In view of the line this discusion is taking, I repeat below
the reply I sent off-list to Dan. The essential point is that
there are infinitely many functions with the property he seeks,
so choice depends on additional criteria. It should not be said
that "this is the transformation that Dan requires.".
The exponential exp(-x) is only one possibility.
========================================================
Hello Dan.
The precise choice will depend on how you want variation
on the original scale the be reflected in variation on
the new scale -- i.e. how should sensitivity vary along
the scale.
That being said, the simplest transformation which achieves
your basic requirement is
Y = 1/(1+X)
where Y is the original, Y is the transformed.
This is monotonic: as X increases, Y decreases.
When X=0, Y=1, as required. When X = Inf, Y=0, as required.
Tuning sensitivity could be approached by a transformation
of Y which takes (0,1) to (0,1), e.g. Z = Y^alpha where alpha
is positive, but there are of course infinitely many others,
for example Z = sin(pi*Y/2) -- a completely arbitrary choice!
Hoping this helps,
Ted.
========================================================
On 23-Jul-11 21:56:18, Andre Francis wrote:
> Martin (and Dan) .....
>
> I clearly didn't read the question as Martin did. I took
> Dan's expression 'inverse relationship' to mean 'inverse
> correlation'.
>
> That is, f(x) is a function mapping [0, +inf) to [0,1]
> such that f(0)=1 and f(+Inf)=0 ... or strictly, f(x)->0
> as x->+Inf
>
> This is satisfied by f(x)=exp(-x) and so this is the
> transformation that Dan requires.
>
> ..... Andre Francis
>
> On 22 July 2011 15:58, Martin Holt <[log in to unmask]> wrote:
>> Hi,
>> I sent the following answer to the SPSS list in error, so am now
>> forwarding it on to ALLSTAT.
>> >Hi Dan,
>> >Tentatively, because playing with infinities and 0 is risky, so
>> >using a calculator:
>> >If first variable = 0, then exp0 = 1 and then 1/exp0 = 1.
>> >If first variable = 20, say, then exp20 = 485165195.4 and
>> >1/exp20 = 0.000000009
>> >So, as first variable 'x' increases towards +Inf, the function
>> >1/exp(x) tends towards 0 and does not go below 0.
>> >So, if x=0, 1/exp(0) =1 and if x=+Inf, 1/exp(+Inf) = 0 as required.
>> >This assumes that x lies between 0 and +Inf: you'd need to check with
>> >any potential -ve x's as to whether the function does what you
>> >want it to.
>> >HTH
>> >Martin Holt
>> >Medical Statistician
>>
>> ----- Forwarded Message -----
>> From: Martin Holt <[log in to unmask]>
>> To: Dan Abner <[log in to unmask]>; "[log in to unmask]" <
>> [log in to unmask]>
>> Sent: Thursday, 21 July 2011, 20:29
>> Subject: Re: Transformation Question XXXX
>>
>>
>> Hi Dan,
>>
>> Tentatively, because playing with infinities and 0 is risky, so using
>> a
>> calculator:
>>
>> If first variable = 0, then exp0 = 1 and then 1/exp0 = 1.
>>
>> If first variable = 20, say, then exp20 = 485165195.4 and 1/exp20 =
>> 0.000000009
>> So, as first variable 'x' increases towards +Inf, the function
>> 1/exp(x)
>> tends towards 0 and does not go below 0.
>>
>> So, if x=0, 1/exp(0) =1 and if x=+Inf, 1/exp(+Inf) = 0 as required.
>>
>> This assumes that x lies between 0 and +Inf: you'd need to check with
>> any
>> potential -ve x's as to whether the function does what you want it to.
>>
>> HTH
>>
>> Martin Holt
>> Medical Statistician
>>
>> From: Dan Abner <[log in to unmask]>
>> To: [log in to unmask]
>> Sent: Thursday, 21 July 2011, 7:27
>> Subject: Transformation Question XXXX
>>
>> Hello everyone,
>>
>> I have a variable that ranges from
>> 0 to +Inf and need to map it to a space
>> from 0 to 1 with an inverse relationship with the original variable
>> (in
>> other words, 0 of the new variable corresponds with max(old variable)
>> and 1
>> of new variable corresponds with min(old variable).
>>
>> Can anyone suggest a transformation?
>>
>> Thank you!
>> Dan
--------------------------------------------------------------------
E-Mail: (Ted Harding) <[log in to unmask]>
Fax-to-email: +44 (0)870 094 0861
Date: 24-Jul-11 Time: 01:46:54
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