I have the PDF of a Gumbel Distribution which goes like: $$ f(x) = \frac{e^{-e^{\frac{x-a}{b}} + \frac{x-a}{b}}}{b}; b > 0 $$

The parameters of this distribution are a and b. I wish to estimate these parameters using simulated annealing (to find the maximum of the likelihood function), but nowhere in the algorithm for simulated annealing does it require the data or the sample I have.

If that's the case, wouldn't the parameter estimates be the same no matter what the data is? Is there any way to sort of include the sample or data in the simulated annealing algorithm for it to find parameter estimates of my sample?

  • $\begingroup$ Re "nowhere in the algorithm:" the likelihood function, of which you are trying to find the maximum, explicitly depends on the data! $\endgroup$
    – whuber
    Commented Jan 11, 2022 at 18:54

1 Answer 1


Simulated annealing is a general-purpose optimization algorithm, technically a "metaheuristic", that requires a supplemental function (the "energy function") to optimize. A nice description is here: http://webpages.iust.ac.ir/yaghini/Courses/AOR_891/05_Simulated%20Annealing_01.pdf . This energy function is where your data would come into play. A typical energy function might be the log of the likelihood function:

$$ l(a,b;x) = -\exp\left[ \sum_i\left(x_i-a \over b\right)\right] + \sum_i\frac{x_i-a}{b} - n\log b$$

Note that SA is not a deterministic optimization algorithm and will not find the exact optimal solution, although it can get arbitrarily close. If you are just interested in playing around with it, that's fine, but there are other algorithms that will do the job much faster (and better) for the likelihood function, and of course exact solutions for the method-of-moments estimator, as outlined at the bottom of https://www.itl.nist.gov/div898/handbook/eda/section3/eda366g.htm.

  • $\begingroup$ Thanks a lot! I was wondering if I should be using a theoretical estimator or a numeric one like SA. Initially I tried finding the MLE, but I was left with a system of equations I could not solve, but the link you provided has been very helpful. $\endgroup$
    – WiseRohin
    Commented Jan 11, 2022 at 18:43

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