
Re: A guess of the Probability density function from percentile values
Posted:
Dec 30, 2010 1:44 AM


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 >= .05F(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 z1constraints as z1 >= B7 (better written as z1B7 >= 0)and > z1 >= B7 (better written as z1+B7 >= 0). Note that if you use a x > error measure and seek the least absolutedeviation 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 superaccurate 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 illposed optimization problems. Anyway, Solver does not permit strict inequalities as far as I know. You need nonstrict 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 derivativeseven though it does not actually _use_ derivatives! The absolutevalue 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 absolutevalues 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 absolutevalue 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 *nonsmooth* 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

