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These are replies submitted by mmcdara

@acer @tomleslie

Thanks to both of you.
I've been on vacation since yesterday and will be back at the office next Monday, so I can't give you, right now,  the exact version of Maple I used.
All I can say is that it was Maple 2020. ???? on Windows 10.
I will come back to this thread on Monday with more information.


Very disturbing last figure.
Stacking histograms is quite unusual and gives a very wrong idea of the empirical distributions.


Thanks for your reply

@Shameera : the answer of your deleted answer is avaliable here  233822-How--To-Solve-This-Equation-By-Using


(please look to my last question and @acer's answer)

Here is, after a lot of corrections, the solution of your last problem



T = { {}, Z} is a sigma-algebra
T = P(Z) where P(Z) is the set of parts of Z (the output of combinat:-powerset) ia another sigma-algebra
T = { {}, {1}, {0, 2, 3}, Z } is still another sigma-algebra.

What do you want ?

Hi, is there a difference in using


instead of




The integration along x should be -3..1


Thanks Tom, I hadn't had time to check your solution.
I just sent you and Acer a reply that seems to answer my problem.

Thanks again

@acer @tomleslie

Having look to the procedure RandomVariable, I've seen that instead of 

A := RandomVariable(Uniform(0, 1));

it is possible to write 

RandomVariable(A, Uniform(0, 1));

(which is an undocumented feature).

Thus I think that the better way (IMO) to fix my problem is to do this

RandomVariable(A, Uniform(0, 1));
RandomVariable(B, Uniform(0, 2));
RandomVariable(C, Uniform(1, 2));
F := (A+2*B)/C
                            A + 2 B



"which may not be true...." : I agree, it was implicit to me that I would not write something like P:=A :

"I am not a fan of this kind of thing,..." : Too bad, it seemed to me a good alternative to tomleslie's answer.

Concerning your last proposal, what annoys me is that the introduction of new names (a, b, c) can make the evaluation "unsafe" in the sense that we could write inadvertently [A=a, B=b, C=b].
I had the feeling that tomleslie's proposal was a good solution: what is your opinion about it? Has his method any drawbacks?

What is the format of all these dates?


Very good idea Tom.

What do you expect for the number 123000321 ???

@Carl Love 

A very astute way.

Nevertheless, here is a problem with the question the OP asked, please look at this

x := 123000321;
ListTools:-Reverse(convert(x, base, 1000))
                         [123, 0, 321]

I suppose the OP would like to get  [123, 000, 321] instead, which in some sense is stupid.


I vote up.

BTW, expressing your solution in terms of hypergeometric functions and comparing the result to the one got by the OP could mabe (or not) help understanding why your change of variable leads to a solution?

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