Thread Subject:

Random number for triangular distribution

I want generate random numbers from a triangular distribution. Can anyone help me?
Thanks in advance

"Partha Roy" <p_roy_007@yahoo.com> wrote in message <k43g98$ksh$1@newscl01ah.mathworks.com>...
> I want generate random numbers from a triangular distribution. Can anyone help me?
> Thanks in advance

% function that generates triangular distribution between (0,1)
trand = @(varargin) 0.5*(rand(varargin{:})+rand(varargin{:}));

% Test
r=trand(1,100000);
hist(r,30)

% Bruno

Thanks Bruno Luong, but may be i was not specific what i wanna know. I want to determine random numbers from a triangular distribution like (5,7,10). how can i write the program for this in the M file. Could you help me with this?
Thanks again.

Asymmetrical triangular distribution is more complicated to generated:

p = [5 7 10];

c = (p(2)-p(1))/(p(3)-p(1));
a = (p(3)-p(1))*(p(2)-p(1));
b = (p(3)-p(1))*(p(3)-p(2));
l = @(r) r <= c;
f = @(r) p(1) + sqrt(a*r);
g = @(r) p(3) - sqrt(b*(1-r));
h = @(r) l(r).*f(r) + (1-l(r)).*g(r);
trand = @(varargin) h(rand(varargin{:}));

r = trand(1,100000);
hist(r,30)

% Bruno

"Partha Roy" <p_roy_007@yahoo.com> wrote in message <k43hpv$plh$1@newscl01ah.mathworks.com>...
> ..... I want to determine random numbers from a triangular distribution like (5,7,10).
- - - - - - - - -
  Here's a slightly different method:

 n = 100000;
 a = 5; b = 10; c = 7;
 r = c+sqrt(rand(n,1)).*(a-c+rand(n,1)*(b-a));

Roger Stafford

"Roger Stafford" wrote in message <k463ad$dm5$1@newscl01ah.mathworks.com>...
>
> n = 100000;
> a = 5; b = 10; c = 7;
> r = c+sqrt(rand(n,1)).*(a-c+rand(n,1)*(b-a));
>

Brilliant!

Bruno

What can I say Roger... Mind Blowing :)

Might I ask you Roger what leads you to this formula?

You don't need to elaborate the calculation details, just explaining the overall approach will satisfy me. Thanks.

Bruno

"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <k46k48$7s3$1@newscl01ah.mathworks.com>...
> Might I ask you Roger what leads you to this formula?
> You don't need to elaborate the calculation details, just explaining the overall approach will satisfy me. Thanks.
> Bruno
- - - - - - - - - -
  The triangular distribution f(x|a,b,c) as defined in Wikipedia:

 http://en.wikipedia.org/wiki/Triangular_distribution

is the same as the distribution of the x-coordinates of a statistically uniform distribution of (x,y) points in the interior of the triangle (a,1),(b,1),(c,0), and that is what my method is based on. The y-coordinates would be given by the sqrt(rand(n,1)) factor.

  This approach admittedly has the disadvantage of requiring twice as many calls on 'rand' to achieve the same result as yours, Bruno, though it has fewer lines of code.

Roger Stafford

"Roger Stafford" wrote in message <k46mdc$equ$1@newscl01ah.mathworks.com>...
>
> is the same as the distribution of the x-coordinates of a statistically uniform distribution of (x,y) points in the interior of the triangle (a,1),(b,1),(c,0), and that is what my method is based on. The y-coordinates would be given by the sqrt(rand(n,1)) factor.
>

Got it. Thanks.

Bruno