[R] Ignoring the domain of RV in punif()

Ted Harding ted@h@rd|ng @end|ng |rom w|@ndre@@net
Tue Oct 23 12:52:19 CEST 2018


Sorry -- stupid typos in my definition below!
See at ===*** below.

On Tue, 2018-10-23 at 11:41 +0100, Ted Harding wrote:
Before the ticket finally enters the waste bin, I think it is
necessary to explicitly explain what is meant by the "domain"
of a random variable. This is not (though in special cases
could be) the space of possible values of the random variable.

Definition of (real-valued) Random Variable (RV):
Let Z be a probability space, i.e. a set {z} of entities z
on which a probability distribution is defined. The entities z
do not need to be numeric. A real-valued RV X is a function
X:Z --> R defined on Z such that, for any z in Z, X(z) is a
real number. The set Z, in tthis context, is (by definitipon)
the *domain* of X, i.e. the space on which X is defined.
It may or may not be (and usually is not) the same as the set
of possible values of X.

Then. given any real value x0, the CDF of X at x- is Prob[X <= X0].
The distribution function of X does not define the domain of X.

As a simple exam[ple: Suppose Q is a cube of side A, consisting of
points z=(u,v,w) with 0 <= u,v,w <= A. Z is the probability space
of points z with a uniform distribution of position within Q.
Define the random variable X:Q --> [0,1] as
===***
  X[u,v,w) = x/A

Wrong! That should  have been:

  X[u,v,w) = w/A
===***
Then X is uniformly distributed on [0,1], the domain of X is Q.
Then for x <= 0 _Prob[X <= x] = 0, for 0 <= x <= 1 Prob(X >=x] = x,
for x >= 1 Prob(X <= x] = 1. These define the CDF. The set of poaaible
values of X is 1-dimensional, and is not the same as the domain of X,
which is 3-dimensional.

Hopiong this helps!
Ted.

On Tue, 2018-10-23 at 10:54 +0100, Hamed Ha wrote:
> > Yes, now it makes more sense.
> > 
> > Okay, I think that I am convinced and we can close this ticket.
> > 
> > Thanks Eric.
> > Regards,
> > Hamed.
> > 
> > On Tue, 23 Oct 2018 at 10:42, Eric Berger <ericjberger using gmail.com> wrote:
> > 
> > > Hi Hamed,
> > > That reference is sloppy. Try looking at
> > > https://en.wikipedia.org/wiki/Cumulative_distribution_function
> > > and in particular the first example which deals with a Unif[0,1] r.v.
> > >
> > > Best,
> > > Eric
> > >
> > >
> > > On Tue, Oct 23, 2018 at 12:35 PM Hamed Ha <hamedhaseli using gmail.com> wrote:
> > >
> > >> Hi Eric,
> > >>
> > >> Thank you for your reply.
> > >>
> > >> I should say that your justification makes sense to me.  However, I am in
> > >> doubt that CDF defines by the Pr(x <= X) for all X? that is the domain of
> > >> RV is totally ignored in the definition.
> > >>
> > >> It makes a conflict between the formula and the theoretical definition.
> > >>
> > >> Please see page 115 in
> > >>
> > >> https://books.google.co.uk/books?id=FEE8D1tRl30C&printsec=frontcover&dq=statistical+distribution&hl=en&sa=X&ved=0ahUKEwjp3PGZmJzeAhUQqxoKHV7OBJgQ6AEIKTAA#v=onepage&q=uniform&f=false
> > >> The
> > >>
> > >>
> > >> Thanks.
> > >> Hamed.
> > >>
> > >>
> > >>
> > >> On Tue, 23 Oct 2018 at 10:21, Eric Berger <ericjberger using gmail.com> wrote:
> > >>
> > >>> Hi Hamed,
> > >>> I disagree with your criticism.
> > >>> For a random variable X
> > >>> X: D - - - > R
> > >>> its CDF F is defined by
> > >>> F: R - - - > [0,1]
> > >>> F(z) = Prob(X <= z)
> > >>>
> > >>> The fact that you wrote a convenient formula for the CDF
> > >>> F(z) = (z-a)/(b-a)  a <= z <= b
> > >>> in a particular range for z is your decision, and as you noted this
> > >>> formula will give the wrong value for z outside the interval [a,b].
> > >>> But the problem lies in your formula, not the definition of the CDF
> > >>> which would be, in your case:
> > >>>
> > >>> F(z) = 0 if z <= a
> > >>>        = (z-a)/(b-a)   if a <= z <= b
> > >>>        = 1 if 1 <= z
> > >>>
> > >>> HTH,
> > >>> Eric
> > >>>
> > >>>
> > >>>
> > >>>
> > >>> On Tue, Oct 23, 2018 at 12:05 PM Hamed Ha <hamedhaseli using gmail.com> wrote:
> > >>>
> > >>>> Hi All,
> > >>>>
> > >>>> I recently discovered an interesting issue with the punif() function.
> > >>>> Let
> > >>>> X~Uiform[a,b] then the CDF is defined by F(x)=(x-a)/(b-a) for (a<= x<=
> > >>>> b).
> > >>>> The important fact here is the domain of the random variable X. Having
> > >>>> said
> > >>>> that, R returns CDF for any value in the real domain.
> > >>>>
> > >>>> I understand that one can justify this by extending the domain of X and
> > >>>> assigning zero probabilities to the values outside the domain. However,
> > >>>> theoretically, it is not true to return a value for the CDF outside the
> > >>>> domain. Then I propose a patch to R function punif() to return an error
> > >>>> in
> > >>>> this situations.
> > >>>>
> > >>>> Example:
> > >>>> > punif(10^10)
> > >>>> [1] 1
> > >>>>
> > >>>>
> > >>>> Regards,
> > >>>> Hamed.
> > >>>>
> > >>>>         [[alternative HTML version deleted]]
> > >>>>
> > >>>> ______________________________________________
> > >>>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > >>>> https://stat.ethz.ch/mailman/listinfo/r-help
> > >>>> PLEASE do read the posting guide
> > >>>> http://www.R-project.org/posting-guide.html
> > >>>> and provide commented, minimal, self-contained, reproducible code.
> > >>>>
> > >>>
> > 
> > 	[[alternative HTML version deleted]]
> > 
> > ______________________________________________
> > R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> 
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> and provide commented, minimal, self-contained, reproducible code.




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