you sincerly, I was completely confused. It's the solution I was looking for.

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更多回答（2 个）

kowshik Thopalli
kowshik Thopalli 2017-3-5
I dont understand.
two_var =@(x) 242*log(1-x(1)-x(2))+120*log(2*x(1)-x(1)^2-2*x(1)*x(2))+79*log(2*x(2)-x(2)^2-2*x(1)*x(2))+33*log(2*x(1)*x(2));
v = [0.5,0.5];
x = fminsearch(fun,v)
this gives me x =
0.525000000000000 0.500000000000000
You forgot to attach the link. Changing v would give a different answer. What is it that you exactly want to do?
  1 个评论
Roberto López
Roberto López 2017-3-5
https://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-fall-2006/lecture-notes/lecture12.pdf
Sorry this is the link. Is for finding numerically the Maximum likelihood estimator for a chi square test. The sum of p+q+r= 1, r is described by p and q as r=1-p-q. The solution =.5 and 0.52 is incorrect because they must sum less than 1. With other values for example [.3, .2] The function does not work even if I increase MaxFunEvals option as Matlab suggest. Do you know how can I solve it?

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Roger Stafford
Roger Stafford 2017-3-5
It is quite possible for functions such as yours to have more than one point of local minimum. If that is the case, the one that fminsearch will arrive at will depend on the given initial estimate. You need to try a different initial estimate if you are to arrive at the value you refer to in the MIT article.
  1 个评论
Roberto López
Roberto López 2017-3-5
Yeah, I've tried with values close to that given by the article but message appears about MaxFunEvals must be increase even If I do it. It's still not working, maybe the problem looks to be singular but I don't know how these values in the article were obtain.

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