I need to code a Maximum Likelihood Estimator to estimate the mean and variance of some toy data. I have a vector with 100 samples, created with numpy.random.randn(100). The data should have zero mean and unit variance Gaussian distribution.
I checked Wikipedia and some extra sources, but I am a little bit confused since I don't have a statistics background.
Is there any pseudo code for a maximum likelihood estimator? I get the intuition of MLE but I cannot figure out where to start coding.
Wiki says taking argmax of log-likelihood. What I understand is: I need to calculate log-likelihood by using different parameters and then I'll take the parameters which gave the maximum probability. What I don't get is: where will I find the parameters in the first place? If I randomly try different mean & variance to get a high probability, when should I stop trying?
I just came across this, and I know its old, but I'm hoping that someone else benefits from this. Although the previous comments gave pretty good descriptions of what ML optimization is, no one gave pseudo-code to implement it. Python has a minimizer in Scipy that will do this. Here's pseudo code for a linear regression.
# import the packages
import numpy as np
from scipy.optimize import minimize
import scipy.stats as stats
import time
# Set up your x values
x = np.linspace(0, 100, num=100)
# Set up your observed y values with a known slope (2.4), intercept (5), and sd (4)
yObs = 5 + 2.4*x + np.random.normal(0, 4, 100)
# Define the likelihood function where params is a list of initial parameter estimates
def regressLL(params):
# Resave the initial parameter guesses
b0 = params[0]
b1 = params[1]
sd = params[2]
# Calculate the predicted values from the initial parameter guesses
yPred = b0 + b1*x
# Calculate the negative log-likelihood as the negative sum of the log of a normal
# PDF where the observed values are normally distributed around the mean (yPred)
# with a standard deviation of sd
logLik = -np.sum( stats.norm.logpdf(yObs, loc=yPred, scale=sd) )
# Tell the function to return the NLL (this is what will be minimized)
return(logLik)
# Make a list of initial parameter guesses (b0, b1, sd)
initParams = [1, 1, 1]
# Run the minimizer
results = minimize(regressLL, initParams, method='nelder-mead')
# Print the results. They should be really close to your actual values
print results.x
This works great for me. Granted, this is just the basics. It doesn't profile or give CIs on the parameter estimates, but its a start. You can also use ML techniques to find estimates for, say, ODEs and other models, as I describe here.
I know this question was old, hopefully you've figured it out since then, but hopefully someone else will benefit.
If you do maximum likelihood calculations, the first step you need to take is the following: Assume a distribution that depends on some parameters. Since you generate your data (you even know your parameters), you "tell" your program to assume Gaussian distribution. However, you don't tell your program your parameters (0 and 1), but you leave them unknown a priori and compute them afterwards.
Now, you have your sample vector (let's call it x, its elements are x[0] to x[100]) and you have to process it. To do so, you have to compute the following (f denotes the probability density function of the Gaussian distribution):
f(x[0]) * ... * f(x[100])
As you can see in my given link, f employs two parameters (the greek letters µ and σ). You now have to calculate the values for µ and σ in a way such that f(x[0]) * ... * f(x[100]) takes the maximum possible value.
When you've done that, µ is your maximum likelihood value for the mean, and σ is the maximum likelihood value for standard deviation.
Note that I don't explicitly tell you how to compute the values for µ and σ, since this is a quite mathematical procedure I don't have at hand (and probably I would not understand it); I just tell you the technique to get the values, which can be applied to any other distributions as well.
Since you want to maximize the original term, you can "simply" maximize the logarithm of the original term - this saves you from dealing with all these products, and transforms the original term into a sum with some summands.
If you really want to calculate it, you can do some simplifications that lead to the following term (hope I didn't mess up anything):
Now, you have to find values for µ and σ such that the above beast is maximal. Doing that is a very nontrivial task called nonlinear optimization.
One simplification you could try is the following: Fix one parameter and try to calculate the other. This saves you from dealing with two variables at the same time.
You need a numerical optimisation procedure. Not sure if anything is implemented in Python, but if it is then it'll be in numpy or scipy and friends.
Look for things like 'the Nelder-Mead algorithm', or 'BFGS'. If all else fails, use Rpy and call the R function 'optim()'.
These functions work by searching the function space and trying to work out where the maximum is. Imagine trying to find the top of a hill in fog. You might just try always heading up the steepest way. Or you could send some friends off with radios and GPS units and do a bit of surveying. Either method could lead you to a false summit, so you often need to do this a few times, starting from different points. Otherwise you may think the south summit is the highest when there's a massive north summit overshadowing it.
As joran said, the maximum likelihood estimates for the normal distribution can be calculated analytically. The answers are found by finding the partial derivatives of the log-likelihood function with respect to the parameters, setting each to zero, and then solving both equations simultaneously.
In the case of the normal distribution you would derive the log-likelihood with respect to the mean (mu) and then deriving with respect to the variance (sigma^2) to get two equations both equal to zero. After solving the equations for mu and sigma^2, you'll get the sample mean and sample variance as your answers.
See the wikipedia page for more details.
Related
I'm currently using the curve_fit function of the scipy.optimize package in Python, and know that if you take the square root of the diagonal entries of the covariance matrix that you get from curve_fit, you get the standard deviation on the parameters that curve_fit calculated. What I'm not sure about, is what exactly this standard deviation means. It's an approximation using a Hesse matrix as far as I understand, but what would the exact calculation be? Standard deviation on the Gaussian Bell Curve tells you what percentage of area is within a certain range of the curve, so I assumed for curve_fit it tells you how many datapoints are between certain parameter values, but apparently that isn't right...
I'm sorry if this should be basic knowledge for curve fitting, but I really can't figure out what the standard deviations do, they express an error on the parameters, but those parameters are calculated as the best possible fit for the function, it's not like there's a whole collection of optimal parameters, and we get the average value of that collection and consequently also have a standard deviation. There's only one optimal value, what is there to compare it with? I guess my question really comes down to this: how can I manually and accurately calculate these standard deviations, and not just get an approximation using a Hesse matrix?
The variance in the fitted parameters represents the uncertainty in the best-fit value based on the quality of the fit of the model to the data. That is, it describes by how much the value could change away from the best-fit value and still have a fit that is almost as good as the best-fit value.
With standard definition of chi-square,
chi_square = ( ( (data - model)/epsilon )**2 ).sum()
and reduced_chi_square = chi_square / (ndata - nvarys) (where data is the array of the data values, model the array of the calculated model, epsilon is uncertainty in the data, ndata is the number of data points, and nvarys the number of variables), a good fit should have reduced_chi_square around 1 or chi_square around ndata-nvary. (Note: not 0 -- the fit will not be perfect as there is noise in the data).
The variance in the best-fit value for a variable gives the amount by which you can change the value (and re-optimize all other values) and increase chi-square by 1. That gives the so-called '1-sigma' value of the uncertainty.
As you say, these values are expressed in the diagonal terms of the covariance matrix returned by scipy.optimize.curve_fit (the off-diagonal terms give the correlations between variables: if a value for one variable is changed away from its optimal value, how would the others respond to make the fit better). This covariance matrix is built using the trial values and derivatives near the solution as the fit is being done -- it calculates the "curvature" of the parameter space (ie, how much chi-square changes when a variables value changes).
You can calculate these uncertainties by hand. The lmfit library (https://lmfit.github.io/lmfit-py/) has routines to more explicitly explore the confidence intervals of variables from least-squares minimization or curve-fitting. These are described in more detail at
https://lmfit.github.io/lmfit-py/confidence.html. It's probably easiest to use lmfit for the curve-fitting rather than trying to re-implement the confidence interval code for curve_fit.
I want to fit some data to a Pareto distribution using the scipy.stats library. I am not sure if the issue might be numerical, so just to be safe; I have values measured for the dependent variable (let's call them 'pushes') for the independent variable ('minutes') starting at a few thousand minutes and every ten minutes thereafter (with the exception of a few points that were removed during data cleaning).
e.g.
2780.0 362.0
2800.0 376.0
2810.0 393.0
...
The best info I can find says something like
from scipy.stats import pareto
result = pareto.fit(data)
and I have no idea how this data is to be formatted in this case. I've tried the following but all result in errors.
result = pareto.fit(zip(minutes, pushes))
result = pareto.fit(pushes)
The error is usually
Warning: invalid value encountered in double_scalars
would greatly appreciate some guidance, thank you.
As I mentioned in the comments above, pareto.fit() is not what you're looking for.
The .fit() methods of the continuous distributions in scipy.stats obtain an estimate of the parameters of the distribution that maximise the probability of observing some particular set of sample values. Therefore, pareto.fit() wants only a single array argument containing the samples you want to fit the distribution to. The other keyword arguments control various aspects of the fitting process, for example by specifying initial values for the distribution parameters.
What you're actually trying to do is to fit the relationship between some independent variable x and some dependent variable y, i.e.
y_fit = f(x, params)
What you need to do is:
Choose some functional form for f. From your description, the plot of y vs x resembles the probability density function for a Pareto distribution, so perhaps either this or a decaying exponential might be appropriate.
Find the set of params that minimize some measure of the difference between y and y_fit (usually the sum of squared differences). You could use scipy.optimize.curve_fit or scipy.optimize.minimize to do this.
I have a function compare_images(k, a, b) that compares two 2d-arrays a and b
Inside the funcion, I apply a gaussian_filter with sigma=k to a My idea is to estimate how much I must to smooth image a in order for it to be similar to image b
The problem is my function compare_images will only return different values if k variation is over 0.5, and if I do fmin(compare_images, init_guess, (a, b) it usually get stuck to the init_guess value.
I believe the problem is fmin (and minimize) tends to start with very small steps, which in my case will reproduce the exact same return value for compare_images, and so the method thinks it already found a minimum. It will only try a couple times.
Is there a way to force fmin or any other minimizing function from scipy to take larger steps? Or is there any method better suited for my need?
EDIT:
I found a temporary solution.
First, as recommended, I used xtol=0.5 and higher as an argument to fmin.
Even then, I still had some problems, and a few times fmin would return init_guess.
I then created a simple loop so that if fmin == init_guess, I would generate another, random init_guess and try it again.
It's pretty slow, of course, but now I got it to run. It will take 20h or so to run it for all my data, but I won't need to do it again.
Anyway, to better explain the problem for those still interested in finding a better solution:
I have 2 images, A and B, containing some scientific data.
A looks like a few dots with variable value (it's a matrix of in which each valued point represents where a event occurred and it's intensity)
B looks like a smoothed heatmap (it is the observed density of occurrences)
B looks just like if you applied a gaussian filter to A with a bit of semi-random noise.
We are approximating B by applying a gaussian filter with constant sigma to A. This sigma was chosen visually, but only works for a certain class of images.
I'm trying to obtain an optimal sigma for each image, so later I could find some relations of sigma and the class of event showed in each image.
Anyway, thanks for the help!
Quick check: you probably really meant fmin(compare_images, init_guess, (a,b))?
If gaussian_filter behaves as you say, your function is piecewise constant, meaning that optimizers relying on derivatives (i.e. most of them) are out. You can try a global optimizer like anneal, or brute-force search over a sensible range of k's.
However, as you described the problem, in general there will only be a clear, global minimum of compare_images if b is a smoothed version of a. Your approach makes sense if you want to determine the amount of smoothing of a that makes both images most similar.
If the question is "how similar are the images", then I think pixelwise comparison (maybe with a bit of smoothing) is the way to go. Depending on what images we are talking about, it might be necessary to align the images first (e.g. for comparing photographs). Please clarify :-)
edit: Another idea that might help: rewrite compare_images so that it calculates two versions of smoothed-a -- one with sigma=floor(k) and one with ceil(k) (i.e. round k to the next-lower/higher int). Then calculate a_smooth = a_floor*(1-kfrac)+a_ceil*kfrac, with kfrac being the fractional part of k. This way the compare function becomes continuous w.r.t k.
Good Luck!
Basin hopping may do a bit better, as it has a high chance of continuing anyway when it gets stuck at the plateau's.
I found on this example function that it does reasonably well with a low temperature:
>>> opt.basinhopping(lambda (x,y): int(0.1*x**2 + 0.1*y**2), (5,-5), T=.1)
nfev: 409
fun: 0
x: array([ 1.73267813, -2.54527514])
message: ['requested number of basinhopping iterations completed successfully']
njev: 102
nit: 100
I realize this is an old question but I haven't been able to find many discussion of similar topics. I am facing a similar issue with scipy.optimize.least_squares. I found that xtol did not do me much good. It did not seem to change the step size at all. What made a big difference was diff_step. This sets the step size taken when numerically estimating the Jacobian according to the formula step_size = x_i*diff_step, where x_i is each independent variable. You are using fmin so you aren't calculating Jacobians, but if you used another scipy function like minimize for the same problem, this might help you.
I had the same problem and got it to work with the 'TNC' method.
res = minimize(f, [1] * 2, method = 'TNC', bounds=[(0,15)] * 2, jac = '2-point', options={'disp': True, 'finite_diff_rel_step': 0.1, 'xtol': 0.1, 'accuracy': 0.1, 'eps': 0.1})
The combination between 'finite_diff_rel_step' and setting 'jac' to one of {‘2-point’, ‘3-point’, ‘cs’} did the trick for the jacobian calculation step, and the 'accuracy' did the trick for the step size. The 'xtol' and 'eps' I don't think are needed, I just added them just in case.
In the example, I have 2 variables that are initialized to 1 and with boundaries [0,15] because I'm approximating the parameters of beta distribution, but it should apply to your case also.
I was wondering if someone could please explain what the following functions in scipy.stats do:
rv_continuous.expect
rv_continuous.pdf
I have read the documentation but I am still confused.
Here is my task, quite simple in theory, but I am still confused with what these functions do.
So, I have a list of areas, 16383 values. I want to find the probability that the variable area takes any value between a smaller value , called "inf" and a larger value "sup".
So, what I thought is:
scipy.stats.rv_continuous.pdf(a) #a being the list of areas
scipy.stats.rv_continuous.expect(pdf, lb = inf, ub = sup)
So that i can get the probability that any area is between sup and inf.
Can anyone help me by explaining in a simple way what the functions do and any hint on how to compute the integral of f(a) between inf and sup, please?
Thanks
Blaise
rv_continuous is a base class for all of the probability distributions implemented in scipy.stats. You would not call methods on rv_continuous yourself.
Your question is not entirely clear about what you want to do, so I will assume that you have an array of 16383 data points drawn from some unknown probability distribution. From the raw data, you will need to estimate the cumulative distribution, find the values of that cumulative distribution at the sup and inf values and subtract to find the probability that a value drawn from the unknown distribution.
There are lots of ways to estimate the unknown distribution from the data depending on how much modelling you want to do and how many assumptions you want to make. At the more complicated end of the spectrum, you could try to fit one of the standard parametric probability distributions to the data. For example, if you had a suspicion that your data were lognormally distributed, you could use scipy.stats.lognorm.fit(data, floc=0) to find the parameters of the lognormal distribution that fit your data. Then you could use scipy.stats.lognorm.cdf(sup, *params) - scipy.stats.lognorm.cdf(inf, *params) to estimate the probability of the value being between those values.
In the middle are the non-parametric forms of distribution estimation like histograms and kernel density estimates. For example, scipy.stats.gaussian_kde(data).integrate_box_1d(inf, sup) is an easy way to make this estimate using a Gaussian kernel density estimate of the unknown distribution. However, kernel density estimates aren't always appropriate and require some tweaking to get right.
The simplest thing you could do is just count the number of data points that fall between inf and sup and divide by the total number of data points that you have. This only works well with a largish number of points (which you have) and with bounds that aren't too far in the tails of the data.
The cumulative density function might give you what you want.
Then the probability P of being between two values is
P(inf < area < sup) = cdf(sup) - cdf(inf)
There's a tutorial about probabilities here and here
They are all related. The pdf is the "density" of the probabilities. They must be greater than zero and sum to 1. I think of it as indicating how likely something is. The expectation is is a generalisation of the idea of average.
E[x] = sum(x.P(x))
I have a graph between 2 functions f and g.
I know it follows a power law function with exponential cutoff.
f(x) = x**(-alpha)*e**(-lambda*x)
How do I find the value of exponent alpha?
If you have sufficiently close x points (for example one every 0.1), you can try the following:
ln(f(x)) = -alpha ln(x) - lambda x
ln(f(x))' = - alpha / x - lambda
So depending on where you have your points:
If you have a lot of points near 0, you can try:
h(x) = x ln(f(x))' = -alpha - lambda x
So the limit of the function h when x goes to 0 is -alpha
If you have large values of x, the function x -> ln(f(x))' tends toward lambda when x goes to infinity, so you can guess lambda and use pwdyson's expression.
If you don't have close x points, the numerical derivative will be very noisy, so I would try to guess lambda as the limit of -ln(f(x)/x for large x's...
If you don't have large values, but a large number of x's, you can try a minimization of
sum_x_i (ln(y_i) + alpha ln(x_i) + lambda x_i) ^2
on both alpha and lambda (I guess It would be more precise than the initial expression)...
It is a simple least square regression (numpy.linalg.lstsq will do the job).
So you have plenty of methods, the one to chose really depends on you inputs.
The usual and general way of doing what you want is to perform a non-linear regression (even though, as noted in another response, it is possible to linearize the problem). Python can do this quite easily with the help of the SciPy package, which is used by many scientists.
The routine you are looking for is its least-square optimization routine (scipy.optimize.leastsq). Once you wrap your head around the way this general optimization procedure works (see the example), you will probably find many other opportunities to use it. Basically, you calculate the list of differences between your measurements and their ideal value f(x), and you ask SciPy to find the parameters that make these differences as small as possible, so that your data fits the model as well as possible. This then gives you the parameter you are looking for.
It sounds like you might be trying to fit a power-law to a distribution with an exponential cutoff at the low end due to incompleteness - but I may be reading too far into your problem.
If that is the problem you're dealing with, this website (and accompanying publication) addresses the issue: http://tuvalu.santafe.edu/~aaronc/powerlaws/. I wrote the python implementation of the power-law fitter on that page; it is linked from there.
If you know that the points follow this law exactly, then invert the equation and put in an x and its corresponding f(x) value:
import math
alpha = -(lambda*x + math.log(f(x)))/math.log(x)
But the if the points do not exactly fit the equation you will need to do some sort of regression to determine alpha.
EDIT: Ok, so they don't fit exactly. This is getting beyond a Python question, but there may be something in numpy that can handle it. Here is a numpy linear regression recipe but your equation can't be rearranged into a linear form, so you'll have to look into non-linear regression.