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<font face="Courier New">Stéphane,<br>
<br>
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. <br>
<br>
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)<br>
<br>
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.<br>
<br>
Regards,<br>
<br>
Federico Miyara<br>
<br>
</font><br>
<div class="moz-cite-prefix">On 19/02/2020 17:14, Stéphane Mottelet
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:a12739b1-a830-ce31-8622-d240136f5f2e@utc.fr">
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<p><tt>Hi all,</tt></p>
<p><tt>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:</tt><tt><br>
</tt></p>
<p><tt>* covar()</tt></p>
<p><tt>Let us start with a definition of covariance in general:</tt><tt><br>
</tt></p>
<tt> </tt>
<p><tt><a class="moz-txt-link-freetext"
href="https://fr.wikipedia.org/wiki/Covariance#D%C3%A9finition_de_la_covariance"
moz-do-not-send="true">https://fr.wikipedia.org/wiki/Covariance#D%C3%A9finition_de_la_covariance</a></tt></p>
<tt> </tt>
<p><tt>and with an example there:</tt><tt><br>
</tt></p>
<p><tt><a class="moz-txt-link-freetext"
href="https://en.wikipedia.org/wiki/Covariance#Example"
moz-do-not-send="true">https://en.wikipedia.org/wiki/Covariance#Example</a></tt></p>
<tt> </tt>
<p><tt>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:</tt></p>
<p><tt>--> x=[1 2];y=[1 2 3];fre = [1/4 1/4 0;0 1/4
1/4];covar(x,y,fre)</tt><tt><br>
</tt><tt> ans =</tt><tt><br>
</tt><tt><br>
</tt><tt> 0.25</tt></p>
<p><tt>Please note that covar() output is always a scalar. Now let
us consider cov():</tt><tt><br>
</tt></p>
<tt> </tt>
<p><tt>* cov()</tt></p>
<p><tt>Here is a definition of the covariance matrix:</tt><tt><br>
</tt> </p>
<tt> </tt>
<p><tt><a class="moz-txt-link-freetext"
href="https://en.wikipedia.org/wiki/Covariance_matrix"
moz-do-not-send="true">https://en.wikipedia.org/wiki/Covariance_matrix</a></tt></p>
<tt> </tt>
<p><tt>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:</tt></p>
<p><tt><a class="moz-txt-link-freetext"
href="https://en.wikipedia.org/wiki/Estimation_of_covariance_matrices"
moz-do-not-send="true">https://en.wikipedia.org/wiki/Estimation_of_covariance_matrices</a></tt><tt><br>
</tt></p>
<p><tt>You can verify in actual code of cov() that this macro
computes the same estimation (sums are vectorized). </tt><tt><br>
</tt></p>
<p><tt>We can summarize these facts this way:</tt><tt><br>
</tt></p>
<p><tt>* 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 </tt><tt><br>
</tt></p>
<p><tt>* 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(:)]).</tt><tt><br>
</tt></p>
<p><tt>To me, the main difference is that covar(x,y,fre) does not
compute an </tt><tt><u>estimator</u></tt><tt> but a </tt><tt><u>exact
value</u></tt><tt>. 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 :</tt></p>
<p><tt>N=1000;</tt><tt><br>
</tt><tt>x=ceil(rand(N,1)*2);</tt><tt><br>
</tt><tt>y=x+floor(rand(N,1)*2);</tt><tt><br>
</tt><tt><br>
</tt><tt>[pairs,k]=unique(gsort([x y],'lr','i'),'r');</tt><tt><br>
</tt><tt>f=diff([k;N+1])/N;</tt><tt><br>
</tt></p>
<p><tt>freq=sparse(pairs,f)</tt><tt><br>
</tt><tt>N/(N-1)*covar(1:2,1:3,freq)</tt><tt><br>
</tt><tt>cov(x,y)</tt><tt><br>
</tt></p>
<p><tt>If you have a look to the results</tt><tt>,<br>
</tt></p>
<p><tt>--> freq</tt><tt><br>
</tt><tt> freq = </tt><tt><br>
</tt><tt><br>
</tt><tt> 0.2526 0.2489 0. </tt><tt><br>
</tt><tt> 0. 0.2453 0.2532</tt><tt><br>
</tt><tt><br>
</tt><tt>--> N/(N-1)*covar(1:2,1:3,freq)<br>
ans =<br>
<br>
0.249769<br>
<br>
--> cov(x,y)<br>
ans =<br>
<br>
0.2500182 0.249769 <br>
0.249769 0.4995447</tt><tt><br>
</tt></p>
<p><tt>you can see that <br>
</tt></p>
<p><tt>1. we have considered the same random variables as in the
example </tt><tt><a class="moz-txt-link-freetext"
href="https://en.wikipedia.org/wiki/Covariance#Example"
moz-do-not-send="true">https://en.wikipedia.org/wiki/Covariance#Example</a><br>
2. covar's output (up to the normalization to correct the
bias) gives the off diagonal term of cov(x,y)<br>
</tt></p>
<p><tt>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.<br>
</tt></p>
<p><tt>If you agree or disagree, feel free to continue the
discussion in this thread.</tt></p>
<p><tt>S.</tt><tt><br>
</tt></p>
<pre class="moz-signature" cols="72">--
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
<a class="moz-txt-link-freetext" href="http://www.utc.fr/~mottelet" moz-do-not-send="true">http://www.utc.fr/~mottelet</a>
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