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# optimization routine for function that is not smooth and also has a lot of local minimum

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xueqi
on 18 Feb 2013

### Answers (5)

Alan Weiss
on 19 Feb 2013

If you want to minimize a nonsmooth function, the general recommendation is to try patternsearch. If there are many local minima, the general recommendation is to use a variety of start points. If you have finite bounds on all components, lb and ub, then you can obtain random start points as follows:

x0 = lb + rand(size(lb)).*(ub - lb);

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

##### 4 Comments

xueqi
on 20 Feb 2013

Hi. These are the result I have got when using fmincon...Firstly I thought that my function was not smooth, but now by looking at the first condition (always very big for several iterations...) I begin to think maybe it is just somehow that matlab just can't find the minimum...Am I right?

sumllh =

1.7499e+003

sumllh =

1.7499e+003

sumllh =

1.7499e+003

sumllh =

1.7499e+003

sumllh =

1.7499e+003

sumllh =

1.7499e+003

sumllh =

1.7499e+003

sumllh =

1.7499e+003

sumllh =

1.7499e+003

First-order Norm of

Iter F-count f(x) Feasibility optimality step

0 9 1.749900e+003 0.000e+000 3.433e+005

sumllh =

Inf

sumllh =

Inf

sumllh =

Inf

sumllh =

398.8533

sumllh =

398.8533

sumllh =

398.8533

sumllh =

398.8533

sumllh =

398.8533

sumllh =

398.8533

sumllh =

398.8533

sumllh =

398.8533

sumllh =

398.8533

User objective function returned Inf; trying a new point... 1 21 3.988533e+002 0.000e+000 8.908e+003 1.242e-001

sumllh =

4.0347e+003

sumllh =

1.4260e+003

sumllh =

647.5076

sumllh =

437.0004

sumllh =

4.0665e+003

sumllh =

1.4419e+003

sumllh =

643.5412

sumllh =

418.2376

sumllh =

365.8173

sumllh =

365.8173

sumllh =

365.8173

sumllh =

365.8173

sumllh =

365.8174

sumllh =

365.8173

sumllh =

365.8173

sumllh =

365.8173

sumllh =

365.8173

2 38 3.658173e+002 0.000e+000 7.766e+002 7.457e-002

sumllh =

597.1229

sumllh =

434.6045

sumllh =

368.9302

sumllh =

356.8667

sumllh =

356.8667

sumllh =

356.8667

sumllh =

356.8667

sumllh =

356.8667

sumllh =

356.8667

sumllh =

356.8667

sumllh =

356.8667

sumllh =

356.8667

3 50 3.568667e+002 0.000e+000 4.334e+002 5.845e-002

sumllh =

889.8367

sumllh =

516.4006

sumllh =

311.5622

sumllh =

311.5622

sumllh =

311.5622

sumllh =

311.5622

sumllh =

311.5622

sumllh =

311.5622

sumllh =

311.5622

sumllh =

311.5622

sumllh =

311.5622

4 61 3.115622e+002 0.000e+000 4.625e+002 5.322e-001

sumllh =

Inf

sumllh =

235.7194

sumllh =

235.7194

sumllh =

235.7194

sumllh =

235.7194

sumllh =

235.7194

sumllh =

235.7194

sumllh =

235.7194

sumllh =

235.7194

sumllh =

235.7194

User objective function returned Inf; trying a new point... 5 71 2.357194e+002 0.000e+000 3.726e+003 3.923e+000

sumllh =

Inf

sumllh =

206.5185

sumllh =

206.5185

sumllh =

206.5185

sumllh =

206.5185

sumllh =

206.5185

sumllh =

206.5185

sumllh =

206.5185

sumllh =

206.5185

sumllh =

206.5185

User objective function returned Inf; trying a new point... 6 81 2.065185e+002 0.000e+000 5.962e+002 2.708e+000

sumllh =

183.2263

sumllh =

183.2263

sumllh =

183.2263

sumllh =

183.2263

sumllh =

183.2263

sumllh =

183.2263

sumllh =

183.2263

sumllh =

183.2263

sumllh =

183.2263

7 90 1.832263e+002 0.000e+000 5.587e+002 6.503e-001

sumllh =

248.7550

sumllh =

190.8685

sumllh =

182.2106

sumllh =

182.2106

sumllh =

182.2106

sumllh =

182.2106

sumllh =

182.2106

sumllh =

182.2106

sumllh =

182.2106

sumllh =

182.2106

sumllh =

182.2106

8 101 1.822106e+002 0.000e+000 5.133e+002 6.867e-002

sumllh =

210.8628

sumllh =

183.1427

sumllh =

179.3117

sumllh =

179.3117

Alan Weiss
on 20 Feb 2013

It seems that your objective function returns its value whenever it is evaluated. The reason there are so many function evaluations between each line of the iterative display is that fmincon is taking small perturbations of the curent point and evaluating the objective at these new points. You see there are many times where the small perturbation in the objective leads to an Inf or a big departure from the current function value.

I conclude that your objective function is not finite at points very close to the current point after the first iteration.

Can you put some constraints on your problem so that the objective doesn't get so large? Or can you check to ensure that the objective function is defined correctly?

And one more thing. I see that you have tried to use GA to optimize your function. I strongly recommend that you use patternsearch instead--it is much faster and more reliable.

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

xueqi
on 20 Feb 2013

xueqi
on 1 Mar 2013

Mohsen Davarynejad
on 18 Feb 2013

##### 5 Comments

xueqi
on 18 Feb 2013

Mohsen Davarynejad
on 18 Feb 2013

Mohsen Davarynejad
on 18 Feb 2013

xueqi
on 20 Feb 2013

I have 8 independent variables to estimate. And this is the code I use for coding ga

if true

% A=[1 1 1 0 0 0 0 0];

b=[1];

lb=[0.1;0.1;0.1;0.0001;1.1;0.001;1.1;0.001];

ub=[1;1;1;1;Inf;1;Inf;1];

options=optimset('Algorithm','interior-point','Display','iter','MaxIter',50,'TolFun', 1e-1);

[x,fval,exitflag,output,lambda,grad] = fmincon(@beta,[0.2,0.2,0.2,0.01,100,0.1,100,0.1],A,b,[],[],lb,ub,[],options)

%x= ga(@beta,8,A,b,[],[],lb,ub,[],options)

end

xueqi
on 20 Feb 2013

Juan Camilo Medina
on 2 Mar 2013

Edited: Juan Camilo Medina
on 2 Mar 2013

Based on my experience, the best way is to use simulated annealing, which is less heuristic than GA and it's powered by a Markov chain. fmincon() or any other gradient-based algorithm won't work well since the function is non-smooth.

[x,fval,exitFlag,output] = simulannealbnd(ObjectiveFunction,X0)

you can also try

x = fminsearch(fun,x0)

##### 1 Comment

Matt J
on 3 Mar 2013

Edited: Matt J
on 3 Mar 2013

I mentioned this to you in another post, but I'll document it here as well. I think the best way to deal with an ill-behaved objective function is to replace it with a better one. I.e., question your initial reasoning behind this function and see if there are alternative formulations that you haven't considered.

If nothing else, it is suspicious that you think you absolutely must accept the non-smoothness of the function. If you've thought about the problem formulation carefully enough, there are almost always smooth approximations that you can find. The local minima are another matter.

If you can't see an alternative yourself, it may be time to walk us through the development of the function you're thinking of so we can see where you might have taken a wrong turn.

##### 41 Comments

xueqi
on 3 Mar 2013

Matt J
on 3 Mar 2013

xueqi
on 3 Mar 2013

Edited: xueqi
on 3 Mar 2013

This is the objective function

if true

% function [ sumllh] = beta1 (x)

DD=xlsread('parameters.xlsx');

rn=size(DD,1);

edw=DD(:,7);

lb1=x(1);

lb2=x(2);

lb3=x(3);

r=x(4);

sub =[lb1,lb2,lb3,r];

s1=x(5);

b1=x(6);

s2=x(7);

b2=x(8);

mu = maxmin(sub,DD); save mu; hatmu=xlsread('hatmu.xlsx'); % assign the simulated mu to hatmu %hatmu=simbeta(39);

for i=1:rn alpha1(i)= (b1*edw(i)/2+(1-b1)*mu(i,1))*(s1-1)/edw(i); beta1(i)= (edw(i)-b1*edw(i)/2-(1-b1)*mu(i,1))*(s1-1)/edw(i); alpha2(i)= (b2*edw(i)/2+(1-b2)*mu(i,2))*(s2-1)/edw(i); beta2(i)= (edw(i)-b2*edw(i)/2-(1-b2)*mu(i,2))*(s2-1)/edw(i); llh1(i,:)=betacdf((hatmu(i,1)+0.5)/edw(i,:),alpha1(i),beta1(i))-betacdf((hatmu(i,1)-0.5)/edw(i,:),alpha1(i),beta1(i)); llh2(i,:)=betacdf((hatmu(i,2)+0.5)/edw(i,:),alpha2(i),beta2(i))-betacdf((hatmu(i,2)-0.5)/edw(i,:),alpha2(i),beta2(i)); %llh(i,:)=log(llh1(i,:)*llh2(i,:)) llh(i,:)=llh1(i)*llh2(i); save llh end %llh(i,:)=log( mvnpdf(hatmu(i,:),mu(i,:),[s1 s12; s12 s2])) % mallh(i,:)=mvnpdf(hatmu(i,:),mu(i,:),[s1^2 s12; s12 s2^2]) % save mallh %save llh

sumllh_=sum(log(llh));

if (sumllh_==-Inf)

sumllh_=-1000;

end

%because we need to maximize this but fmincon is for minimization so we %need to minus it sumllh=-sumllh_ end

xueqi
on 3 Mar 2013

and this is all the constaint for all the parameters

if true

% A=[1 1 1 0 0 0 0 0];

b=[1];

lb=[0.1;0.1;0.1;0.0001;500;0.001;500;0.001];

ub=[1;1;1;1;Inf;0.6;Inf;0.6];

options=optimset('Algorithm','interior-point','Display','iter','MaxIter',20);

x0=[0.3,0.3,0.3,0.08,2000,0.05,2000,0.05];

[x,fval,exitflag,output,lambda,grad] = fmincon(@beta1,x0,A,b,[],[],lb,ub,[],options)

end

xueqi
on 3 Mar 2013

The objective funtion is for calculating the sum of log likelihood. I am assuming two variables as independent beta distribution(with some modification). hatmu is the simulated data(or experimental data). maxmin is a function used to calulate the two variables for different parameters-sub, which contains 4 parameters. So basically beta1 is to calculate the likelihood for the 4 parameters plus anther 4 parameters of noise. I want to maximize the sum of log likelihood to get the 8 parameters.

When fmincon is search for different values, it is easily to get a extremely small likelihood, so the log likelihood is going to Inf.

Matt J
on 3 Mar 2013

I'm still waiting for the formula for the objective function, as opposed to the code. Regardless, you haven't formatted the for-loop, so your objective function code is hard to read.

In any case, I have 2 remarks

- It looks like you are minimizing a loglikelihood when (if maximum likelihood estimation is the goal) you should be minimizing the negative of the loglikelihood.
- The Infs are probably coming from taking the log of the likelihood in regions of low probability. One way to eliminate them would be to minimize minus the likelihood, instead of minus the loglikelihood.

xueqi
on 3 Mar 2013

Edited: xueqi
on 3 Mar 2013

Yes I am using maximum likelihood estimation. And I add a minus to the sum of log likelihood and then minimize it. I am trying your remarks 2 now. The formulae is

if true

% llh1=betacdf(hatmu1+0.5)-betacdf(hatmu1-0.5);

llh2=betacdf(hatmu2+0.5)-betacdf(hatmu2-0.5)

llh=llh1*llh2;

sumllh= multiplication of llh for all the 39 set of data

end

Matt J
on 3 Mar 2013

Edited: Matt J
on 3 Mar 2013

I didn't think you were still using FMINCON. Everyone's been telling you to use simulated annealing and pattern search.

In any case, you should be constraining hatmu2>= -0.5, because

- That's the region where probabilities are >0 and so where the solution lies.
- When working with loglikelihoods, instead of likelihoods, that's where the objective function is defined.

Matt J
on 3 Mar 2013

BTW you've shown llh1 really commented out

% llh1=betacdf(hatmu1+0.5)-betacdf(hatmu1-0.5);

So if this is not the formula for llh1 that you're using, what formula are you using instead?

xueqi
on 3 Mar 2013

Edited: xueqi
on 3 Mar 2013

If I change the log of likelihood to likelihood, then fmincon just stops at the start point since all the point around the function value is equal to 0 so the gradient is equal to 0.

Is there anyway to improve the precision of matlab? I think it might assume a extremely small number to 0.

Matt J
on 3 Mar 2013

Edited: Matt J
on 3 Mar 2013

Keep up with my posts! I told you that you need to be using constraints that keep the algorithm away from this region.

Incidentally, is there any reason why you're using a finite difference of betacdf

llh=betacdf(hatmu+0.5)-betacdf(hatmu-0.5);

to approximate the beta distribution, when the density function is given exactly and trivially by

hatmu.^(a-1)*(1-hatmu)^(b-1)/beta(a,b)

Matt J
on 3 Mar 2013

Edited: Matt J
on 3 Mar 2013

Reprinting your function with more readable formatting

for i=1:rn

alpha1(i)= (b1*edw(i)/2+(1-b1)*mu(i,1))*(s1-1)/edw(i);

beta1(i)= (edw(i)-b1*edw(i)/2-(1-b1)*mu(i,1))*(s1-1)/edw(i);

alpha2(i)= (b2*edw(i)/2+(1-b2)*mu(i,2))*(s2-1)/edw(i);

beta2(i)= (edw(i)-b2*edw(i)/2-(1-b2)*mu(i,2))*(s2-1)/edw(i);

llh1(i,:)=betacdf((hatmu(i,1)+0.5)/edw(i,:),alpha1(i),beta1(i))-... betacdf((hatmu(i,1)-0.5)/edw(i,:),alpha1(i),beta1(i));

llh2(i,:)=...

betacdf((hatmu(i,2)+0.5)/edw(i,:),alpha2(i),beta2(i))-

betacdf((hatmu(i,2)-0.5)/edw(i,:),alpha2(i),beta2(i));

end

xueqi
on 3 Mar 2013

Edited: xueqi
on 3 Mar 2013

- I am trying patternsearch now
- hatmu2 is >=0.5 in my data set. Both hatmu1 and hatmu2 >=0 and <= edw(a specific number). Is this what you meaning here?
- I am using llh1=betacdf(hatmu1+0.5)-betacdf(hatmu1-0.5). Some of the comments are nonsense I just forget to delete it...
- Patternsearch seems a a little bit slow. Is there anyway to speed it up?

Matt J
on 3 Mar 2013

Edited: Matt J
on 3 Mar 2013

Now that I can read your function, I can make more complete recommendations.

First, I think you should go back to loglikelihoods.

Second, you also have to make sure that you have no hatmu that would cause betacdf to be zero. Remember, the beta distribution only has non-zero probability for data values between 0 and 1.

Third, it appears that the dependence on your unknown parameters are entirely through alpha1(i), alpha2(i), beta1(i), beta2(i) and this dependence is linear. You need to ensure that the alphas and betas are constrained to be non-negative, which is the assumption of the beta distribution. Since their dependence on the unknowns is linear, you can do this by specifying linear constraints.

Matt J
on 3 Mar 2013

Edited: Matt J
on 3 Mar 2013

hatmu2 is >=0.5 in my data set. Both hatmu1 and hatmu2 >=0 and <= edw(a specific number). Is this what you meaning here?

But that means that hatmu can be equal to or close to edw. When this happens

(hatmu+0.5)/edw >1

and the beta distribution is zero at such values for all parameter choices.

What is the purpose of the 0.5? Are you trying to compute finite difference derivatives of betacdf? As I said before, this is unnecessary and also might be creating problems.

Patternsearch seems a a little bit slow. Is there anyway to speed it up?

Forget patternsearch. Now that I can read your code and understand your problem, I think that FMINCON should be fine, once you apply the constraints that I've described.

xueqi
on 3 Mar 2013

What is the purpose of the 0.5?

Because in the experiment, subject can only choose integer. For example, if he wants to choose 76.4, but limited by the experiment requirement, he then needs to choose 76 instead. So the likelihood for 76 is actually from 76-0.5 to 76+0.5.

beta distribution

I am not using beta distribution exactly but use a modified version just because of the points 0 and edw. For beta distributin, at the point 0 and 1 the error should be 0, but this is not an proper distribution set up for the experiment. So I introduce the extra two parameters to solve this problem. I am not quite clear how to explain it now. I will think about now.

Second, you also have to make sure that you have no hatmu that would cause betacdf to be zero

When hatmu=edw, (hatmu(i,1)+0.5)/edw will be more than 1. But this is the cdf not the pdf, so the result should be equal to 1. And for hatmu=0, (hatmu(i,1)-0.5)/edw<0 so the result should be equal to 0. But since llh=(hatmu(i,1)+0.5)/edw-(hatmu(i,1)+0.5)/edw the total result should not be 0. So I think it is impossible for llh to get an exact 0 but may be a quite near 0 value.

Matt J
on 3 Mar 2013

xueqi
on 5 Mar 2013

Matt J
on 5 Mar 2013

Edited: Matt J
on 5 Mar 2013

It doesn't sound like a precision problem. If you have a loglikelihood term that is equal to 0, it means that you are operating in a flat region of BETACDF, which means one of the following is true

- That (hatmu+0.5)/edw <=0 or equivalently that hatmu<=-0.5,
- That (hatmu-0.5)/edw >= 1 or equivalently that hatmu>=edw+.5

You must ensure that all the hatmu lie in the "legal" open interval (-.5, edw+.5). Otherwise, it's bad data, plain and simple.

xueqi
on 5 Mar 2013

Oh But I really don't understand. See this is my hatmu1 and hatmu2

if true

% 12 8

13 9

12 8

32 25

11 14

26 30

27 28

27 30

19 44

19 42

17 40

8 19

14 22

14 22

24 30

19 18

8 9

32 41

3 3

4 5

2 3

7 7

4 4

5 4

5 4

5 6

4 4

21 15

5 6

6 3

5 3

4 4

2 21

4 4

40 4

15 17

3 20

5 22

3 22

end

Matt J
on 5 Mar 2013

xueqi
on 5 Mar 2013

Edited: xueqi
on 5 Mar 2013

x1=[0.3,0.3,0.26,0.08,2000,0.05,2000,0.05](the first 4 parameters decides mu and the next 4 is for add some noise for mu), I get mu=

6.8795 5.1597

7.6101 5.7076

7.9938 5.9954

27.2074 21.7659

7.9180 9.5016

22.5482 26.3062

22.5482 26.3062

24.2338 26.9264

17.0101 42.5253

16.3487 40.8717

15.5410 38.8526

6.6654 14.9972

13.0653 20.5312

13.2404 20.8064

24.1087 30.1359

16.1521 16.1521

1.6451 1.6451

20.8231 28.6318

0 0

0.2595 0.1946

0 0

1.5615 1.7350

0 0

0.6682 0.5011

0.4296 0.2685

1.5193 1.5193

0 0

17.5306 10.5184

2.1790 1.4752

1.8606 1.1629

0.2595 0.1946

0.2595 0.1946

0 17.0691

0.2595 0.1946

33.5384 2.2172

10.0319 12.1661

0 12.3521

3.0461 16.9758

0 17.0691

and hence alph1=

180.6209

194.4948

201.7820

566.6565

200.3422

478.1758

478.1758

510.1867

373.0055

360.4444

345.1069

176.5549

298.0915

301.4169

507.8112

356.7106

81.2166

445.4166

49.9750

54.9032

49.9750

79.6290

49.9750

62.6644

58.1336

78.8267

49.9750

382.8900

91.3545

85.3085

54.9032

54.9032

49.9750

54.9032

686.8855

240.4863

49.9750

107.8224

49.9750

and beta1=

1.0e+003 *

1.8184

1.8045

1.7972

1.4323

1.7987

1.5208

1.5208

1.4888

1.6260

1.6386

1.6539

1.8224

1.7009

1.6976

1.4912

1.6423

1.9178

1.5536

1.9490

1.9441

1.9490

1.9194

1.9490

1.9363

1.9409

1.9202

1.9490

1.6161

1.9076

1.9137

1.9441

1.9441

1.9490

1.9441

1.3121

1.7585

1.9490

1.8912

1.9490

and alpha2=

147.9594

158.3649

163.8303

463.3202

230.4157

549.5426

549.5426

561.3213

857.5512

826.1484

787.8046

334.7799

439.8723

445.0980

622.2702

356.7106

81.2166

593.7072

49.9750

53.6712

49.9750

82.9238

49.9750

59.4920

55.0741

78.8267

49.9750

249.7240

77.9891

72.0584

53.6712

53.6712

374.1265

53.6712

92.0810

281.0146

284.5482

372.3539

374.1265

and* beta2*

1.0e+003 *

1.8510

1.8406

1.8352

1.5357

1.7686

1.4495

1.4495

1.4377

1.1414

1.1729

1.2112

1.6642

1.5591

1.5539

1.3767

1.6423

1.9178

1.4053

1.9490

1.9453

1.9490

1.9161

1.9490

1.9395

1.9439

1.9202

1.9490

1.7493

1.9210

1.9269

1.9453

1.9453

1.6249

1.9453

1.9069

1.7180

1.7145

1.6266

1.6249

and the corresponding likelihood is llh1=

0.0002

0.0000

0.0210

0.0010

0.2188

0.0465

0.0039

0.1354

0.4009

0.2439

0.4309

0.2913

0.2709

0.2153

0.1532

0.1918

0.0000

0

0.4773

0.0259

0.5176

0.0000

0.0046

0.0008

0.0001

0.1032

0.0046

0.0603

0.5188

0.0055

0.0000

0.0259

0.5176

0.0259

0.0000

0.0006

0.4773

0.5614

0.4773

and llh2=

0.3898

0.1624

0.5693

0.0740

0.0039

0.0217

0.3372

0.0711

0.2187

0.2933

0.3066

0.0191

0.4106

0.3968

0.2163

0.4319

0.0000

0

0.4773

0.0000

0.4773

0.0000

0.0046

0.0886

0.0273

0.0006

0.0046

0.0044

0.0004

0.4127

0.6666

0.0175

0.0212

0.0175

0.4167

0.0013

0.0000

0.0007

0.0010

and the *sum of loglikehood *is equal to Inf.

xueqi
on 5 Mar 2013

Edited: xueqi
on 5 Mar 2013

It is not surprised that sum of loglikelihood is equal to Inf since llh1(18)=llh2(18)=0. But there is nothing abnormal for alpha1(18) beta1(18) alpha2(18) beta2(18) .Actually if I insert everything in the formulae for calculating llh1(18) I get

llh1(18)=betacdf((32+0.5)/100,445.4416,1553.6)-betacdf((32-0.5)/100,445.4416,1553.6)=0!

Matt J
on 5 Mar 2013

Edited: Matt J
on 5 Mar 2013

But there is nothing abnormal for alpha1(18) beta1(18) alpha2(18) beta2(18)

They look pretty abnormal to me. With alpha and beta on the order of 1000, it means you're working with a beta distribution that is a polynomial of degree 1000+. That's a pretty ill-behaved thing. As I said before, I think you need to be constraining your alphas and betas in some way

Also, I now see that your earlier claim, that the objective function is smooth, is not true.

mu = maxmin(sub,DD)

is not a differentiable function of x, assuming

maxmin(v)=[max(v), min(v)]

xueqi
on 5 Mar 2013

I think you need to be constraining your alphas and betas in some way

Yes I will look into alpha and beta.

Also, I now see that your earlier claim, that the objective function is smooth, is not true

Here maxmin is a just a function name given by me.

xueqi
on 5 Mar 2013

Ha...It is a long story. It finds the minimum for objective function in terms of several variables and then maximize objective function from the minimum in terms of other several variables.

I have plotted beta1 in terms of different variables. Except the fact sumllh is Inf when it is away from optimal, it seems quite smooth around the optimal.

Matt J
on 5 Mar 2013

it seems quite smooth around the optimal.

It needs to be smooth throughout the feasible region, not just at the optimum. Also, determining smoothness by inspection of plots can be tricky. For example, discontinuities can be hidden by undersampling. Also, FMINCON algorithms generally require functions to be twice continuously differentiable. Smoothness of 1st and higher derivatives are generally invisible graphically.

xueqi
on 7 Mar 2013

xueqi
on 9 Mar 2013

Edited: xueqi
on 9 Mar 2013

Not everything. But at least from the aspect of the Inf problem. Now I don't have inf when optimization routine is working. But for some simulation data it works perfectly, but for some it fails...I guess it is because of there are a lot of local minimum, so I am trying using multistart and set 1000 start points. Is there any suggestion for that?

I still think the function is smooth so still using fmincon...Though I tried pattersearch, it clearly only returns local minimum and multistart is not compatible with patternsearch.

P.S. how can i upload a graph? I think it might be easier for you to understand the objective function if I show you the plot....

Matt J
on 9 Mar 2013

xueqi
on 9 Mar 2013

Edited: xueqi
on 9 Mar 2013

No, the only thing that would help is to see maxmin itself Well, the maxmin is not my objective function(but it is the model I want to fit in), my objective function is the one calculating the likelihood. Do you suggest if maxmin is smooth then the likelihood function will also be smooth?

how did you end up getting rid of the Infs

just assign a very small number(1e-10)if one of the likelihood is equal to 0...

multi start start point Currently I am using a loop to choose different start point for fmincon. Is there anyway matlab can do the optimization for all the points at the same time. It would be much efficient...

Matt J
on 9 Mar 2013

Edited: Matt J
on 10 Mar 2013

Do you suggest if maxmin is smooth then the likelihood function will also be smooth

Yes. All other operations in your likelihood function look smooth. maxmin is the only question mark.

just assign a very small number(1e-10)if one of the likelihood is equal to 0

I doubt that's really solving anything. Sure, it replaces the infs with some large number, but it won't change the fact that the likelihood will be locally flat to within numerical precision for certain alpha, beta. I suspect that's why you're still getting stuck so often.

Matt J
on 10 Mar 2013

Edited: Matt J
on 10 Mar 2013

One idea would be, for large alpha, beta to write the loglikelihood in terms of Stirling's approximation of Beta(alpha,beta),

llh(x,alpha, beta)= alpha*log(x)+beta*log(1-x) + 0.5*log(2*pi)...

(alpha+beta-.5)*log(alpha+beta) - ...

(alpha-.5)*log(alpha) - ...

(beta-.5)*log(beta) - ...

xueqi
on 12 Mar 2013

There are something wrong with the maxmin function. Maybe it is the casue of all the problems...

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