[SciPy-dev] stats.distributions.poisson loc parameter : is it wise ?
Thu Aug 6 17:11:04 CDT 2009
On Thu, Aug 6, 2009 at 17:02, Pierre GM<email@example.com> wrote:
> On Aug 6, 2009, at 5:49 PM, Robert Kern wrote:
>> On Thu, Aug 6, 2009 at 16:43, Pierre GM<firstname.lastname@example.org> wrote:
>>> Even if
>>> the scale is simply discarded already, using a location will probably
>>> NOT give the expected result
>> It depends on what your expectations are. For the discrete
>> distributions, all the loc parameter means is this, as documented:
>> pmf(x; loc) -> pmf(x-loc)
>> That's it. I don't know why you would expect anything else.
> Because using a location parameter, you change the support domain.
> Back to the example of a Poisson distribution with loc=1, the support
> domain is now x>=1, which amounts to truncating the zeroes.
I don't understand why you go through all of these contortions. It
does not amount to truncation at all. It just shifts the distribution.
> The mean
> of a zero-truncated Poisson with parameter pr should be pr/(1-exp(-
> pr)), but we end up with pr+1. Not the expected result.
Because you are expecting that the operation is equivalent to
something that it is not.
pmf(x; loc) -> pmf(x-loc)
Nothing more. It is definitely *not* the same thing as setting all
x<loc to 0 and renormalizing.
"I have come to believe that the whole world is an enigma, a harmless
enigma that is made terrible by our own mad attempt to interpret it as
though it had an underlying truth."
-- Umberto Eco
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