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Re: TransformedDistribution -- odd problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg120497] Re: TransformedDistribution -- odd problem
  • From: Paul von Hippel <paulvonhippel at yahoo.com>
  • Date: Tue, 26 Jul 2011 07:06:40 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <j0gh7s$bd7$1@smc.vnet.net> <201107251129.HAA25540@smc.vnet.net> <4E2EA04E.813B.006A.0@newcastle.edu.au>
  • Reply-to: Paul von Hippel <paulvonhippel at yahoo.com>

Thanks -- that fixes it!

Bonus question: if we don't specify v2>2, why doesn't Mathematica return a two part solution:
 k v2 / (v2-2) v2>2

 Indeterminate True

That's what it does if we request the mean of F -- why doesn't it do the same if we request the mean of k*F.




________________________________
From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
To: mathgroup at smc.vnet.net; paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Sent: Monday, July 25, 2011 8:09 PM
Subject: [mg120497] Re: TransformedDistribution -- odd problem

Hi Paul

There are conditions on the v1 and v2, the degrees of freedom of the F distribution:

Assuming[v2 > 2, 
Mean[TransformedDistribution[F , 
  F \[Distributed] FRatioDistribution[v1, v2]]]]

{Assuming[v2 > 2, 
 Mean[TransformedDistribution[k*F , 
  F \[Distributed] FRatioDistribution[v1, v2]]]], k*v2/(-2 + v2)}

{Assuming[v2 > 2, 
  Mean[TransformedDistribution[k + F , 
   F \[Distributed] FRatioDistribution[v1, v2]]]], 
 k + v2/(-2 + v2)} // FullSimplify

which shows precisely what you expect for k*F and k+F.

Cheers

Barrie

>>> On 25/07/2011 at 9:29 pm, in message <201107251129.HAA25540 at smc.vnet.net>,
paulvonhippel at yahoo <paulvonhippel at yahoo.com> wrote:
> A little more experimenting shows that the TransformedDistribution
> function will also not provide the mean of k+F where k is a constant
> and F has an F distribution -- i.e.,
>  Mean[TransformedDistribution[k*F ,  F \[Distributed]
> FRatioDistribution[v, v]]]
> 
> If I changce the distribution of F to NormalDistribution or
> ChiSquareDistribution, I can get a mean for k*F or k+F. So the problem
> only occurs when I define a simple function of an F variable using the
> TransformedDistribution function.
> This all strikes me as very strange, and I'd be curious to know if
> others can reproduce my results. If you can't reproduce my results,
> I'd be interested in theories about why my results differ from yours.
> E.g., is there a setting I should change in the software?
> 
> I am using version 8.0.0.0 and looking to upgrade to 8.0.1, if that
> makes a difference.
> 
> Many thanks for any pointers.
> 
> On Jul 24, 2:22 am, paulvonhippel at yahoo <paulvonhip... at yahoo.com>
> wrote:
>> I'm having a very strange problem with TransformedDistribution, where
>> I can calculate the mean of an F distribution but I cannot calculate
>> the mean of a constant multiplied by an F distribution. That is, if I
>> type
>>
>>  Mean[TransformedDistribution[F, F \[Distributed]
>> FRatioDistribution[v, v]]]
>>
>> Mathematica gives me an answer. But if I type
>>
>>  Mean[TransformedDistribution[k*F ,  F \[Distributed]
>> FRatioDistribution[v, v]]]
>>
>> Mathematica just echoes the input. I swear I got an answer for the
>> second expression earlier today. What am I doing wrong?



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