I know that this issue was already discussed here but I faced with the problem I can't solve. I have list of persons, each represented with some time series consisting from 4-8 points. I want to approximate them all with the function $y=a\cdot x^2\cdot exp(-bx)+c$. Thus for each person I am going to find his own "a", "b" and "c". For most of them next code works very good:

res=nls(p2[,2] ~ c+a*I(p2[,1]^2)*exp(b*p2[,1]),start=list(a=0.005,b=-0.005,c=5))

However for some persons these starting values don't work, R returned "Missing value or an infinity produced when evaluating the model" or "singular gradient matrix at initial parameter estimates". For some of these people these starting values worked:

res=nls(p2[,2] ~ c+a*I(p2[,1]^2)*exp(b*p2[,1]),start=list(a=0.1,b=-0.02,c=5))

Could anybody give any clear suggestion how to choose starting points for all the people I consider? I tried to use tryCatch to try different staring values and find those which work but another problem appeared: this code nls(p2[,2] ~ c+a*I(p2[,1]^2)*exp(b*p2[,1]),start=list(a=5,b=0,c=5)) led to:

        a         b         c 
 -0.00166  -0.00269 140.87366 

while nls(p2[,2] ~ c+a*I(p2[,1]^2)*exp(b*p2[,1]),start=list(a=0.1,b=-0.02,c=5)) led to

      a       b       c 
 0.2024 -0.0251 47.7811 

So by choosing different starting values we have different answers. How can this happen? I thought that since NLS function is quadratic, it can't have more than 1 extremum... Do you have any suggestions about how should I proceed in this situation?

  • $\begingroup$ You might find the discussion (look at all the answers) at stats.stackexchange.com/questions/7308/… to be illuminating: it concerns a situation where multiple, strikingly-different optima of a nonlinear function were found. An example function studied there is qualitatively similar to yours (it has no constant term and minimizes a sum of fourth powers instead of squares). $\endgroup$ – whuber Nov 6 '13 at 21:48

Non-linear least squares solves $min_\beta \sum (y_i-f(x_i;\beta))^2$. This is quadratic in $\beta$ if $f$ is linear in $\beta$. Your $f$ is not linear in $\beta$, so the NLS objective function is not quadratic in $\beta$. Of course, you don't need the function to be quadratic to guarantee convergence to a unique minimum, rather you need $min_\beta \sum (y_i-f(x_i;\beta))^2$ to be convex in $\beta$. Presumably, with your $f$, the NLS objective function is not convex. It doesn't look, to me, like the kind of $f$ which generates a convex objective function. That's pretty much the explanation. You can have lots of minima or one minimum.

If I were fitting the function that you are, I would use an entirely different approach. I would not just blindly use NLS. If you look carefully at your function, $f(x_i;\beta)=a*x_i^2exp(-bx_i)+c$ it is almost linear in the parameters. If you fixed $b$ at some value, say 0.1, then you could fit $a$ and $c$ by OLS: \begin{align} y_i &= a*x_i^2exp(-0.1x_i)+c \\ &= a*z_i+c \end{align} The variable $z_i$ is defined $z_i=x_i^2exp(-0.1x_i)$. This means that, once you have picked $b$, the optimal value of $a=\widehat{Cov}(y,z)/\hat{V}(z)$ and the optimal value of $c=\overline{y}-a*\overline{z}$.

So what, right? At the very least, this is how you should pick starting values for $a$ and $c$. But, really, this reduces the search for optimal parameters to a one dimensional search over $b$. With a modern computer, one dimensional searches are fast and easy. If you have some idea of what reasonable values for $b$ are, then you can just define an interval $[b_{low},b_{high}]$ and grid search for the b which gives the lowest sum of squared errors. Then use that $b$ and its associated optimal $a$ and $c$ to start NLS from.

Or, you could do something more sophisticated. Suppose you are searching over $b$, using the optimal $a(b)$ and $c(b)$ from OLS. Then the NLS objective function is $\sum \left(y_i - f(x_i;a(b),b,c(b))\right)^2$. The envelope theorem makes the derivative of this very easy to calculate: \begin{align} \frac{d}{d b} \sum \left(y_i - f(x_i;\beta)\right)^2 &= \sum 2\left(y_i - f(x_i;\beta)\right)\frac{d}{d b}f(x_i;\beta)\\ &= \sum 2\left(y_i - f(x_i;\beta)\right)(-abx_i^2exp(-bx_i)) \end{align}

So, you can easily write a function to calculate the NLS objective function for any given $b$ and you can easily write a function to calculate the derivative of the NLS objective function for any $b$. These two ingredients are enough to get a optimizer going on your function. Then, after you find the optimal $b$, just run NLS with that $b$ and its associated optimal $a$ and $c$. It will converge in one iteration.


Bill has given a great answer. But maybe some practical things are worth adding.

  1. nls is sensitive to starting values. Depending on your staring values you're going to get different answers or your model might not converge. That's life. No reason to be surprised. An important part of nls is working out a methodology of deriving your starting values.
  2. nls is particularly sensitive. If you're having troubles you should always minpack.lm before reassessing your approach. Using an optimizing function that is more robust to starting values.
  3. Try what Bill recommends.

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