If I estimate the entropy of a vector of standard normal random variables using the Matlab entropy() function, I get an answer somewhere in the region of 4, whereas the actual entropy should be 0.5 * log(2*pi*e*sigma^2) which is approximately equal to 1.4.
Does anyone know where the discrepancy is coming from?
Note: To save time here is the Matlab code
for i = 1:1000
X(i) = randn();
end
'The entropy of X is'
entropy(X)
Please read the help (help entropy) or documentation for entropy. You'll see that it's designed for images and uses a histogram technique rather than calculating the it analytically. You'll need to create your own function if you want the formula from Wikipedia, but as the formula is so simple, that should be no problem.
I believe that the reason that you're getting such divergent answers is that entropy scales the bins of the histogram by the number of elements. If you want to uses such an estimation technique you'll want to use hist and scale the bins by area. See this StackOverflow question.
Related
Is it possible to generate random numbers with a distribution that depends on empirical probability data? I imagine this is possible by taking the inverse cumulative distribution function. I have seen some examples where this is done in MATLAB (the software that I'm using) but all of those examples have an underlying analytic form for the probability. Here I have only the PDF. For instance, I have data of probabilities for a particular event. Most of the probabilities are zero and hence not unique, but not all.
My goal is to generate the random numbers and then figure out what the distribution is. I'd really appreciate if people can help clear up my thinking here.
EDIT:
I think I want something like:
cdf=cumsum(pdf); % calculate pdf from empirical pdf
M=length(cdf);
xq=linspace(0,1,M);
invcdf=interp1(cdf,xq,xq); % calculate inverse cdf, i.e., x
but how do I take into account that a lot of the values of the pdf are zero and not unique? Is this even the right approach?
I am basing my answer on Inverse empirical cumulative distribution function from the MathWorks File Exchange. See that link for other suggestions to solving your problem.
% y - input: data set
% q - input: desired quantile (can be a scalar or a vector)
% xq - output: ICDF at specified quantile
[f, x] = ecdf(y);
xq = zeros(size(q));
for ii = 1:length(q)
xq(ii) = min(x(q(ii) <= f));
end
I'd eliminate the for loop if you're only using scalars. Also, there may be a more efficient way to vectorize the for loop, but this should at least get you started.
I am trying to obtain an equation for a function fitted to some histogram data, I was thinking of trying to do this by fitting a rational function as the data doesn't resemble any distribution recognisable by myself.
The data is experimental, and I want to be able to generate a random number according to its distribution. Hence I am hoping to be able to fit it to some sort of PDF from which I can obtain a CDF, which can be rearranged to a function into which a uniformly distributed random number between 0 and 1 can be substituted in order to obtain the desired result.
I have attempted to use the histfit function, which has worked but I couldn't figure out how to obtain an equation for the curve it fitted. Is there something stupid I have missed?
Update: I have discovered the function rationalfit, however I am struggling to figure out what the inputs need to be.
Further Update: Upon exploring the histfit command further I have discovered the option to fit it to a kernal, the figure for which looks promising, however I am only able to obtain a set of x and y values for the curve, not its equation as a I wanted.
From the documentation on histfit:
Algorithms
histfit uses fitdist to fit a distribution to data. Use fitdist
to obtain parameters used in fitting.
So the answer to your question is to use fitdist to get the parameters you're after. Here's the example from the documentation:
rng default; % For reproducibility
r = normrnd(10,1,100,1);
histfit(r)
pd = fitdist(r,'Normal')
pd =
NormalDistribution
Normal distribution
mu = 10.1231 [9.89244, 10.3537]
sigma = 1.1624 [1.02059, 1.35033]
I`m learning a paper
the paper presents a figure
the figure shows CDF of buildings height
and the paper also gives details about this figure
Building height statistics: The present model uses the statistics of
building heights in typical built-up areas as input data. A suitable
form was sought by comparing with geographical data for the city of
Guildford, United Kingdom. The probability density function that was
selected to fit the data was the log-normal distribution with unknown
parameters: mean value p and standard deviation t. As can be noted
from Fig. 3, it was found to be a good fit to the geographical data
values with parameters p = 7.3m, t= 0.26.
it tells the mean value is 7.3 and the standard deviation is 0.26 right?
however, when I try them in matlab by adding codes
x=0:0.01:20;
meanValue = 7.3;
standardDeviation = 0.26;
y1 = logncdf(x,meanValue,standardDeviation);
plot(x,y1);
what the result showed is different from the figure 3
I tried to re-read the paper to make sure parameters are correct.
and check the document on matlab about how to use this method.
everything seem all right except the simulation result.
please help me fix it ! thanks
As mentioned in the comments, the parameters mu and sigma are the mean and standard derivation of the associated normal distribution, not of the log normal distribution. The details, especially the connection between both is explained in the Wikipedia article.
To calculate mu and sigma from the mean and variance, the formulas are given in the Wikipedia article or here in the matlab syntax:
m=7.3
t=0.26
v=t.^2;
%A lognormal distribution with mean m and variance v has parameters
mu = log((m^2)/sqrt(v+m^2));
sigma = sqrt(log(v/(m^2)+1));
%finally your code:
x=0:0.01:20;
y1 = logncdf(x,mu,sigma);
plot(x,y1);
Which is much closer to the graph in your question, but the graph in your question seems to be the CDF for a much higher standard derivation. Visually guessing the parameters form your plot, I would say it's roughly t=5
I have a non-negative function f defined on a unit square S = [0,1] x [0,1] such that
My question is, how can I use MATLAB to generate a 2D random vector from S according to the probability density function f?
Rejection Sampling
The suggestion Luis Mendo made is very good because it applies to nearly all distribution functions. Based on this answer I wrote code for m.
An important point when using rejection sampling this way is that you must know the maximum of your pdf within the range. If you over-estimate the maximum your code will only run slower. If you under-estimate it it will create wrong numbers!
The idea is that you sample many uniform distributed points and accept depending on the probability density for the points.
pdf=#(x).5.*x(:,1)+3./2.*x(:,2);
maximum=2; %Right maximum for THIS EXAMPLE.
%If you are unable to determine the maximum of your
%function within the [0,1]x[0,1] range, please give an example.
result=[];
n=10;
while (size(result,1)<n)
%1. sample random point:
val=rand(1,2);
%2. Accept with probability pdf(val)/maximum
if rand<pdf(val)/maximum
%append to solution
result(end+1,:)=val;
end
end
I know that this solution is not a fast implementation, but I wanted to start with an implementation as simple as possible to make sure that the concept of rejection sampling becomes clear.
ICDF
Besides rejection sampling there is a different approach to address this issue on a more mathematical level, but you need to sit down and do some math first to end up with a better solution. For 1 dimensional distributions you typically sample using the ICDF (inverted cumulative density function) function simply using ICDF(rand(n,1)) to get random samples.
If you manage to do the math, you could instead for your PDF function define two functions ICDF1 (ICDF for the first dimension) and ICDF2 (ICDF for the second dimension) in matlab.
The first ICDF1 would map unifrom random distributed samples to sample values for the first dimension of your random distribution.
The second ICDF2 would map the output if ICDF1 and uniform distributed samples to your intended solution.
Here is some matlab code assuming you already defined ICDF1 and ICDF2
samples=ICDF1(rand(n,1));
samples(:,2)=ICDF2(samples,rand(n,1));
The great advantage of this solution is, that it does not reject any samples, being potentially much faster.
i have question how to calculate weighted correlations for matrices,from wikipedia i have created three following codes
1.weighted mean calculation
function [y]= weighted_mean(x,w);
n=length(x);
%assume that weight vector and input vector have same length
sum=0.0;
sum_weight=0.0;
for i=1:n
sum=sum+ x(i)*w(i);
sum_weight=sum_weight+w(i);
end
y=sum/sum_weight;
end
2.weighted covariance
function result=cov_weighted(x,y,w)
n=length(x);
sum_covar=0.0;
sum_weight=0;
for i=1:n
sum_covar=sum_covar+w(i)*(x(i)-weighted_mean(x,w))*(y(i)-weighted_mean(y,w));
sum_weight=sum_weight+w(i);
end
result=sum_covar/sum_weight;
end
and finally weighted correlation
3.
function corr_weight=weighted_correlation(x,y,w);
corr_weight=cov_weighted(x,y,w)/sqrt(cov_weighted(x,x,w)*cov_weighted(y,y,w));
end
now i want to apply weighted correlation method for matrices,related to this link
http://www.mathworks.com/matlabcentral/fileexchange/20846-weighted-correlation-matrix/content/weightedcorrs.m
i did not understand anything how to apply,that why i have created my self,but need in case of input are matrices,thanks very much
#dato-datuashvili Maybe I am providing too much information...
1) I would like to stress that the evaluation of Weighted Correlation matrices are very uncommon. This happens because you have to provide beforehand the weights. Unless you have a clear reason to choose the weights, there is no clear way to provide them.
How can you tell that a measurement of your sample is more or less important than another measurement?
Having said that, the weights are up to you! Yo have to choose them!
So, people usually consider just the correlation matrix (no weights or all weights are the same e.g w_i=1).
If you have a clear way to choose good weights, just do not consider this part.
2) I understand that you want to test your code. So, in order to that, you have to have correlated random variables. How to generate them?
Multivariate normal distributions are the simplest case. See the wikipedia page about them: Multivariate Normal Distribution (see the item "Drawing values from the distribution". Wikipedia shows you how to generate the random numbers from this distribution using Choleski Decomposition). The 2-variate case is much simpler. See for instance Generate Correlated Normal Random Variables
The good news is that if you are using Matlab there is a function for you. See Matlab: Random numbers from the multivariate normal distribution.]
In order to use this function you have to provide the desired means and covariances. [Note that you are making the role of nature here. You are generating the data! In real life, you are going to apply your function to the real data. What I am trying to say is that this step is only useful for tests. Furthermore, pay attencion to the fact that in the Matlab function you are providing the variances and evaluating the correlations (covariances normalized by standard errors). In the 2-dimensional case (that is the case of your function it is possible to provide directly the correlation. See the page above that I provided to you of Math.Stackexchange]
3) Finally, you can apply them to your function. Generate X and Y from a normal multivarite distribution and provide the vector of weights w to your function corr_weight_correlation and you are done!
I hope I provide what you need!
Daniel
Update:
% From the matlab page
mu = [2 3];
SIGMA = [1 1.5; 1.5 3];
n=100;
[x,y] = mvnrnd(mu,SIGMA,n);
% Using your code
w=ones(n,1);
corr_weight=weighted_correlation(x,y,w); % Remember that Sigma is covariance and Corr_weight is correlation. In order to calculate the same thing, just use result=cov_weighted instead.