I am trying to compute the covariance between two vectors in matlab:
x = [1:10]
y = [21:30]
cov(x,y)
This returns the matrix of covariance and variance. I just want 1 number: the covariance between the two vectors. How does one get this in matlab?
If you only have two one-dimensional vectors, the number you're looking for is the (1,2) element of the output of cov. By definition, the covariance matrix contains variances on its diagonal and covariance values on off-diagonal values.
I guess x in one realization of a random variable, so is y. Then cov(x',y') will give you the covariance matrix, where the diagonal entries are the variance of x and y, and the off diagonal element is the their covariance. Note cov(x',y') is symmetric matrix.
Related
If X is a multivariate t random variable with mean=[1,2,3,4,5] and a covariance matrix C, how to simulate points in matlab? I try mvtrnd in matlab, but clearly the sample mean does not give mean close to [1,2,3,4,5]. Also, when I test three simple examples, say X1 with mean 0 and C1=[1,0.3;0.3,1], X2 with mean 0 and C2=[0.5,0.15;0.15,0.5] and X3 with mean 0 and C3=[0.4,0.12;0.12,0.4] and use mvtrnd(C1,3,1000000), mvtrnd(C2,3,1000000) amd mvtrnd(C2,3,1000000) respectively, I find the sample points in each case give nearly the correlation matrix [1,0.3;0.3,1] but the sample covariance computed all give near [3,1;1,3]. Why and how to fix it?
The Mean
The t distribution has a zero mean unless you shift it. In the documentation for mvtrnd:
the distribution of t is that of a vector having a multivariate normal
distribution with mean 0, variance 1, and covariance matrix C, divided
by an independent chi-square random value having df degrees of
freedom.
Indeed, mean(X) will approach [0 0] for X = mvtrnd(C,df,n); as n gets larger.
The Correlation
Matching the correlation is straightforward as it addresses a part of the relationship between the two dimensions of X.
% MATLAB 2018b
df = 5; % degrees of freedom
C = [0.44 0.25; 0.25 0.44]; % covariance matrix
numSamples = 1000;
R = corrcov(C); % Convert covariance to correlation matrix
X = mvtrnd(R,df,numSamples); % X ~ multivariate t distribution
You can compare how well you matched the correlation matrix R using corrcoef or corr().
corrcoef(X) % Alternatively, use corr(X)
The Covariance
Matching the covariance is another matter. Admittedly, calling cov(X) will reveal that this is lacking. Recall that the diagonal of the covariance is the variance for the two components of X. My intuition is that we fixed the degrees of freedom df, so there is no way to match the desired variance (& covariance).
A useful function is corrcov which converts a covariance matrix into a correlation matrix.
Notice that this is unnecessary as the documentation for mvtrnd indicates
C must be a square, symmetric and positive definite matrix. If its
diagonal elements are not all 1 (that is, if C is a covariance matrix
rather than a correlation matrix), mvtrnd rescales C to transform it
to a correlation matrix before generating the random numbers.
In Matlab, I have created a matrix A with size (244x2014723)
and a matrix B with size (244x1)
I was able to calculate the correlation matrix using corr(A,B) which yielded in a matrix of size 2014723x1. So, every column of matrix A correlates with matrix B and gives one row value in the matrix of size 2014723x1.
My question is when I ask for a covariance matrix using cov(A,B), I get an error saying A and B should be of same sizes. Why do I get this error? How is the method to find corr(A,B) any different from cov(A,B)?
The answer is pretty clear if you read the documentation:
cov:
If A and B are matrices of observations, cov(A,B) treats A and B as vectors and is equivalent to cov(A(:),B(:)). A and B must have equal size.
corr
corr(X,Y) returns a p1-by-p2 matrix containing the pairwise correlation coefficient between each pair of columns in the n-by-p1 and n-by-p2 matrices X and Y.
The difference between corr(X,Y) and the MATLAB® function corrcoef(X,Y) is that corrcoef(X,Y) returns a matrix of correlation coefficients for the two column vectors X and Y. If X and Y are not column vectors, corrcoef(X,Y) converts them to column vectors.
One way you could get the covariances of your vector with each column of you matrix is to use a loop. Another way (might be in-efficient depending on the size) is
C = cov([B,A])
and then look at the first row (or column) or C.
See link
In the more about section, the equation describing how cov is computed for cov(A,B) makes it clear why they need to be the same size. The summation is over only one variable which enumerates the elements of A,B.
I want to make similar graphs to this given on the picture:
I am using Fisher Iris data and employ PCA to reduce dimensionality.
this is code:
load fisheriris
[pc,score,latent,tsquare,explained,mu] = princomp(meas);
I guess the eigenvalues are given in Latent, that shows me only four features and is about reduced data.
My question is how to show all eigenvalues of original matrix, which is not quadratic (150x4)? Please help! Thank you very much in advance!
The short (and useless) answer is that the [V, D] eig(_) function gives you the eigenvectors and the eigenvalues. However, I'm afraid I have bad news for you. Eigenvalues and eigenvectors only exist for square matrices, so there are no eigenvectors for your 150x4 matrix.
All is not lost. PCA actually uses the eigenvalues of the covariance matrix, not of the original matrix, and the covariance matrix is always square. That is, if you have a matrix A, the covariance matrix is AAT.
The covariance matrix is not only square, it is symmetric. This is good, because the singular values of a matrix are related to the eigenvalues of it's covariance matrix. Check the following Matlab code:
A = [10 20 35; 5 7 9]; % A rectangular matrix
X = A*A'; % The covariance matrix of A
[V, D] = eig(X); % Get the eigenvectors and eigenvalues of the covariance matrix
[U,S,W] = svd(A); % Get the singular values of the original matrix
V is a matrix containing the eigenvectors, and D contains the eigenvalues. Now, the relationship:
SST ~ D
U ~ V
I use '~' to indicate that while they are "equal", the sign and order may vary. There is no "correct" order or sign for the eigenvectors, so either is valid. Unfortunately, though, you will only have four features (unless your array is meant to be the other way around).
I am having some difficulties resolving this:
Given eigen values for a matrix in n dimension how one can generate a corresponding covariance matrix which result in having those eigen values.
Any advice is much appreciated.
Ali
Take any orthogonal matrix R and construct
covariance = R*diag(eigenvalues)*R'
Is it possible to reverse the following in matlab:
[U,S,V]=svds(fulldata,columns);
Quoting MathWorks:
[U,S,V] = svd(X) produces a diagonal matrix S of the same dimension as X, with nonnegative diagonal elements in decreasing order, and unitary matrices U and V so that X = U*S*V'.
In the case of svds, one will lose some information unless columns is equal to the size of the square matrix fulldata. In this case, I believe the original matrix cannot be reconstructed uniquely.