I want to share this with the list.
----- Original Message -----
From: "Richard Hennessy" <rich.hennessy at verizon.net>
To: "Jack Schmidt" <jack.schmidt at sbcglobal.net>
Sent: Monday, August 24, 2009 9:00 PM
Subject: Re: [Maxima] Ideas welcome
I have a new, faster pw.mac which isn't finished yet but it works in some cases.
define(boxcar(x), 1/(2*a)*(signum(x+a)-signum(x-a))/2)$
dc:convolution(boxcar(x),boxcar(x),x);
Evaluation took 0.0600 seconds (0.0600 elapsed)
((x+2*a)*signum(x+2*a)+(x-2*a)*signum(x-2*a)-2*x*signum(x))/(8*a^2)
for i : 1 thru 40 do dc:convolution(dc,boxcar(x),x);
Evaluation took 84.9800 seconds (84.9800 elapsed)
done
(s:at(pwdefint(dc,x,40/100,inf),a=1/100),float(s));
Evaluation took 2.0300 seconds (2.0300 elapsed)
7.117406731291439*10^-52
It still has a lot of bugs, but this case worked like a charm.
Rich
----- Original Message -----
From: "Jack Schmidt" <jack.schmidt at sbcglobal.net>
To: "Richard Hennessy" <rich.hennessy at verizon.net>
Sent: Sunday, August 09, 2009 1:43 PM
Subject: Re: [Maxima] Ideas welcome
The problem is:
Given two ball bearings whose diameters are random and uniformly
distributed, what is the probability distribution of the total width of
the two bearings together when they are in contact (ie., sum of the two
diameters)?
The answer is the convolution of the two bearing distributions. In the
specific case of uniform distribution of the individual bearings, the
lower bound on the total width is the lower bound of bearing A plus the
lower bound of bearing B; and the upper bound on the total is the upper
bound of A plus the upper bound of B. And in between, the distribution
function is triangular. This is like Example 7.3 in
http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf
Here's some stuff on wikipedia:
http://en.wikipedia.org/wiki/Probability_distribution#Some_propertieshttp://en.wikipedia.org/wiki/Probability_density_function#Sums_of_independent_random_variables
Richard Hennessy wrote:
> Thanks but I don't quite get the relationship between convolutions and the answer. Oh well, I guess I am not that smart.
>
> Rich
>
>
> ----- Original Message -----
> From: "Jack Schmidt" <jack.schmidt at sbcglobal.net>
> To: "Richard Hennessy" <rich.hennessy at verizon.net>
> Sent: Sunday, August 09, 2009 12:28 PM
> Subject: Re: [Maxima] Ideas welcome
>
>
> Hi Rich,
>
> I don't have a specific case. I would suggest starting with something
> that can be easily worked out on paper, like the convolution of two
> uniformly distributed random variables, each with a different width.
> The convolution is a triangle function if the widths are equal, or a
> trapezoid if not.
>
> Jack
>
>
>
>
> Richard Hennessy wrote:
>
>> Jack,
>>
>> Can you give me the answers you get for this specific case given some real tolerance and cutoff values and the real numerical
>> values. I just want to check my answer with yours. This kind of stuff is right up my alley.
>>
>> Rich
>>
>
>
>
>