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

[Hamed Ha]

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.
>>
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>>
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