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scott wrote:
>> So are you *really* trying to estimate coin flips? Or is there some
>> more complex problem you're trying to solve here?
>
> Not really more complex, just a game that is played repeatedly and you
> have a certain % chance of winning (say 57%). I just want to know if
> you play 10,100, 1000 times what is the *likely* number of wins in some
> kind of worst case and best case scenario. ie it's obviously possible
> to win every single game, but that is *really* unlikely, I want to know
> how many you'd win with 90% or 99% confidence.
Right. So it's ye olde "2SD = 95% confidence" rule then.
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Invisible wrote:
> I presume it's relatively easy to work out the inverse function for a
> normal distribution...
Actually it's impossible to analytically compute. Fortunately many
pieces of software have a built in function to numerically compute it.
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>> I presume it's relatively easy to work out the inverse function for a
>> normal distribution...
>
> Actually it's impossible to analytically compute.
Define "analytically".
Last time I checked, it's impossible to compute the square root of a
number "exactly" - and yet it's trivially easy to approximate it to any
desired accuracy.
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Invisible wrote:
>>> I presume it's relatively easy to work out the inverse function for a
>>> normal distribution...
>>
>> Actually it's impossible to analytically compute.
>
> Define "analytically".
As a closed-form expression. As I said, it's still possible to compute
numerically.
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Kevin Wampler wrote:
> Invisible wrote:
>>>> I presume it's relatively easy to work out the inverse function for
>>>> a normal distribution...
>>>
>>> Actually it's impossible to analytically compute.
>>
>> Define "analytically".
>
> As a closed-form expression. As I said, it's still possible to compute
> numerically.
And Sqrt[2] erf^-1 (2z - 1) doesn't count because...?
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Invisible wrote:
> Kevin Wampler wrote:
>> Invisible wrote:
>>>>> I presume it's relatively easy to work out the inverse function for
>>>>> a normal distribution...
>>>>
>>>> Actually it's impossible to analytically compute.
>>>
>>> Define "analytically".
>>
>> As a closed-form expression. As I said, it's still possible to
>> compute numerically.
>
> And Sqrt[2] erf^-1 (2z - 1) doesn't count because...?
erf isn't an elementary function. See:
http://en.wikipedia.org/wiki/Closed-form_expression
http://en.wikipedia.org/wiki/Error_function#Taylor_series
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>>>>>> I presume it's relatively easy to work out the inverse function
>>>>>> for a normal distribution...
>>>>>
>>>>> Actually it's impossible to analytically compute.
>>>>
>>>> Define "analytically".
>>>
>>> As a closed-form expression. As I said, it's still possible to
>>> compute numerically.
>>
>> And Sqrt[2] erf^-1 (2z - 1) doesn't count because...?
>
> erf isn't an elementary function. See:
>
> http://en.wikipedia.org/wiki/Closed-form_expression
>
> http://en.wikipedia.org/wiki/Error_function#Taylor_series
Right. So what you're actually saying is that you can't compute the
confidence interval exactly using a finite number of applications of a
particular arbitrarily chosen set of functions - the elementary functions.
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Invisible wrote:
>>>> As a closed-form expression.
>
> Right. So what you're actually saying is that you can't compute the
> confidence interval exactly using a finite number of applications of a
> particular arbitrarily chosen set of functions - the elementary functions.
That is indeed the definition.
This obviously doesn't matter for scott's case since he just wants a
numeric answer and Excel seems to have erf built in, I just wanted to
point out that far from being "relatively easy" to work out, the
cumulative distribution function of a Gaussian is generally the first
function people encounter which is impossible to actually compute in
closed form (normally this will come up in first or second year calculus
at some point). I'll agree that at first glance it certainly *looks*
like it would be simple to compute though, which is part of why it's
interesting.
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>>>>> As a closed-form expression.
>>
>> Right. So what you're actually saying is that you can't compute the
>> confidence interval exactly using a finite number of applications of a
>> particular arbitrarily chosen set of functions - the elementary
>> functions.
>
> That is indeed the definition.
>
> This obviously doesn't matter for scott's case since he just wants a
> numeric answer and Excel seems to have erf built in, I just wanted to
> point out that far from being "relatively easy" to work out, the
> cumulative distribution function of a Gaussian is generally the first
> function people encounter which is impossible to actually compute in
> closed form (normally this will come up in first or second year calculus
> at some point). I'll agree that at first glance it certainly *looks*
> like it would be simple to compute though, which is part of why it's
> interesting.
Well, the sine and cosine functions can't be computed by a finite number
of additions, subtractions, multiplications, divisions and exponents.
But since sine and cosine are included in the arbitrary "set of
permissible functions", they don't count. But erf is not, so it does count.
To me, it seems that approximating erf (or its inverse) is no harder
than approximating a sine or cosine - and we do that all day.
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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Orchid XP v8 <voi### [at] dev null> wrote:
> Well, the sine and cosine functions can't be computed by a finite number
> of additions, subtractions, multiplications, divisions and exponents.
They can if you allow the parameters to be complex numbers (which is why
trigonometric functions are included in the set of elementary functions).
--
- Warp
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