# How to deal with unstable estimates during curve fitting?

First of all, I understand this isn't a strictly statistical question, but I've seen other questions involving optim() here. Please feel free to suggest a better SE subdomain for this if you know one.

The problem: I'm trying to recover latent components from a signal. The functional form of the components is assumed known, though the amount actually present could be anything from 2 to 5. There is also some noise.

If I initialize the parameters near seemingly reasonable points, I get pretty good results:

However, slight changes to the initial conditions ($\pm 10$ for starting position in the x-axis) make the optimization settle for clearly suboptimal fits:

The estimated parameters are obviously different:

            A           B           C           D           E           F
Good 0.2437936   0.8574553   0.2551376 311.4988629 356.2413314 410.4340460
Meh1 0.1968331   0.8300569   0.3587093 300.0033490 350.0018268 399.9951828
Meh2 0.3160437   0.8076175   0.1806510 324.6438328 362.8249570 420.1755116


I noticed that the final error size it settles for is also higher in the bad fits, so I figured that optimizing on the initial parameters to minimize the final error could work. But that seems a bit forced, so I was wondering if there's a more "natural" way to make the optimization routines explore the parameter space better.

Here's the code and the data I've used to get the estimates above:

# DATA
structure(list(nm = c(290, 291.0700073, 292, 293.0700073, 294,
295.0700073, 296, 297.0700073, 298, 299.0700073, 300, 301.0700073,
302, 303.0700073, 304, 305.0700073, 306, 307.0700073, 308, 309.0700073,
310, 310.9299927, 312.0299988, 312.9599915, 314.0599976, 315,
315.9299927, 317.0299988, 317.9599915, 319.0599976, 320, 321.0700073,
322, 323.0700073, 324, 325.0700073, 326, 327.0700073, 328, 329.0700073,
330, 331.0700073, 332, 333.0700073, 334, 335.0700073, 336, 337.0700073,
338, 339.0700073, 340, 341.0700073, 342, 343.0700073, 344, 345.0700073,
346, 347.0700073, 348, 349.0700073, 350, 351.0599976, 351.9599915,
353.0299988, 353.9299927, 355, 356.0599976, 356.9599915, 358.0299988,
358.9299927, 360, 361.0700073, 362, 363.0700073, 364, 365.0700073,
366, 367.0700073, 368, 369.0700073, 370, 371.0700073, 372, 373.0700073,
374, 375.0700073, 376, 377.0700073, 378, 379.0700073, 380, 381.0599976,
381.9599915, 383.0299988, 383.9299927, 385, 386.0599976, 386.9599915,
388.0299988, 388.9299927, 390, 391.0700073, 392, 393.0700073,
394, 395.0700073, 396, 397.0700073, 398, 399.0700073, 400, 401.0599976,
401.9599915, 403.0299988, 403.9299927, 405, 406.0599976, 406.9599915,
408.0299988, 408.9299927, 410, 411.0599976, 411.9599915, 413.0299988,
413.9299927, 415, 416.0599976, 416.9599915, 418.0299988, 418.9299927,
420, 421.0599976, 421.9599915, 423.0299988, 423.9299927, 425,
426.0599976, 426.9599915, 428.0299988, 428.9299927, 430, 431.0599976,
431.9599915, 433.0299988, 433.9299927, 435, 436.0599976, 436.9599915,
438.0299988, 438.9299927, 440, 441.0599976, 441.9599915, 443.0299988,
443.9299927, 445, 446.0599976, 446.9599915, 448.0299988, 448.9299927,
450),
Irel = c(0.117806361618286, 0.124055360135408, 0.132286087317653,
0.134765173276003, 0.141416137595884, 0.154651050395524, 0.150792836007325,
0.1564751297397, 0.168489707784141, 0.179055956196472, 0.182165493262127,
0.197649148327743, 0.205262794893577, 0.214227392088028, 0.229183782751442,
0.244643054198938, 0.253658311323034, 0.256158450913007, 0.279918545689736,
0.292411259981127, 0.298011575703029, 0.30800050219483, 0.308153514083128,
0.324290067808231, 0.323991856500551, 0.34613575945743, 0.376828983513388,
0.376172574407897, 0.405651374778084, 0.409478976390944, 0.42516737006287,
0.447803209783957, 0.459725364616002, 0.497083173184919, 0.492693254698212,
0.521438933418449, 0.528993505602943, 0.574070496055267, 0.592562867551184,
0.599977161734153, 0.616551241235139, 0.618316074083678, 0.642397163265336,
0.670244422495287, 0.681992262150335, 0.726539768487631, 0.750815856559603,
0.728690744532417, 0.76931865595805, 0.77320961687876, 0.793517997428088,
0.81044222137464, 0.826698988737789, 0.863562451258101, 0.871270035330893,
0.858135039696234, 0.885867075272038, 0.890256099017011, 0.899116950151812,
0.882963567297772, 0.952403820552076, 0.930567111505424, 0.944340792149357,
0.949783209073671, 0.964888292257969, 0.962151218200197, 0.97105811774725,
0.946144789965987, 0.988312112100969, 0.991161862945315, 0.9892146960761,
1, 0.994246259414727, 0.972130508456595, 0.978568637828816, 0.977238543005258,
0.95938736887762, 0.94203322502379, 0.941573570009061, 0.938253426572537,
0.961694178844629, 0.92750240070744, 0.970346815661228, 0.939917490571128,
0.912161501764443, 0.875776829146493, 0.870000856247766, 0.88240348761658,
0.855000878264457, 0.865616375454144, 0.856034172797298, 0.845213007931437,
0.836370190342225, 0.805299908541629, 0.791224127722616, 0.80136338142642,
0.777883739578135, 0.810225747103884, 0.759593422057342, 0.73576225902955,
0.723087606776591, 0.695577001172421, 0.682645332946674, 0.685600739775804,
0.676688609135976, 0.671682788737244, 0.63731514682292, 0.639013144471281,
0.647606104698215, 0.630829936713537, 0.608760302508152, 0.601968449272337,
0.587685348651311, 0.57670249919507, 0.572182283125727, 0.56294110495427,
0.550330809825504, 0.5585902481355, 0.522153539305056, 0.520661484724767,
0.512877842191466, 0.495941090958452, 0.491016801221881, 0.491587618480521,
0.483935099480003, 0.462098149550021, 0.486031457936156, 0.458126587217451,
0.459458678124788, 0.451513936067923, 0.442474536479693, 0.444839784336694,
0.431150387371712, 0.439101530654984, 0.427179134939608, 0.423819551143095,
0.407499562280818, 0.394993443102741, 0.409101161713293, 0.394138731306351,
0.380156423143598, 0.388180217786638, 0.380508185206435, 0.358726368914768,
0.351223557776416, 0.345344888510005, 0.350888556050928, 0.34390456046729,
0.328386696405115, 0.33055680756308, 0.315991257929834, 0.336901601862216,
0.328079743378219, 0.3185103779083, 0.318298687246679, 0.292512613897891,
0.307027159643554, 0.30604015418075, 0.290402867922108, 0.282963484657648,
0.300103460224965)), class = "data.frame", row.names = c(NA, -161L)) -> ds

# TARGET FUNCTION
Im <- function(v,ivm,inv=F){
if(inv){v<-(10^7)/v;ivm<-(10^7)/ivm}
vneg <- 1.177*ivm - 7780
vpos <- 0.831*ivm + 7070
ir <- (ivm - vneg)/(vpos - ivm)
ia <- ivm + ir*(vpos - vneg)/(ir^2 - 1)
exp(-log(2)*(log(ia - v)-log(ia - ivm))^2/(log(ir)^2))
}
estI01 <- function(pars,refd){
n <- length(pars)/2
outer(refd$nm, pars[n+1:n], Im, inv=T) -> Im.j Im.j%*%pars[1:n] -> Iest return(mean(abs(refd$Irel - Iest)))
}

# OPTIMIZATION CONFIG
c(rep(0,3),rep(290,3)) -> lowb
c(rep(1,3),rep(450,3)) -> uppb
list(maxit=10**4) -> conl

# EXAMPLES
initp <- c(rep(0.5,3),300,350,400)

optim(par=initp,estI01,refd=ds,
method="L-BFGS-B",
lower=lowb,
upper=uppb,
control=conl) -> res1

initp <- c(rep(0.5,3),310,360,410)

optim(par=initp,estI01,refd=ds,
method="L-BFGS-B",
lower=lowb,
upper=uppb,
control=conl) -> res2

initp <- c(rep(0.5,3),320,370,420)

optim(par=initp,estI01,refd=ds,
method="L-BFGS-B",
lower=lowb,
upper=uppb,
control=conl) -> res3

• Would there be value in using a two-stage approach: first fitting a single peak equation, and using those values as initial parameters as part of a second stage using three peak equations? Jul 27, 2018 at 1:05
• I just tried this - the single curve does find a reasonable estimate, but that still leaves two free initial values for the second stage, which as I just verified, greatly influence the results. Jul 27, 2018 at 1:29
• I agree that this is certainly on topic here. It looks like your optimization gets stuck in a local optimum. It's always a good idea to re-run optimization problems multiple times to guard against this (and to check the dependence on initial values). Jul 27, 2018 at 6:42
• It might be worthwhile to try the following test: first fit a single peak, and using those values fit two peaks, and if that seems OK then use those values for finally fitting three peaks. Jul 27, 2018 at 14:14
• I assume you mean that the two curve fit should be initialized using the same values that were found in the single fit for both components? I tried this but the results were not good. Interesting to note that using the optimization routines from Stan via rstan I managed to get a good and stable fit, so I suppose that settles it for now. Jul 28, 2018 at 1:52

I believe that your problem occurs because the algorithm stops too early (another issue would be ending up in a local minimum) and you can "solve" this by working on the stopping rule.

For the L-BFGS-B algorithm in the optim the algorithm stops when the change of the objective function is smaller than a certain limit.

### Zigzagging

Note that the optimum is not in the direction of the slope.

Even when there is a single (global) maximum, what you may end up with is the situation that the change of the function is in certain directions more extreme than in other directions. The result is that the algorithm selects only a small step size and mostly determined by those dominant directions. You will only get a small change of the objective function, possibly resulting in the termination of the algorithm.

The way that the function will approach the optimum is in a zigzagging pattern which is only slowly converging and possibly early terminating.

Below are three ways/solutions too 'help' the algorithm. Another "solution" might be too use a different (smarter) algorithm.

### Solution 1: scaling parameters

You can debug this by observing the Hessian matrix (the second order partial derivatives)

> optim(par=initp,estI01,refd=ds,
+       method="L-BFGS-B",
+       lower=lowb,
+       upper=uppb,
+       control=conl, hessian = 1) -> res3
> res3$hessian [,1] [,2] [,3] [,4] [,5] [,6] [1,] 7.609540375 5.339149352 1.253786410 2.902051e-02 -9.718628e-02 -4.618742e-03 [2,] 5.339149352 11.231282671 7.121692787 8.657414e-02 -4.019626e-03 -2.007495e-02 [3,] 1.253786410 7.121692787 11.868611589 3.210269e-02 1.689158e-01 -8.289745e-03 [4,] 0.029020509 0.086574137 0.032102688 -6.388602e-05 0.000000e+00 0.000000e+00 [5,] -0.097186278 -0.004019626 0.168915754 0.000000e+00 7.534015e-05 -2.602085e-14 [6,] -0.004618742 -0.020074953 -0.008289745 0.000000e+00 -2.602085e-14 -8.705671e-07  and you see that the change of the parameters 1-3 has more effect on the slope than the parameters 4-6. If you scale your parameters (which changes the gradient and puts more weight on changes in the direction of the parameters 4-6) then you get the same results for the three starting conditions. conl <- list(maxit = 10^4, parscale = c(rep(10^0,3),rep(10^2,3)) )  ### Solution 2: Changing objective function and convergence limits You can change the objective function such that you will not reach the machine limit so easily. For instance with your function you can change the mean (which involves a division of your objective function by 161) into the sum. #return(mean(abs(refd$$Irel - Iest)) return(sum(abs(refd$$Irel - Iest)))  and also change the conditions for convergence. conl <- list(maxit=10^4, factr = 1 )  The algorithm stops when the change of the function is below factr multiplied with the machine tolerance (the default is $$10^7$$ and setting it at $$1$$ is the most extreme you can go) ### Solution 3: Segreated solving for parameters (this works most effectively in your situation) You can solve the first three parameters separately from the other three parameters. This can be done in various ways. For instance if you use this function # I am putting the estimation in a seperate function # such that you call this function seperately, e.g. for plotting Iest <- function(pars,refd, coefout = 0){ n <- length(pars)/2 outer(refd$nm, pars[n+1:n], Im, inv=T) -> Im.j

# use fitting to estimate the first three parameter values
fit <- L1pack::l1fit(x = Im.j, y = refd$$Irel, intercept = 0) #Iest <- Im.j%*%pars[1:n] Iest <- fit$$fitted.values

# the stuff with coefout allows you to
# use this function in optim but also outside optim
# when you want to get the coefficients
if (coefout == 0) {
Iest
} else {
fit$coefficients } } estI01 <- function(pars,refd){ Iest <- Iest(pars,refd) return(mean(abs((refd$Irel - Iest))^1))
}


Now optim only optimizes for three parameters. The optimization of the other three parameters is nested inside the prediction of the values. In this example this nested prediction is done with the function l1fit from the L1pack package because you seek to optimize the L1-norm. But this method of splitting up the variables is especially useful when you seek to optimize the L2-norm because then the optimization of the first three parameters can be done with an explicit function.

### Comparison of output from solution 1, 2 and 3

plotting the solutions in the colors red green and blue.