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Re: A guess of the Probability density function from percentile values
Posted:
Dec 30, 2010 1:44 AM
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On Dec 29, 5:31 pm, Allamarein <matteo.diplom...@gmail.com> wrote:
> Thanks Ray.You wrote:
>
> "You might be much better off looking at the problem of minimizing L =
> |
> F(77.8)-.05| + |F(80.3) - .5| + |F(82.1)-.95|, which you can
> implement
> as follows: minimize (z1 + z2 + z3), subject to the constraints z1 >=
> F(77.8)-.05, z1 >= .05-F(77.8), etc. In this formulation the
> variables
> are m,s,z1,z2,z3, and if you have the quantity F(77.8)-.05 =
> NORMDIST(77.8,m,s,TRUE)-0.05 in some cell such as B7, you just have
> the two z1-constraints as z1 >= B7 (better written as z1-B7 >= 0)and
> z1 >= -B7 (better written as z1+B7 >= 0). Note that if you use a x-
> error measure and seek the least absolute-deviation fit, you would
> have m - 3.2365s - .05 in cell B7, so you would have a purely linear
> optimization problem. Solver would handle this by a much better
> algorithm if you go to the "choose linear model" menu option.
> However,
> in your F(77.8)-.05 type of error, the resulting problem is not
> linear, so EXCEL might not get a super-accurate solution."
>
> I didn't understand. In particular what is z and why I need that.
> If you have time, please give me details about this procedure.
> (it also might be that tomorrow morning it will be clearer..anyway I
> wanted to post also to regreat you.)
> Let's define L as:
> L(s,m) =
> abs(NORMDIST(77.8,m,s,TRUE)-0.05)+abs(NORMDIST(80.3,m,s,TRUE)-0.5)+abs(NORMDIST(82.1,m,s,TRUE)-0.95)
> and solve that by Excel solver, handling m and s.
See remarks below.
> My costraints will be s>0 in order to avoid error that NORMDIST should
> give me.
This is a bad idea: *strict* inequality constraints lead to ill-posed
optimization problems. Anyway, Solver does not permit strict
inequalities as far as I know. You need non-strict inequalities, such
as s >= 0.001 or something similar.
> How you have suggested, I should handle on Solver options.
> Probably my statement of the problem is based on not linear approach.
> Apart that, T = 0.0012 could be a sufficient accurate zero for my
> purposes.
> In 240 computated cases the highest values that I found is T =
> 0.01159.
> Like I wrote, I will read better your post tomorrow.
> In the meanwhile, very thanks, Ray.
You cannot just enter
abs(NORMDIST(77.8,m,s,TRUE)-0.05)+abs(NORMDIST(80.3,m,s,TRUE)-0.5)+abs(NORMDIST(82.1,m,s,TRUE)-0.95)
into EXCEL and ask Solver to minimize it: Solver wants smooth
functions having at least continuous derivatives---even though it does
not actually _use_ derivatives! The absolute-value function is not
differentiable, because it fails to have a derivative at zero. In
other words, you cannot just solve the problem by setting the
derivatives to zero! Therefore, you need some way to represent
absolute-values in the minimization, but in terms of differentiable
functions. That is where the zi come in.
Look at this little example. How can we represent an absolute-value
such as |-9|, but in terms of smooth functions and smooth constraints?
Consider the problem min z, subject to z >= -9 and z >= -(-9) = 9. Its
solution is z = 9, which is just |-9|. In general, min z subject to z
>= f(x) and z >= -f(x) has the solution z = |f(x)|. The objective z is
certainly a smooth function, as are each of the separate constraints z
>= f(x) and z >= -f(x), at least if f(x) is itself a smooth function.
So, even though |f(x)| is not a smooth function, we can "smoothify" it
using the variable z and the two constraints. Variable z will
represent the absolute value at the optimal solution. So if we want to
minimize |f1(x)| + |f2(x)| + ... + |fm(x)|, which is a *non-smooth*
problem, we can convert it to a smooth problem: min z1 + z2 + ... +
zm, subject to z1 >= f1(x), z1 >= -f1(x), z2 >= f2(x), z2 >= -
f2(x), ...., zm >= fm(x), zm >= -fm(x). We can now use a standard
constrained optimization package (such as the Solver) to handle the
problem.
R.G. Vickson
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