Phil Steitz wrote:
> J.Pietschmann wrote:
>
>> Phil Steitz wrote:
>>
>>> 1) Decide what to do about inverse cumulative probabilities where p =
>>> 1 (easy solution is to document and throw)
>>
>>
>>
>> Nearly +1
>>
>
> My own "nearly +1" on this just turned to 1. After looking some more
> at the code and thinking some more, I think that both p=1 and p=0 should
> be handled correctly in all cases. The difficult cases are when the
> probability density function has unbounded support. Here is what I
> propose for the values of inverseCumulativeProbability() at p=0 and p=1
> for current distributions. Unless otherwise noted, these values are
> intented to be independent of distribution parameters.
>
> Distribution p=0 p=1
> 
> Binomial 0 Integer.MAX_VALUE
> Chisquare 0 Double.POSITIVE_INFINITY
> Exponential 0 Double.POSITIVE_INFINITY
> F 0 Double.POSITIVE_INFINITY
> Gamma 0 Double.POSITIVE_INFINITY
> HyperGeometric 0 finite, parameterdependent
> Normal Double.NEGATIVE_INFINITY Double.POSITIVE_INFINITY
> T Double.NEGATIVE_INFINITY Double.POSITIVE_INFINITY
>
> Other than the value for Chisquare with p=1 (which causes R to hang),
> these values are consistent with what R returns using the q* functions.
> It might be more convenient to return Double.MAX_VALUE,
> Double.MAX_VALUE in place of the INFINITY's (since then we could just
> use getDomainLowerBound at 0 and 1) but this would not be correct
> mathematically. If there are no objections, I will find a way to get
> the values above returned.
I have committed changes and tests to ensure that the values in the table
above are returned, modulo correcting the following mistakes:
Both of the discrete distributions (Binomial and Hypergeometric) should
return 1 for the inverseCumulativeProbability(0). The definition that we
are using is that inverseCumulativeProbability(p) = the largest x such that
P(X <= x) <= p.
Since 0 has positive probability for both the Binomial and Hypergeometric
distributions, and the function is integervalued, the correct value to
return in these cases is actually 1, not 0.
>
> Phil
>
>
>
>
>
>
> 
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