A basic statistics question

On 12-Aug-2014 21:41:52 Rolf Turner wrote:

On 13/08/14 07:57, Ron Michael wrote:

Hi,

I would need to get a clarification on a quite fundamental statistics
property, hope expeRts here would not mind if I post that here.

I leant that variance-covariance matrix of the standardized data is
equal to the correlation matrix for the unstandardized data. So I
used following data.

<SNIP>

(t(Data_Normalized) %*% Data_Normalized)/dim(Data_Normalized)[1]

Point is that I am not getting exact CORR matrix. Can somebody point
me what I am missing here?

You are using a denominator of "n" in calculating your "covariance" 
matrix for your normalized data.  But these data were normalized using 
the sd() function which (correctly) uses a denominator of n-1 so as to 
obtain an unbiased estimator of the population standard deviation.

If you calculated

    (t(Data_Normalized) %*% Data_Normalized)/(dim(Data_Normalized)[1]-1)

then you would get the same result as you get from cor(Data) (to within 
about 1e-15).

cheers,
Rolf Turner

One could argue about "(correctly)"!

From the "descriptive statistics" point of view, if one is given a single

number x, then this dataset has no variation, so one could say that
sd(x) = 0. And this is what one would get with a denominator of "n".

But if the single value x is viewed as sampled from a distribution
(with positive dispersion), then the value of x gives no information
about the SD of the distribution. If you use denominator (n-1) then
sd(x) = NA, i.e. is indeterminate (as it should be in this application).

The important thing when using pre-programmed functions is to know
which is being used. R uses (n-1), and this can be found from
looking at

  ?sd

or (with more detail) at

  ?cor

Ron had assumed that the denominator was n, apparently not being aware
that R uses (n-1).

Just a few thoughts ...
Ted.
-------------------------------------------------
E-Mail: (Ted Harding) <Ted.Harding at wlandres.net>
Date: 12-Aug-2014  Time: 23:22:09

After some hesitation (not wanting to prolong a thread whose original
query has been effectively settled), nevertheless I think it may be
worth stating the general picture, for the sake of users who might
be confused or misled regarding the use of  the functions var(), cov(),
cor(), sd() etc.

The distinction to become aware of is between "summary" statistics
and statistics which will be used for estimation/inference.

Given a set of numbers, x[1], x[2], ... , x[N], they have a mean

  MEAN = sum(x)/N

and their variance VAR is the mean of the deviations between the x[i]
and MEAN:

  VAR = (sum(x - MEAN)^2)/N

If a random value X is drawn from {x[1], x[2], ... , x[N]}, with
uniform probability, then the expectation of X is again

  E(X) = MEAN

and the variance of X is again

  Var(X) = E((X - MEAN)^2) = VAR

with MEAN and VAR as given above. And the R function mean(x) will
return MEAN is its value. However, the R function var(x) will not
return VAR -- it will return (N/(N-1))*VAR

The above definitions of MEAN and VAR use division by N, i.e.
"denominator = N". But the R functions var(x), sd(x) etc. divide by
(N-1), i.e. use "denominator = N-1", as explained in
  ?var
  ?sd
etc.

The basic reason for this is that, given a random sample X[1], ... ,X[n]
of size n from {x[1], x[2], ... , x[N]}, the expectation of

  sum((X - mean(X))^2)

is (n-1)*VAR, so to obtain an unbiased estimate of VAR from the X
sample one must use the "bias-corrected" sample variance

  var(X) = sum((X - mean(X))^2)/(n-1)

i.e. "denominator = (n-1)" as described in the help pages.

So the function var(), with denominator = (n-1), is "correct" for
obtaining an unbiased estimator. But it will not be correct for the
variance sum((X - mean(X))^2)/n of the numbers X[1], ... ,X[n].

Since sd() also uses denominator = (n-1), sd(X) is the square root
of var(X). But, while var(X) is unbiased for VAR, sd(X) is not
unbiased for SD = sqrt(VAR), since, for a positive ransom variable Y,
in general

  E(sqrt(Y)) < sqrt(E(Y))

i.e. E(sd(X)) = E(sqrt(var(X))) < sqrt(E(var(X))) = sqrt(VAR) = SD.

The R functions var() etc., which use denominator = (n-1), do not
have an option which allows the user to choose denominator = n.

Therefore, in particular, a user who has a set of numbers
  {x[1], x[2], ... , x[N]}
(e.g. a record of a popuation) and wants the SD of these (e.g. for
use in summary statistics Mean and SD), could inadvertently use
R's sd(x), expecting SD, without being alerted to the fact that it
will give the wrong answer. And the only way round it is to explicitly
write one's own correction, e.g.

  SD <- function(x){n<-length(x); sd(x)*sqrt((n-1)/n)}

Indeed, this topic has got me wondering how many times I may have
blindly used sd(x) in the past, as if it were going to give me the
standard (sum(x - mean(x))^2)/length(x) result!

As I wrote earlier, when there is more than one definition which
might be used, the important thing is to know which one is being used,
and to correct accordingly if it is not the one you want.

Best wishes to all,
Ted.

-------------------------------------------------
E-Mail: (Ted Harding) <Ted.Harding at wlandres.net>
Date: 13-Aug-2014  Time: 19:49:02
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