Stéphane,
My first argument in favor of keeping covar and cov as separate
functions is that often what one needs is the covariance between two
potentially correlated signals regardless of their individual variances,
so it seems somewhat inefficient to compute essentially three
covariances (two of them between two equal signals) when it is only one
of them what one wants to calculate.
However, the syntax of covar should have an option to process the
signals directly instead of their statistics (values and joint
frequencies): covar(x,y)
My second argument seems to be the opposite of my pevious request:
covar(x,y,fre) is a quite different function since the input information
is presented in a different way, which is valuable when one happens to
have the information in such fashion.
Regards,
Federico Miyara
On 19/02/2020 17:14, Stéphane Mottelet wrote:
Hi all,
Within the development team we recently had a discussion about the
improvement of cov() in terms of speed and memory requirement and
about the opportunity of merging cov() and covar() wich are two
disctinct macros. Since we did not manage to reach a consensus we
thought it could be the occasion to have the opinion of members of
this list which have a recognized academical/research knowledge in
probability and statistics. Here are some elements to start the
discussion. Let us start with covar() macro and what it actually computes:
* covar()
Let us start with a definition of covariance in general:
https://fr.wikipedia.org/wiki/Covariance#D%C3%A9finition_de_la_covariance
and with an example there:
https://en.wikipedia.org/wiki/Covariance#Example
In the two above links scalar/real variables are considered and in the
second link discrete random variables are considered. In the example
the covariance is computed knowing the possible values and their joint
density. You can easily check in the source of covar() (type "edit
covar") that, after normalizing the matrix of joint probabilities
(named "frequencies" in the source), the macro computes the same
value, which is confirmed by the result of the following statements:
--> x=[1 2];y=[1 2 3];fre = [1/4 1/4 0;0 1/4 1/4];covar(x,y,fre)
ans =
0.25
Please note that covar() output is always a scalar. Now let us
consider cov():
* cov()
Here is a definition of the covariance matrix:
https://en.wikipedia.org/wiki/Covariance_matrix
Here we consider vectors of random variables (not scalar random
variables) and in this case the covariance is a matrix. When there is
no a priori knowledge on these variables (when the joint density is
not known, typically), the best you can do is, when you have samples
of this random vector, is to compute an estimation of the covariance
matrix, see e.g. he following page:
https://en.wikipedia.org/wiki/Estimation_of_covariance_matrices
You can verify in actual code of cov() that this macro computes the
same estimation (sums are vectorized).
We can summarize these facts this way:
* covar(x,y,fre) computes the scalar covariance of two discrete random
variables knowing their possible values x(:) and y(:) and their joint
probability density
* When x is a matrix, cov(x) computes an estimator of the covariance
matrix of a vector X of size(x,2) random variables by using size(x,1)
samples of this vector (each x(i,:) is a sample). if x and y are
vectors of the same size, cov(x,y) is computed as cov([x(:) y(:)]).
To me, the main difference is that covar(x,y,fre) does not compute an
_estimator_but a _exact value_. Of course, the vectors x and y can be
the unique value of two random variables, gathered from samples (x,y)
and "fre" be the empirical frequency of samples (x_i,y_j). In this
case covar() will compute an estimation. For example, consider the two
random variables X and Y, where X takes values {1,2} with equal
probability, and Y=X+U where U takes values {0,1} with equal
probability. We can use covar() to compute the exact covariance of X
and Y, but if we only have samples, like in the below script, if we
want to estimate the covariance with the same macro, then unique pairs
have to be found and occurences counted in order to estimate the
frequency :
N=1000;
x=ceil(rand(N,1)*2);
y=x+floor(rand(N,1)*2);
[pairs,k]=unique(gsort([x y],'lr','i'),'r');
f=diff([k;N+1])/N;
freq=sparse(pairs,f)
N/(N-1)*covar(1:2,1:3,freq)
cov(x,y)
If you have a look to the results,
--> freq
freq =
0.2526 0.2489 0.
0. 0.2453 0.2532
--> N/(N-1)*covar(1:2,1:3,freq)
ans =
0.249769
--> cov(x,y)
ans =
0.2500182 0.249769
0.249769 0.4995447
you can see that
1. we have considered the same random variables as in the example
https://en.wikipedia.org/wiki/Covariance#Example
2. covar's output (up to the normalization to correct the bias) gives
the off diagonal term of cov(x,y)
So, yes, off diagonal term of cov(x,y) and covar(x,y,fre) (up to
unique pairs determination, computation of "fre" and bias correction)
have the same value, but is it a reason to merge the two functions. I
think that the answer is NO.
If you agree or disagree, feel free to continue the discussion in this
thread.
S.
--
Stéphane Mottelet
Ingénieur de recherche
EA 4297 Transformations Intégrées de la Matière Renouvelable
Département Génie des Procédés Industriels
Sorbonne Universités - Université de Technologie de Compiègne
CS 60319, 60203 Compiègne cedex
Tel : +33(0)344234688
http://www.utc.fr/~mottelet
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