I am trying to plot Gaussian using Matlab. My code is like this.
a=1/(0.1*sqrt(2*3.14))
y1=a*exp(-1*(((X1-Mu).^2)./(2*(Sigma^2)) ))
plot(X1,y1)
My graph looks like the image on link
It is showing correct shape but values at y axis is going up to 4. As per my knowledge Gaussian is a probability distribution function and thus must always return value between 0 and 1.Thus I am apprehensive if my implementation is correct?
Yes it is a probability distribution function but it is not required to return value between 0 and 1 everytime. As you can see from the picture below, Gaussian graph depends on variance and mean.
Your implementation is correct. The gaussian is a probability DENSITY function, which is different from a probability distribution. The former must only be larger or equal than zero but when integrated over the entire range of posible X1, the result must be equal to 1.
Probability distributions are the ones whos values must be lower or equal to 1.
As a sidenote. Matlab has both the gaussian probability density and distribution functions builtin as normpdf and normcdf respectively.
Related
I've got an arbitrary probability density function discretized as a matrix in Matlab, that means that for every pair x,y the probability is stored in the matrix:
A(x,y) = probability
This is a 100x100 matrix, and I would like to be able to generate random samples of two dimensions (x,y) out of this matrix and also, if possible, to be able to calculate the mean and other moments of the PDF. I want to do this because after resampling, I want to fit the samples to an approximated Gaussian Mixture Model.
I've been looking everywhere but I haven't found anything as specific as this. I hope you may be able to help me.
Thank you.
If you really have a discrete probably density function defined by A (as opposed to a continuous probability density function that is merely described by A), you can "cheat" by turning your 2D problem into a 1D problem.
%define the possible values for the (x,y) pair
row_vals = [1:size(A,1)]'*ones(1,size(A,2)); %all x values
col_vals = ones(size(A,1),1)*[1:size(A,2)]; %all y values
%convert your 2D problem into a 1D problem
A = A(:);
row_vals = row_vals(:);
col_vals = col_vals(:);
%calculate your fake 1D CDF, assumes sum(A(:))==1
CDF = cumsum(A); %remember, first term out of of cumsum is not zero
%because of the operation we're doing below (interp1 followed by ceil)
%we need the CDF to start at zero
CDF = [0; CDF(:)];
%generate random values
N_vals = 1000; %give me 1000 values
rand_vals = rand(N_vals,1); %spans zero to one
%look into CDF to see which index the rand val corresponds to
out_val = interp1(CDF,[0:1/(length(CDF)-1):1],rand_vals); %spans zero to one
ind = ceil(out_val*length(A));
%using the inds, you can lookup each pair of values
xy_values = [row_vals(ind) col_vals(ind)];
I hope that this helps!
Chip
I don't believe matlab has built-in functionality for generating multivariate random variables with arbitrary distribution. As a matter of fact, the same is true for univariate random numbers. But while the latter can be easily generated based on the cumulative distribution function, the CDF does not exist for multivariate distributions, so generating such numbers is much more messy (the main problem is the fact that 2 or more variables have correlation). So this part of your question is far beyond the scope of this site.
Since half an answer is better than no answer, here's how you can compute the mean and higher moments numerically using matlab:
%generate some dummy input
xv=linspace(-50,50,101);
yv=linspace(-30,30,100);
[x y]=meshgrid(xv,yv);
%define a discretized two-hump Gaussian distribution
A=floor(15*exp(-((x-10).^2+y.^2)/100)+15*exp(-((x+25).^2+y.^2)/100));
A=A/sum(A(:)); %normalized to sum to 1
%plot it if you like
%figure;
%surf(x,y,A)
%actual half-answer starts here
%get normalized pdf
weight=trapz(xv,trapz(yv,A));
A=A/weight; %A normalized to 1 according to trapz^2
%mean
mean_x=trapz(xv,trapz(yv,A.*x));
mean_y=trapz(xv,trapz(yv,A.*y));
So, the point is that you can perform a double integral on a rectangular mesh using two consecutive calls to trapz. This allows you to compute the integral of any quantity that has the same shape as your mesh, but a drawback is that vector components have to be computed independently. If you only wish to compute things which can be parametrized with x and y (which are naturally the same size as you mesh), then you can get along without having to do any additional thinking.
You could also define a function for the integration:
function res=trapz2(xv,yv,A,arg)
if ~isscalar(arg) && any(size(arg)~=size(A))
error('Size of A and var must be the same!')
end
res=trapz(xv,trapz(yv,A.*arg));
end
This way you can compute stuff like
weight=trapz2(xv,yv,A,1);
mean_x=trapz2(xv,yv,A,x);
NOTE: the reason I used a 101x100 mesh in the example is that the double call to trapz should be performed in the proper order. If you interchange xv and yv in the calls, you get the wrong answer due to inconsistency with the definition of A, but this will not be evident if A is square. I suggest avoiding symmetric quantities during the development stage.
I'm trying to fit a multivariate normal distribution to data that I collected, in order to take samples from it.
I know how to fit a (univariate) normal distribution, using the fitdist function (with the 'Normal' option).
How can I do something similar for a multivariate normal distribution?
Doesn't using fitdist on every dimension separately assumes the variables are uncorrelated?
There isn't any need for a specialized fitting function; the maximum likelihood estimates for the mean and variance of the distribution are just the sample mean and sample variance. I.e., compute the sample mean and sample variance and you're done.
Estimate the mean with mean and the variance-covariance matrix with cov.
Then you can generate random numbers with mvnrnd.
It is also possible to use fitmgdist, but for just a multivariate normal distribution mean and cov are enough.
Yes, using fitdist on every dimension separately assumes the variables are uncorrelated and it's not what you want.
You can use [sigma,mu] = robustcov(X) function, where X is your multivariate data, i.e. X = [x1 x2 ... xn] and xi is a column vector data.
Then you can use Y = mvnpdf(X,mu,sigma) to get the values of the estimated normal probability density function.
https://www.mathworks.com/help/stats/normfit.html
https://www.mathworks.com/help/stats/mvnpdf.html
Is this a good way of plotting a Normal Distribution? On occasion, I get a pdf value (pdf_x) which is greater than 1.
% thresh_strain contains a Normally Distributed set of numbers
[mu_j,sigma_j] = normfit(thresh_strain);
x=linspace(mu_j-4*sigma_j,mu_j+4*sigma_j,200);
pdf_x = 1/sqrt(2*pi)/sigma_j*exp(-(x-mu_j).^2/(2*sigma_j^2));
plot(x,pdf_x);
The integral of a pdf is 1, at any point the values can be higher. Your plot is corect.
As #Daniel points out in his answer, with continuous random variables the PDF is a derivative of a probability (or a measure of intensity) so it can be greater than one. The CDF is a probability and must always be on [0, 1].
As an example, take the distributions marked below. The area under each curve is 1 (they are valid distributions) yet the density can be above 1.
Related StackExchange posts: here and here
I have the following code, which I use to obtain the graph below. How can I determine the probability density of the values as I want my Y-Axis label to be Probability density or,do I have to normalise the Y-values?
Thanks
% thresh_strain contains a Normally Distributed set of numbers
[mu_j,sigma_j] = normfit(thresh_strain);
x=linspace(mu_j-4*sigma_j,mu_j+4*sigma_j,200);
pdf_x = 1/sqrt(2*pi)/sigma_j*exp(-(x-mu_j).^2/(2*sigma_j^2));
plot(x,pdf_x);
Your figure as it stands is correct - the area under the curve is 1. It does not need to be normalised.
You can check this by plotting the cumulative distribution function:
plot(x,(x(2)-x(1)).*cumsum(pdf_x));
The y-axis in your figure needs to be relabeled as it is not "number of dents". "Probability density" is an acceptable label.
I have a set of samples, S, and I want to find its PDF. The problem is when I use ksdensity I get values greater than one!
[f,xi] = ksdensity(S)
In array f, most of the values are greater than one! Would you please tell me what the problem can be? Thanks for your help.
For example:
S=normrnd(0.3035, 0.0314,1,1000);
ksdensity(S)
ksdensity, as the name says, estimates a probability density function over a continuous variable. Probability densities can be larger than 1, they can actually have arbitrary values from zero upwards. The constraint on probabilities is that their sum over an exhaustive range of possibilities has to be 1. For probability densities, the constraint is that the integral over the whole range of values is 1.
A crude approximation of an integral of the pdf estimated by ksdensity can be obtained in Matlab like this:
sum(f) * min(diff(xi))
assuming that the values in xi are equally spaced. The value of this expression should be approximately 1.
If in your application you believe this approximation is not close enough to 1, you might want to specify the grid of estimation points (second parameter pts) such that the spacing is finer or the range is wider than the one automatically generated by ksdensity.