@Carl Love 

Right Carl, good point (beta distribution) and nice work (the completely formal solution).

It would be interesting to try to generalize this formal derivation to non uniform pdf.

For some of them order distribution are known and I think it could be possible to obtain interesting results.

By the way, my last mistake was that I write Ecost as a sum of two terms invoking the integral of Q-x as I should have written that it was defined as the value work Wich these two integrals are equal.

I think it would be better to avoid integrating Q-x between a and Q and x-Q between Q and b: fundamentally this is the same loss function Q-x (or x-Q) that we use.

Hope it's clear, I send this from my smartphone

@cinderella look to my code and your code.

The only difference is the expression of N. I define N as the integral of Q-x against the measure f, as you define  this same N as the integral of x-Q against the measure f.

If you are sure of what you did change the sign of N in my code.

You can do this in the 'numerical' code l sent you yesterday or in the formal code I sent you nine hours ago (in this matter change the sign of MAX in the expression of Qstar) 

I have no doubt you will be capable to fix this by yourself. Now I'm in weekend, see you in two days if necessary.

@Magma 

Then you can use this
reference : Maple V, release 5, Monagan et al, 1996  page 55

uniform := proc(r::constant..constant)
  local intrange, f:
  intrange := map(x -> round(x*10^Digits), evalf(r) ):
  f := rand(intrange)
  (evalf @ eval(f)) / 10^Digits)
end proc:

# usage

N := 100:
U := uniform(1..N):
s := [seq(U(), n=1..N)]:
t := sort(s, output=permutation)

@tomleslie 

This is indeed a fine way to address the problem.
Personally, despite an intensive use of the Statistics package, I'd never payed attention to this EmpiricalDistribution feature.
Maybe you have given a definitive answer to the question, I'm interested in reading acer's opinion on this point.

 

@acer 

Je ne suis pas certain que demander à ce qu'on vous fasse vos TD/TP de licence ou d'école d'ingé soit une bonne manière de vous faire apprendre Maple ?
Vous auriez pu, au moins ne pas reproduire mot à mot le sujet de ce TP.

==============================================
I'm not sure that asking for someone to do your master's or  engineering school's tutorials (or practical work) is a good way to get you to learn Maple?
You could have at least not reproduced word for word the subject of your practical work..
===============================================


As a starting point, for question Aa:
First of all you need a stopping criterion, for instance a measure of the difference between the exact value (if known) and its iterated approximations x[n], or the difference between x[n] and x[n-1].
Here is an example :

babylon := proc(x0, k, epsilon)
  local x:
  x := copy(x0):
  while abs(x^2-k) > epsilon do
    x := (x+k/x)/2
  end do:
  return x;
end proc:

# usage
babylon(2, 1, 0.001)

For point Ab: search list, or vector, or Array, ... in the help pages

For point B: search "thisproc" in the help page and take inspiration from the recursive calculus of the factorial of an integer

@tomleslie 

Of course!

I'm stupid, I didn't even think to use Sample~ !!!

Thanks Tom

@mvonhippel 

Just read the help pages about Grid:-Launch.
You will find an example (primetest) that  I've just transposed to your test case.

Unfortunately I can give you detailed explanations, beyond that of saying "respect the content of the help pages".
If you are interesting in knowing why you must write babyGrid := proc() instead of babyGrid := proc(X) and later Grid:-Launch(babyGrid, imports=['X']) instead of Grid:-Launch(babyGrid, X), I'm not qualified to answer.

All you have to do is wait for someone else to give you this information.

@Carl Love  @tomleslie

Hi,

Before sending my answer to Erik, I had tried to sample a function of two RV with values in R^2 (typically the "f" function used by Tom).

Remember I used RandomVariables Dice 1 and Dice2.
So I wrote, in a natural way,  f := (Dice1, Dice2) -> [Dice1, Dice2] and Sample(f(Dice1, Dice2), N).
As it generated an error, I tried to lure Maple by writting instead  f := (Dice1, Dice2) -> Dice1+I*Dice2
... which generated the same error as before.

All of this seems related to the impossibility (?) to define a multivariate random variable.
Am I right ?
Is it possible (without sampling independantly Dice1 and Dice 2 as Tom did) to define multivariate RV in Maple?
 

@Josolumoh 

@acer @tomleslie

What version are you using?

• It was Maple 2016 (usually I use Maple 2018 but I had have a long simulation running on with it and, in these case, I prefer to open a new worksheet in an older version).
  I guess that If I had done the test with Maple 2018 I would not sent this question

Does it change if you set interface(typesetting=extended): ?

• Yes, it works perfectly well in Maple 2016

Thanks to both of you, sorry for the inconvenience.

@Josolumoh 

In my previous reply, please change

@Josolumoh 

A spreviously quoted by Carl the case d=0 is obvious.
More precisely it corresponds to a negative binomial distribution with parameters r and b/(2+b)
(in Maple  NegativeBinomial(r, b/(2+b))  )

Carl also pointed the meaningless of the case d<0

I describe here the way to sample your  "pseudo distribution" B (which does not sum to 1 and then need to me normalized).
The last plot superimposes the probability function of B for r=3, b=2, d=1 (red line) and the histogram in discrete form.
I considered that only integers values of x have a sense.

For other valuers of r, b, d you will have to verify that the value of the variable Xmax is large enough (in this case the variable ShouldBe1 must have reached a "converged" value; to verify this increase the value of Xmax and look to what happened to ShouldBe1).

Let me know ih this suits you

 

Download B_Distribution.mw

@tomleslie

Sorry, you're right, I forgot the "1/3" in the coefficient of x^3.
 

@tomleslie 

Not applicable: see tomleslie's answer

You got a wrong result with Optimization:-NLPSolve (?)

Given fexact is monotone decreasing over [0, 2] and its values are between [0, 1], given fapp is monotone increasing on [0, 2] with fapp(2) > 6789*2, it is obvious that the the maximum M of abs(fapp-fexact) is reached for x=2.

Then M = subs(x=2, fapp-fexact) =  - sin(2)/2 which is about  30543.54535 and not 2827.54535.

Using instead Optimization:-Maximize( abs(fapp-fexact), x=0..2) returns the good value.

More of this: your two calculations (L2 norm and L1 norm) give strangely the same result: could it be that Optimization:-Maximize(  abs(fapp-fexact), x=0..2, maximize) does not do what you were expecting for?