I know for a random variable x that P(x=i) for each i=1,2,...,100. Then how may I sample x by a multinomial distribution, based on the given P(x=i) in Matlab?
I am allowed to use the Matlab built-in commands rand and randi, but not mnrnd.
In general, you can sample numbers from any 1 dimensional probability distribution X using a uniform random number generator and the inverse cumulative distribution function of X. This is known as inverse transform sampling.
random_x = xcdf_inverse(rand())
How does this apply here? If you have your vector p of probabilities defining your multinomial distribution, F = cumsum(p) gives you a vector that defines the CDF. You can then generate a uniform random number on [0,1] using temp = rand() and then find the first row in F greater than temp. This is basically using the inverse CDF of the multinomial distribution.
Be aware though that for some distributions (eg. gamma distribution), this turns out to be an inefficient way to generate random draws because evaluating the inverse CDF is so slow (if the CDF cannot expressed analytically, slower numerical methods must be used).
Related
I have a correlation matrix for N random variables. Each of them is uniformly distributed within [0,1]. I am trying to simulate these random variables, how can I do that? Note N > 2. I was trying to using Cholesky Decomposition and below is my steps:
get the lower triangle of the correlation matrix (L=N*N)
independently sample 10000 times for each of the N uniformly distributed random variables (S=N*10000)
multiply the two: L*S, and this gives me correlated samples but the range of them is not within [0,1] anymore.
How can I solve the problem?
I know that if I only have 2 random variables I can do something like:
1*x1+sqrt(1-tho^2)*y1
to get my correlated sample y. But if you have more than two variables correlated, not sure what should I do.
You can get approximate solutions by generating correlated normals using the Cholesky factorization, then converting them to U(0,1)'s using the normal CDF. The solution is approximate because the normals have the desired correlation, but converting to uniforms is a non-linear transformation and only linear xforms preserve correlation.
There's a transformation available which will give exact solutions if the transformed Var/Cov matrix is positive semidefinite, but that's not always the case. See the abstract at https://www.tandfonline.com/doi/abs/10.1080/03610919908813578.
Suppose I have a continuous probability distribution, e.g., Normal, on a support A. Suppose that there is a Matlab code that allows me to draw random numbers from such a distribution, e.g., this.
I want to build a Matlab code to "approximate" this continuous probability distribution with a probability mass function spanning over r points.
This means that I want to write a Matlab code to:
(1) Select r points from A. Let us call these points a1,a2,...,ar. These points will constitute the new discretised support.
(2) Construct a probability mass function over a1,a2,...,ar. This probability mass function should "well" approximate the original continuous probability distribution.
Could you help by providing also an example? This is a similar question asked for Julia.
Here some of my thoughts. Suppose that the continuous probability distribution of interest is one-dimensional. One way to go could be:
(1) Draw 10^6 random numbers from the continuous probability distribution of interest and store them in a column vector D.
(2) Suppose that r=10. Compute the 10-th, 20-th,..., 90-th quantiles of D. Find the median point falling in each of the 10 bins obtained. Call these median points a1,...,ar.
How can I construct the probability mass function from here?
Also, how can I generalise this procedure to more than one dimension?
Update using histcounts: I thought about using histcounts. Do you think it is a valid option? For many dimensions I can use this.
clear
rng default
%(1) Draw P random numbers for standard normal distribution
P=10^6;
X = randn(P,1);
%(2) Apply histcounts
[N,edges] = histcounts(X);
%(3) Construct the new discrete random variable
%(3.1) The support of the discrete random variable is the collection of the mean values of each bin
supp=zeros(size(N,2),1);
for j=2:size(N,2)+1
supp(j-1)=(edges(j)-edges(j-1))/2+edges(j-1);
end
%(3.2) The probability mass function of the discrete random variable is the
%number of X within each bin divided by P
pmass=N/P;
%(4) Check if the approximation is OK
%(4.1) Find the CDF of the discrete random variable
CDF_discrete=zeros(size(N,2),1);
for h=2:size(N,2)+1
CDF_discrete(h-1)=sum(X<=edges(h))/P;
end
%(4.2) Plot empirical CDF of the original random variable and CDF_discrete
ecdf(X)
hold on
scatter(supp, CDF_discrete)
I don't know if this is what you're after but maybe it can help you. You know, P(X = x) = 0 for any point in a continuous probability distribution, that is the pointwise probability of X mapping to x is infinitesimal small, and thus regarded as 0.
What you could do instead, in order to approximate it to a discrete probability space, is to define some points (x_1, x_2, ..., x_n), and let their discrete probabilities be the integral of some range of the PDF (from your continuous probability distribution), that is
P(x_1) = P(X \in (-infty, x_1_end)), P(x_2) = P(X \in (x_1_end, x_2_end)), ..., P(x_n) = P(X \in (x_(n-1)_end, +infty))
:-)
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
How would you normalize a histogram A so the sum of each bin is 1
Dividing the histogram by the width of the bin, how do you draw it
I have this
dist = rand(50)
average = mean(dist, 1);
[c,x] = hist(average, 15);
normalized = c/sum(c);
bar(x, normalized, 1)
In this case, n = 50,
What is it the formula to get values
of mean and variance^2? We write N(mean, (variance^2) / 50), but how?
How do you plot both uniform distribution and normal distribution?.
The histogram must be close to the normal distribution.
That is a very unusual way of normalizing a probability density function. I assume you want to normalize such that the area under the curve is 1. In that case, this is what you should do.
[c,x]=hist(average,15);
normalized=c/trapz(x,c);
bar(x,normalized)
Either way, to answer your question, you can use randn to generate a normal distribution. You're now generating a 50x50 uniform distribution matrix and summing along one dimension to approximate a normal Gaussian. This is unnecessary. To generate a normal distribution of 1000 points, use randn(1000,1) or if you want a row vector, transpose it or flip the numbers. To generate a Gaussian distribution of mean mu and variance sigma2, and plot its pdf, you can do (an example)
mu=2;
sigma2=3;
dist=sqrt(sigma2)*randn(1000,1)+mu;
[c,x]=hist(dist,50);
bar(x,c/trapz(x,c))
Although these can be done with dedicated functions from the statistics toolbox, this is equally straightforward, simple and requires no additional toolboxes.
EDIT
I missed the part where you wanted to know how to generate a uniform distribution. rand, by default gives you a random variable from a uniform distribution on [0,1]. To get a r.v. from a uniform distribution between [a, b], use a+(b-a)*rand