The question above says it all. Basically my question is for a generic fitting function (could be arbitrarily complicated) which will be nonlinear in the parameters I am trying to estimate, how does one choose the initial values to initialize the fit? I am trying to do nonlinear least squares. Is there any strategy or method? Has this been studied? Any references? Anything besides ad hoc guessing? Specifically, right now one of the fitting forms I am working with is a Gaussian plus linear form with five parameters I am trying to estimate, like

$$y=A e^{-\left(\frac{x-B}{C}\right)^2}+Dx+E$$

where $x = \log_{10}$(abscissa data) and $y = \log_{10}$(ordinate data) meaning that in log-log space my data looks like a straight line plus a bump which I am approximating by a Gaussian. I have no theory, nothing to guide me about how to initialize the nonlinear fit except perhaps graphing and eyeballing like the slope of the line and what the center/width of the bump is. But I have more than hundred of these fits to do so instead of graphing and guessing, I would prefer some approach which can be automated.

I can't find any references, in the library or online. The only thing I can think of is to just randomly choose initial values. MATLAB offers to choose values randomly from [0,1] uniformly distributed. So with each data set, I run the randomly initialized fit a thousand times and then pick the one with the highest $r^2$? Any other (better) ideas?

Addendum #1

First, here are some visual representations of the data sets just to show you guys what kind of data am I talking about. I am posting both the data in its original form without any kind of transformation and then its visual representation in log-log space as it clarifies some of the data's features while distorting others. I am posting a sample of both good and bad data.

good data log-log good data bad data log-log bad data

Each of the six panels in each figure show four data sets plotted together red, green, blue, and cyan and each data set has exactly 20 data points. I am trying to fit each one of them with a straight line plus a gaussian because of the bumps seen in the data.

The first figure is some of the good data. The second figure is the log-log plot of the same good data from figure one. The third figure is some of the bad data. The fourth figure is the log-log plot of figure three. There is much more data, these are just two subsets. Most of the data (about 3/4) is good, similar to the good data I showed here.

Now some comments, please bear with me as this might get long but I think all of this detail is necessary. I'll try to be as concise as possible.

I had originally expected a simple-power law (meaning straight line in log-log space). When I plotted everything in log-log space, I saw the unexpected bump at around 4.8 mHz. The bump was thoroughly investigated and was discovered in others work as well so its not that we messed up. It is physically there and other published works mention this too. So then I just added a gaussian term to my linear form. Note that this fit was to be done in log-log space (hence my two questions including this one).

Now, after reading the answer by Stumpy Joe Pete to another question of mine (not related to these data at all) and reading this and this and references therein (stuff by Clauset), I realize that I shouldn't fit in log-log space. So now I want to do everything in pre-transformed space.

Question 1: Looking at the good data, I still think that a linear plus a gaussian in pre-transformed space is still a good form. I would love to hear from others who have more data-expreience what they think. Is gaussian+linear reasonable? Should I only do a gaussian? Or an entirely different form?

Questions 2: Whatever the answer to question 1, I would still need (most likely) non-linear least squares fit so still need help with the initialization.

The data where we see two sets, we very heavily prefer to capture the first bump at around 4-5 mHz. So I don't want to add more gaussian terms and our gaussian term should be centered on the first bump which is almost always the bigger bump. We want "more accuracy" between 0.8mHz and around 5mHz. We don't care too much for the higher frequencies but don't want to ignore them completely either. So maybe some sort of weighing? Or B can be initialized around 4.8mHz always?

The abscissa data is frequency in units of millihertz, denote it by $f$. The ordinate data is a coefficient we are computing, denote it by $L$. So no log transformation, and the form is

$$L=A e^{-\left(\frac{f-B}{C}\right)^2}+Df+E.$$

  • $f$ is frequency, is always positive.
  • $L$ is a positive coefficient. So we are working in the first quadrant.
  • $A$, the amplitude should always be positive too I think because we are just dealing with bumps. When I look at the data I always see peaks and no valleys. It looks like in all of the data there are multiple bumps at higher frequencies. The first bump is always much larger than others. In good data, the secondary bumps are very weak but in bad data (panel 2 and 5 for example), the secondary bumps are strong. So we actually don't have a valley, but rather two bumps. Meaning that the amplitude $A>0$. And since we care mostly about the first peak, all the more reason for $A$ to be positive.
  • $B$ is the center of the bump and we always want it on that big bump around 4-5mHz. In our resolved frequency range, it almost always appears at 4.8mHz.
  • $C$ is the width of the bump. I imagine it is symmetric around zero meaning $C$ would have the same effect as $-C$ because it is squared. So we don't care what its value is. Let's say we prefer it positive.
  • $D$ is the slope of the line, seems like it could be slightly negative so not enforcing any restriction on it. The slope is interesting in its own right so instead of enforcing any restrictions on it, we just want to see what it would be. Is it positive or negative? How big/small is it magnitude-wise? and so on.
  • $E$ is the (almost) $L$-intercept. The subtle thing here is that because of the gaussian term, $E$ is not quite the $L$-intercept. The actual intercept (if we were to extrapolate to $f=0$) would be


So the only restriction here is that intercept should be positive as well. The intercept being zero, I don't know what that would mean. But negative would seem nonsensical for sure. I guess here we can allow for $E$ to be slightly negative with a small magnitude if necessary. The reason $E$ and the intercept is important here but some of our colleagues are actually interested in extrapolation as well. The minimum frequency we have is 0.8mHz and they want to extrapolate between 0 and 0.8mHz. My naive idea was to just use the fit to go all the way down to $f=0$.

I know extrapolation is harder/more dangerous than interpolation but using a straight line plus a gaussian (hoping it decays fast enough) seems reasonable to me. Kind of like natural cubic splines with natural boundary conditions, the slope at the left endpoint, just extend the line and see where it crosses the $L$ axis. If it isn't negative then use that line for extrapolation.

Questions 3: What do you guys think extrapolating this way in this case? Any pros/cons? Any other ideas for extrapolation? Again we only care about the lower frequencies so extrapolating between 0 and 1mHz...sometimes very very small frequencies, close to zero. I know this post is already packed. I asked this question here because the answers might be related but if you guys prefer I can separate this question and ask another one later.

Lastly, here are two sample data sets upon request.

0.813010000000000   0.091178000000000   0.012728000000000
1.626000000000000   0.103120000000000   0.019204000000000
2.439000000000000   0.114060000000000   0.063494000000000
3.252000000000000   0.123130000000000   0.071107000000000
4.065000000000000   0.128540000000000   0.073293000000000
4.878000000000000   0.137040000000000   0.074329000000000
5.691100000000000   0.124660000000000   0.071992000000000
6.504099999999999   0.104480000000000   0.071463000000000
7.317100000000000   0.088040000000000   0.070336000000000
8.130099999999999   0.080532000000000   0.036453000000000
8.943100000000001   0.070902000000000   0.024649000000000
9.756100000000000   0.061444000000000   0.024397000000000
10.569000000000001   0.056583000000000   0.025222000000000
11.382000000000000   0.052836000000000   0.024576000000000
12.194999999999999   0.048727000000000   0.026598000000000
13.008000000000001   0.045870000000000   0.029321000000000
13.821000000000000   0.041454000000000   0.067300000000000
14.633999999999999   0.039596000000000   0.081800000000000
15.447000000000001   0.038365000000000   0.076443000000000
16.260000000000002   0.036425000000000   0.075912000000000

The first column is the frequencies in mHz, identical in every single data set. The second column is a good data set (good data figure one and two, panel 5, red marker) and the third column is a bad data set (bad data figure three and four, panel 5, red marker).

Hope this is enough to stimulate some more enlightened discussion. Thanks you everyone.

  • $\begingroup$ +1 for extra information, but now this looks like a new question. Incidentally, if you want to delete the previous one now I think that would be okay, it looks like you have now covered the additional information it had. $\endgroup$
    – Glen_b
    Commented Jul 2, 2013 at 4:06
  • $\begingroup$ @Glen_b Why is that so? Why does it look like a new question? As for the old question, we all whore for points ;-D and the old one has two upvotes, any way to merge it with this so I can keep those two votes too? $\endgroup$ Commented Jul 2, 2013 at 7:02
  • $\begingroup$ Well for starters, you're now asking about what you should fit, rather than specifying what to fit, as before. There are a number of other differences, some of which I'd think of as fairly significant. I'll look at changing my answer, but I think this one could stand as the original question and answer and your new parts where you are asking other things could be a new one. I'll leave that to your judgement for the present. $\endgroup$
    – Glen_b
    Commented Jul 2, 2013 at 7:29
  • $\begingroup$ @Glen_b Fair enough, I struck out the extra questions. So the questions is still, I have some data I want to fit using linear+gaussian form, can do I better than random initialization? $\endgroup$ Commented Jul 2, 2013 at 8:12
  • $\begingroup$ I think my present answer shows that - in at least some circumstances - you can do better, and @whuber suggests something even simpler still than my process. I could go back and see how what I have performs on your data, but even as it stands now it does give some idea how to set such starting points up. $\endgroup$
    – Glen_b
    Commented Jul 2, 2013 at 8:16

5 Answers 5


If there was a strategy that was both good and general -- one that always worked - it would already be implemented in every nonlinear least squares program and starting values would be a non-issue.

For many specific problems or families of problems, some pretty good approaches to starting values exist; some packages come with good start value calculations for specific nonlinear models or with more general approaches that often work but may have to be helped out with more specific functions or direct input of start values.

Exploring the space is necessary in some situations but I think your situation is likely to be such that more specific strategies will likely be worthwhile - but to design a good one pretty much requires a lot of domain knowledge we're unlikely to possess.

For your particular problem, likely good strategies can be designed, but it's not a trivial process; the more information you have about the likely size and extent of the peak (the typical parameter values and typical $x$'s would give some idea), the more can be done to design a good starting value choice.

What are the typical ranges of $y$'s and $x$'s you get? What do the average results look like? What are the unusual cases? What parameter values do you know to be possible or impossible?

One example - does the Gaussian part necessarily generate a turning point (a peak or trough)? Or is it sometimes so small relative to the line that it doesn't? Is $A$ always positive? etc.

Some sample data would help - typical cases and hard ones, if you're able.

Edit: Here's an example of how you can do fairly well if the problem isn't too noisy:

Here's some data that is generated from your model (population values are A = 1.9947, B = 10, C = 2.828, D = 0.09, E = 5):

nls data

The start values I was able to estimate are
(As = 1.658, Bs = 10.001, Cs = 3.053, Ds = 0.0881, Es = 5.026)

The fit of that start model looks like this:


The steps were:

  1. Fit a Theil regression to get a rough estimate of D and E
  2. Subtract the fit of the Theil regression off
  3. Use LOESS to fit a smooth curve
  4. Find the peak to get a rough estimate of A, and the x-value corresponding to the peak to get a rough estimate of B
  5. Take the LOESS fits whose y-values are > 60% of the estimate of A as observations and fit a quadratic
  6. Use the quadratic to update the estimate of B and to estimate C
  7. From the original data, subtract off the estimate of the Gaussian
  8. Fit another Theil regression to that adjusted data to update the estimate of D and E

In this case, the values are very suitable for starting a nonlinear fit.

I wrote this as R code but the same thing could be done in MATLAB.

I think better things than this are possible.

If the data are very noisy, this won't work at all well.

Edit2: This is the code I used in R, if anyone is interested:

gausslin.start <- function(x,y) {

  theilreg <- function(x,y){
    yy <- outer(y, y, "-")
    xx <- outer(x, x, "-")
    z  <- yy / xx
    slope     <- median(z[lower.tri(z)])
    intercept <- median(y - slope * x)

  tr <- theilreg(x,y1)
  Ds = tr[2]
  Es = tr[1]
  yf  <- y1-Ds*x-Es
  yfl <- loess(yf~x,span=.5)

  # assumes there are enough points that the maximum there is 'close enough' to 
  #  the true maximum

  yflf   <- yfl$fitted    
  locmax <- yflf==max(yflf)
  Bs     <- x[locmax]
  As     <- yflf[locmax]

  qs     <- yflf>.6*As
  ys     <- yfl$fitted[qs]
  xs     <- x[qs]-Bs
  lf     <- lm(ys~xs+I(xs^2))
  bets   <- lf$coefficients
  Bso    <- Bs
  Bs     <-  Bso-bets[2]/bets[3]/2
  Cs     <- sqrt(-1/bets[3])
  ystart <- As*exp(-((x-Bs)/Cs)^2)+Ds*x+Es

  y1a <- y1-As*exp(-((x-Bs)/Cs)^2)
  tr  <- theilreg(x,y1a)
  Ds  <- tr[2]
  Es  <- tr[1]
  res <- data.frame(As=As, Bs=Bs, Cs=Cs, Ds=Ds, Es=Es)


# population parameters: A = 1.9947 , B = 10, C = 2.828, D = 0.09, E = 5
# generate some data
x  <- runif(50,1,30)
y  <- dnorm(x,10,2)*10+rnorm(50,0,.2)
y1 <- y+5+x*.09 # This is the data
xo <- order(x)

starts <- gausslin.start(x,y1)
ystart <- with(starts, As*exp(-((x-Bs)/Cs)^2)+Ds*x+Es)
  • 3
    $\begingroup$ +1. Repeating the fit a thousand times and choosing the best (if I understand that correctly) sounds a strange idea: nonlinear least squares should converge if the model is reasonable for the data and there are good initial values. Naturally, the second is what you are asking about. But it seems pessimistic to imply that you might need to choose different starting values for each fit. $\endgroup$
    – Nick Cox
    Commented Jul 1, 2013 at 7:37
  • 1
    $\begingroup$ @NickCox It comes down to the range of problems encountered - if I recall right from earlier posts, the OP gets huge numbers of these problems, but I didn't recall seeing enough detail to make good suggestions before, though I invested a bit of time playing around with potential approaches (which didn't yield anything definitive enough to post). The OP likely has the kind of domain knowledge that could yield good start values that solve his or her problems almost always. $\endgroup$
    – Glen_b
    Commented Jul 1, 2013 at 7:41
  • 1
    $\begingroup$ Quite so. I missed the earlier post at stats.stackexchange.com/questions/61724/… $\endgroup$
    – Nick Cox
    Commented Jul 1, 2013 at 7:44
  • 3
    $\begingroup$ +1 I have had good success using a slightly simplified version of this. Instead of Theil regression, just fit a line using the most extreme points to the right and left (3 at each end does well). After subtracting that line and closely smoothing, the range of the smooth estimates $|A|$, the location of the maximum estimates $B$ (assuming $A\gt 0$), and $C$ can be estimated from the proportion of smoothed values exceeding some sizable fraction of $A$, such as $1/4$. You need to do this twice: once assuming $A\gt 0$ and again (with suitable changes) assuming $A\lt 0$. $\endgroup$
    – whuber
    Commented Jul 1, 2013 at 15:41
  • 2
    $\begingroup$ That was the first thing I thought about, too. My reasoning at the time, which has been borne out by simulations (which include tests where $B$ is near the extremes of the domain), is that in most cases you can still get reasonable starting values for the parameters, regardless of where the bump is. The crux of the matter is identifying $B$, I believe: after that the fitting procedure should have little trouble estimating the other parameters. $\endgroup$
    – whuber
    Commented Jul 1, 2013 at 15:54

There is a general approach to fitting these kind of nonlinear models. It involves reparameterizing the linear parameters with values of the dependent variable at say the first, last frequency value and a good point in the middle say the 6'th point. then you can hold these parameters fixed and solve for the nonlinear parameter in the first phase of the minimization and then minimize overall 5 parameters.

Schnute and I figured this out around 1982 when fitting growth models for fish.


However it is not necessary to read this paper. Due to the fact that the parameters are linear it is simply necessary to set up and solve a 3x3 linear system of equations to use the stable parameterization of the model.

For your model the linear part is determined by a matrix $M$

$$ M= \begin{pmatrix} \exp(-((x(1)-B)/C)^2)& x(1)& 1 \\ \exp(-((x(6)-B)/C)^2)& x(6)& 1 \\ \exp(-((x(n)-B)/C)^2)& x(n)& 1 \\ \end{pmatrix} $$ where $n=20$ in this case. 1 The code is written in AD Model Builder which is now open source. It does automatic differentiation and supports a lot of things which make nonlinear optimization easier, such as phases where different parameters are kept fixed.

  init_int n
  int mid
 !! mid=6;
  init_matrix data(1,n,1,3)
  vector x(1,n)
  vector y(1,n)
 !! x=column(data,1);
 !! y=column(data,3);   //use column 3
  init_number L1(3)     //(3) means estimate in phase 3
  init_number Lmid(3)
  init_number Ln(3)

  vector L(1,3)
  init_number log_B       // estimate in phase 1
  init_number log_C(2)    // estimate in phase 2 
  matrix M(1,3,1,3);
  objective_function_value f
  sdreport_vector P(1,3)
  sdreport_number B
  sdreport_number C
  vector pred(1,n);

  P=solve(M,L);  // solve for standard parameters 
                 // P is vector corresponding to A,D,E

  if (current_phase()<4)
    f+=0.5*n*log(norm2(y-pred))  //concentrated likelihood

The model is fit in three phases. In phase 1 only $B$ is estimated. Here the most important thing is that $C$ is large enough so that the model can "see" where to move $B$ to. In phase 2 both $B$ and $C$ are estimated. Finally in phase 3 all parameters are estimated. In the plot the green line is the fit after phase 1 and the blue line is the final fit.

enter image description here

For your case with the bad data, it fits quite easily and the (usual) parameter estimates are:

         estimate    std dev
A      2.0053e-01 5.8723e-02
D      1.6537e-02 4.7684e-03
E     -1.8197e-01 7.3355e-02
B      3.0609e+00 5.0197e-01
C      5.6154e+00 9.4564e-01]
  • $\begingroup$ Dave, this is interesting, but it raises some questions. Exactly what do you mean by "these kind of nonlinear models"? The question begins by referring to a "generic fitting function," but your description refers only to "overall 5 parameters." $\endgroup$
    – whuber
    Commented Aug 24, 2015 at 14:48
  • $\begingroup$ I mean models like the vonbertalanffy, or logistic or double exponential for example. In all cases the model is linear in some of the parameters and nonlinear in other ones. People generally try to transform them to get more stable parameterizationms by concentrating on the non linear parameters. However this is the wrong approach. It is the linear parameterization which should be modified So for example for a 4 parameter logistic the model is linear in the upper and lower asymptote, but rather than using these parameters one should use the predicted values for smallest and largest ind. var. $\endgroup$ Commented Aug 24, 2015 at 15:01
  • $\begingroup$ @davefournier Thanks for answering and pointing to your paper. Your paper seems a little difficult to get ahold of but the technique sounds interesting so can't wait to read it. $\endgroup$ Commented Aug 25, 2015 at 3:23

If you have to do this many times then I would suggest that you use an Evolutionary Algorithm on the SSE function as a front-end to provide the starting values.

On the other hand you could use GEOGEBRA to create the function using sliders for the parameters and play with them to get starting values.

OR starting values from the data can be estimated by observation.

  1. D and E come from the slope and intercept of the data (ignoring the Gaussian)
  2. A is the vertical distance of the maximum of the Gaussian from the Dx+E line estimate.
  3. B is the x value of the maximum of the Gaussian
  4. C is half the apparent width of the Gaussian

The solution interval method discussed in the following paper could be used to find reasonable initial guesses for your problem. Specifically, since your model has 5 parameters, you can pick 5 data points in your data pool and plug these 5 data points into your model to solve the parameter values. Although your model is nonlinear, the equation you need to solve may be linear equations or nonlinear equations that have analytical solutions. The solutions for these 5 parameters can be used as the reasonable initial guesses for nonlinear least squares fitting. More details of this approach (e.g., how to choose 5 data points from the data pool) can be found in the following paper:


If you want to find the guaranteed optimal parameter estimators that correspond to the global minimum of the squared error of the fit, you can use the focused regions identification method introduced in the following paper:



For starting values you could do a ordinary least squares fit. Its slope and intercept would be the starting values for D and E. The largest residual would be the starting value for A. The position of the largest residual would be the starting value for B. Maybe someone else can suggest a starting value for sigma.

However, non-linear least squares without deriving any sort of mechanistic equation from subject matter knowledge is risky business, and doing a lot of separate fits makes things even more questionable. Is there any subject matter knowledge behind your proposed equation? Are there other independent variables that relate to the differences between the 100 or so separate fits? It might help if you can incorporate those differences into a single equation that will fit all of the data at once.


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