y = gauss(x,s,m)
Y = normpdf(X,mu,sigma)
R = normrnd(mu,sigma)
What are the basic differences between these three functions?
Y = normpdf(X,mu,sigma) is the probability density function for a normal distribution with mean mu and stdev sigma. Use this if you want to know the relative likelihood at a point X.
R = normrnd(mu,sigma) takes random samples from the same distribution as above. So use this function if you want to simulate something based on the normal distribution.
y = gauss(x,s,m) at first glance looks like the exact same function as normpdf(). But there is a slight difference: Its calculation is
Y = EXP(-(X-M).^2./S.^2)./(sqrt(2*pi).*S)
while normpdf() uses
Y = EXP(-(X-M).^2./(2*S.^2))./(sqrt(2*pi).*S)
This means that the integral of gauss() from -inf to inf is 1/sqrt(2). Therefore it isn't a legit PDF and I have no clue where one could use something like this.
For completeness we also have to mention p = normcdf(x,mu,sigma). This is the normal cumulative distribution function. It gives the probability that a value is between -inf and x.
A few more insights to add to Leander good answer:
When comparing between functions it is good to look at their source or toolbox. gauss is not a function written by Mathworks, so it may be redundant to a function that comes with Matlab.
Also, both normpdf and normrnd are part of the Statistics and Machine Learning Toolbox so users without it cannot use them. However, generating random numbers from a normal distribution is quite a common task, so it should be accessible for users that have only the core Matlab. Hence, there is a redundant function to normrnd which is randn that is part of the core Matlab.
Related
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 new in matlab. I didn't understand how to derive a dirac delta function and then shift it using symbolic toolbox.
syms t
x = dirac(t)
why can't i see the dirac delta function using ezplot(x,[-10,10]) for example?
As others have noted, the Dirac delta function is not a true function, but a generalized function. The help for dirac indicates this:
dirac(X) is not a function in the strict sense, but rather a
distribution with int(dirac(x-a)*f(x),-inf,inf) = f(a) and
diff(heaviside(x),x) = dirac(x).
Strictly speaking, it's impossible for Matlab to plot the Dirac delta function in the normal way because part of it extends to infinity. However, there are numerous workarounds if you want a visualization. A simple one is to use the stem plot function and the > operator to convert the one Inf value to something finite. This produces a unit impulse function (or Kronecker delta):
t = -10:10;
x = dirac(t) > 0;
stem(t,x)
If t and x already exist as symbolic variables/expressions rather than numeric ones you can use subs:
syms t
x = dirac(t);
t2 = -10:10;
x2 = subs(x,t,t2)>0;
stem(t2, x2)
You can write your own plot routine if you want something that looks different. Using ezplot is not likely to work as it doesn't offer as much control.
First, I've not met ezplot before; I had to read up on it. For things that are functionals like your x, it's handy, but you still have to realize it's exactly giving you what it promises: A plot.
If you had the job of plotting the dirac delta function, how would you go about doing it correctly? You can't. You must find a convention of annotating your plot with the info that there is a single, isolated, infinite point in your plot.
Plotting something with a line plot hence is unsuitable for anything but smooth functions (that's a well-defined term). Dirac Delta definitely isn't amongst the class of functions that are smooth. You would typically use a vertical line or something to denote the point where your functional is not 0.
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.
I would like to measure the goodness-of-fit to an exponential decay curve. I am using the lsqcurvefit MATLAB function. I have been suggested by someone to do a chi-square test.
I would like to use the MATLAB function chi2gof but I am not sure how I would tell it that the data is being fitted to an exponential curve
The chi2gof function tests the null hypothesis that a set of data, say X, is a random sample drawn from some specified distribution (such as the exponential distribution).
From your description in the question, it sounds like you want to see how well your data X fits an exponential decay function. I really must emphasize, this is completely different to testing whether X is a random sample drawn from the exponential distribution. If you use chi2gof for your stated purpose, you'll get meaningless results.
The usual approach for testing the goodness of fit for some data X to some function f is least squares, or some variant on least squares. Further, a least squares approach can be used to generate test statistics that test goodness-of-fit, many of which are distributed according to the chi-square distribution. I believe this is probably what your friend was referring to.
EDIT: I have a few spare minutes so here's something to get you started. DISCLAIMER: I've never worked specifically on this problem, so what follows may not be correct. I'm going to assume you have a set of data x_n, n = 1, ..., N, and the corresponding timestamps for the data, t_n, n = 1, ..., N. Now, the exponential decay function is y_n = y_0 * e^{-b * t_n}. Note that by taking the natural logarithm of both sides we get: ln(y_n) = ln(y_0) - b * t_n. Okay, so this suggests using OLS to estimate the linear model ln(x_n) = ln(x_0) - b * t_n + e_n. Nice! Because now we can test goodness-of-fit using the standard R^2 measure, which matlab will return in the stats structure if you use the regress function to perform OLS. Hope this helps. Again I emphasize, I came up with this off the top of my head in a couple of minutes, so there may be good reasons why what I've suggested is a bad idea. Also, if you know the initial value of the process (ie x_0), then you may want to look into constrained least squares where you bind the parameter ln(x_0) to its known value.
The radii r is drawn from a cut-off log-normal distribution, which has a following probability density function:
pdf=((sqrt(2).*exp(-0.5*((log(r/rch)).^2)))./((sqrt(pi.*(sigma_nd.^2))...
.*r).*(erf((log(rmax/rch))./sqrt(2.*(sigma_nd.^2)))-erf((log(rmin/rch))./sqrt(2.*(sigma_nd.^2))))));
rch, sigma_nd, rmax, and rmin are all constants.
I found the explanation from the net, but it seems difficult to find its integral and then take inverse in Matlab.
I checked, but my first instinct is that it looks like log(r/rch) is a truncated normal distribution with limits of log(rmin/rch) and log(rmax/rch). So you can generate the appropriate truncated normal random variable, say y, then r = rch * exp(y).
You can generate truncated normal random variables by generating the untruncated values and replacing those that are outside the limits. Alternatively, you can do it using the CDF, as described by #PengOne, which you can find on the wikipedia page.
I'm (still) not sure that your p.d.f. is completely correct, but what's most important here is the distribution.
If your PDF is continuous, then you can integrate to get a CDF, then find the inverse of the CDF and evaluate that at the random value.
If your PDF is not continuous, then you can get a discrete CDF using cumsum, and use that as your initial Y value in interp(), with the initial X value the same as the values the PDF was sampled at, and asking to interpolate at your array of rand() numbers.
it's not clear what's your variable, but i'm assuming it's r.
the simplest way to do this is, as Chris noted, first get the cdf (note that if r starts at 0, pdf(1) is Nan... change it to 0):
cdf = cumtrapz(pdf);
cdf = cdf / cdf(end);
then spawn a uniform distribution (size_dist indicating the number of elements):
y = rand (size_dist,1);
followed by a method to place distribution along the cdf. Any technique will work, but here is the simplest (albeit inelegant)
x = zeros(size_dist,1);
for i = 1:size_dist
x(i) = find( y(i)<= cdf,1);
end
and finally, returning to the original pdf. Use matlab numerical indexing to reverse course. Note: use r and not pdf:
pdfHist = r(x);
hist (pdfHist,50)
Probably an overkill for your distribution - but you can always write a Metropolis sampler.
On the other hand - implementation is straight forward so you'd have your sampler very quick.