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 at 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 at 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 at 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 at 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 at 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]]
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
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 at 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 at
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 at
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 at
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 at 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 at 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.
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 at
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 at
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 at
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
at 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 at 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 at 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 at 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.