Thread Subject: random number generator from logistic distribution

Hi,
I'm trying to create a random number generator from the
logistic distribution. Does the following function look
right or am I completely off track?!

r=lrnd(m,n,mu,sigma)
p=rand(m,n)
r=log(p./(1-p)).*sigma+mu

Thanks for your help

Jess

"Jessica " <jandre@mathworks.com> wrote in message <fnod3l$po
$1@fred.mathworks.com>...
> Hi,
> I'm trying to create a random number generator from the
> logistic distribution. Does the following function look
> right or am I completely off track?!
>
> r=lrnd(m,n,mu,sigma)
> p=rand(m,n)
> r=log(p./(1-p)).*sigma+mu
>
> Thanks for your help
>
> Jess
--------
  That looks good to me, except that of course functions are written this way:

 function r=lrnd(m,n,mu,sigma);
 etc.

You have correctly taken the inverse of the cumulative distribution function:

 p = F(x;mu,sigma) = 1/(1+exp(-(x-mu)/sigma))

  Note: Since sigma is a scalar, you don't need the dot when multiplying by it:

  r=log(p./(1-p))*sigma+mu;

Roger Stafford

"Jessica " <jandre@mathworks.com> wrote in message <fnod3l$po
$1@fred.mathworks.com>...
> Hi,
> I'm trying to create a random number generator from the
> logistic distribution. Does the following function look
> right or am I completely off track?!
>
> r=lrnd(m,n,mu,sigma)
> p=rand(m,n)
> r=log(p./(1-p)).*sigma+mu
>
> Thanks for your help
>
> Jess
--------
  Because you expressed doubt about the validity of the method in your
logistic distribution generator, Jessica, I have decided that it would be
appropriate to show details of the reasoning behind it.

  You defined the random variable p in terms of matlab's 'rand' function:

 p = rand

(forget for a moment that your p is actually an m by n array) and the random
variable r in terms of p as:

 r = f(p) = log(p/(1-p))*s+m

This is easily shown to be mathematically equivalent to saying

 p = f_inverse(r) = 1/(1+exp(-(r-m)/s))

Then we can write the cumulative distribution of r as:

 P{r <= c} = P{f(p) <= c} =
 P{p <= f_inverse(c)} = P{p <= 1/(1+exp(-(c-m)/s)

This holds because f is a monotonically increasing function so that

 f(p) <= c

is equivalent to

 p <= f_inverse(c)

and so have equal probabilities. Next notice that random variable p = rand
has the important special property that P{p <= t} is always equal to t itself for
any t in 0 <= t <= 1, because MathWorks has gone to great lengths to ensure
that rand is uniformly distributed from 0 to 1. Thus, applying this to t = 1/(1
+exp(-(c-m)/s)), we have

 P{r <= c} = P{p <= 1/(1+exp(-(c-m)/s))} = 1/(1+exp(-(c-m)/s))

Therefore r has the required cumulative logistic distribution.

  The same kind of reasoning applies to any random variable which is to be
defined in terms of the special random variable p = rand. One simply applies
the inverse of the desired cumulative distribution function to p.

Roger Stafford