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))
Related
I want to find out how many samples are needed at minimum to more or less correctly fit a probability distribution (In my case the Generalized Extreme Value Distribution from scipy.stats).
In order to evaluate the matched function, I want to compute the KL-divergence between the original function and the fitted one.
Unfortunately, all implementations I found (e.g. scipy.stats.entropy) only take discrete arrays as input. So obviously I thought of approximating the pdf by a discrete array, but I just can't seem to figure it out.
Does anyone have experience with something similar? I would be thankful for hints relating directly to my question, but also for better alternatives to determine a distance between two functions in python, if there are any.
I have fit a series of SciPy continuous distributions for a Monte-Carlo simulation and am looking to take a large number of samples from these distributions. However, I would like to be able to take correlated samples, such that the ith sample takes the e.g., 90th percentile from each of the distributions.
In doing this, I've found a quirk in SciPy performance:
# very fast way to many uncorrelated samples of length n
for shape, loc, scale, in distro_props:
sp.stats.norm.rvs(*shape, loc=loc, scale=scale, size=n)
# verrrrryyyyy slow way to take correlated samples of length n
correlate = np.random.uniform(size=n)
for shape, loc, scale, in distro_props:
sp.stats.norm.ppf(correlate, *shape, loc=loc, scale=scale)
Most of the results about this claim that the slowness on these SciPy distros if from the type-checking etc. wrappers. However when I profiled the code, the vast bulk of the time is spent in the underlying math function [_continuous_distns.py:179(_norm_pdf)]1. Furthermore, it scales with n, implying that it's looping through every elemnt internally.
The SciPy docs on rv_continuous almost seem to suggest that the subclass should override this for performance, but it seems bizarre that I would monkeypatch into SciPy to speed up their ppf. I would just compute this for the normal from the ppf formula, but I also use lognormal and skewed normal, which are more of a pain to implement.
So, what is the best way in Python to compute a fast ppf for normal, lognormal, and skewed normal distributions? Or more broadly, to take correlated samples from several such distributions?
If you need just the normal ppf, it is indeed puzzling that it is so slow, but you can use scipy.special.erfinv instead:
x = np.random.uniform(0,1,100)
np.allclose(special.erfinv(2*x-1)*np.sqrt(2),stats.norm().ppf(x))
# True
timeit(lambda:stats.norm().ppf(x),number=1000)
# 0.7717257660115138
timeit(lambda:special.erfinv(2*x-1)*np.sqrt(2),number=1000)
# 0.015020604943856597
EDIT:
lognormal and triangle are also straight forward:
c = np.random.uniform()
np.allclose(np.exp(c*special.erfinv(2*x-1)*np.sqrt(2)),stats.lognorm(c).ppf(x))
# True
np.allclose(((1-np.sqrt(1-(x-c)/((x>c)-c)))*((x>c)-c))+c,stats.triang(c).ppf(x))
# True
skew normal I'm not familiar enough, unfortunately.
Ultimately, this issue was caused by my use of the skew-normal distribution. The ppf of the skew-normal actually does not have a closed-form analytic definition, so in order to compute the ppf, it fell back to scipy.continuous_rv's numeric approximation, which involved iteratively computing the cdf and using that to zero in on the ppf value. The skew-normal pdf is the product of the normal pdf and normal cdf, so this numeric approximation called the normal's pdf and cdf many many times. This is why when I profiled the code, it looked like the normal distribution was the problem, not the SKU normal. The other answer to this question was able to achieve time savings by skipping type-checking, but didn't actually make a difference on the run-time growth, just a difference on small-n runtimes.
To solve this problem, I have replaced the skew-normal distribution with the Johnson SU distribution. It has 2 more free parameters than a normal distribution so it can fit different types of skew and kurtosis effectively. It's supported for all real numbers, and it has a closed-form ppf definition with a fast implementation in SciPy. Below you can see example Johnson SU distributions I've been fitting from the 10th, 50th, and 90th percentiles.
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.
Can you help me out with these questions? I'm using Python
Sampling Methods
Sampling (or Monte Carlo) methods form a general and useful set of techniques that use random numbers to extract information about (multivariate) distributions and functions. In the context of statistical machine learning, we are most often concerned with drawing samples from distributions to obtain estimates of summary statistics such as the mean value of the distribution in question.
When we have access to a uniform (pseudo) random number generator on the unit interval (rand in Matlab or runif in R) then we can use the transformation sampling method described in Bishop Sec. 11.1.1 to draw samples from more complex distributions. Implement the transformation method for the exponential distribution
$$p(y) = \lambda \exp(−\lambda y) , y \geq 0$$
using the expressions given at the bottom of page 526 in Bishop: Slice sampling involves augmenting z with an additional variable u and then drawing samples from the joint (z,u) space.
The crucial point of sampling methods is how many samples are needed to obtain a reliable estimate of the quantity of interest. Let us say we are interested in estimating the mean, which is
$$\mu_y = 1/\lambda$$
in the above distribution, we then use the sample mean
$$b_y = \frac1L \sum^L_{\ell=1} y(\ell)$$
of the L samples as our estimator. Since we can generate as many samples of size L as we want, we can investigate how this estimate on average converges to the true mean. To do this properly we need to take the absolute difference
$$|\mu_y − b_y|$$
between the true mean $µ_y$ and estimate $b_y$
averaged over many, say 1000, repetitions for several values of $L$, say 10, 100, 1000.
Plot the expected absolute deviation as a function of $L$.
Can you plot some transformed value of expected absolute deviation to get a more or less straight line and what does this mean?
I'm new to this kind of statistical machine learning and really don't know how to implement it in Python. Can you help me out?
There are a few shortcuts you can take. Python has some built-in methods to do sampling, mainly in the Scipy library. I can recommend a manuscript that implements this idea in Python (disclaimer: I am the author), located here.
It is part of a larger book, but this isolated chapter deals with the more general Law of Large Numbers + convergence, which is what you are describing. The paper deals with Poisson random variables, but you should be able to adapt the code to your own situation.
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.