Thread Subject: help with generating skewed distribution?

Hello,
I would like to generate a population of numbers, from 1-
100 for example, distributed almost uniformly. However, I
want the larger numbers to have more chance of popping up,
so the population skews to the left.
how would i be able to do this with matlab 2008? i had
thought of using randsrc, but for some reason it's not in
my version? i have the statistics toolbox, if that helps.
thanks!

"Anna Chen" <icedredtea@yahoo.com> wrote in message
<g8466e$amb$1@fred.mathworks.com>...
> Hello,
> I would like to generate a population of numbers, from 1-
> 100 for example, distributed almost uniformly. However,
I
> want the larger numbers to have more chance of popping
up,
> so the population skews to the left.
> how would i be able to do this with matlab 2008? i had
> thought of using randsrc, but for some reason it's not in
> my version? i have the statistics toolbox, if that helps.
> thanks!

randsrc is part of the Communications Toolbox.

"Anna Chen" <icedredtea@yahoo.com> wrote in message
<g8466e$amb$1@fred.mathworks.com>...
> Hello,
> I would like to generate a population of numbers, from 1-
> 100 for example, distributed almost uniformly. However,
I
> want the larger numbers to have more chance of popping
up,
> so the population skews to the left.
> how would i be able to do this with matlab 2008? i had
> thought of using randsrc, but for some reason it's not
in
> my version? i have the statistics toolbox, if that
helps.
> thanks!


Hi Anna,
this is how you can do it. It's actually a general way of
drawing random numbers from any distribution. Start by
defining what kind of distribution you want. You said you
used a uniform distribution, p(x) = 1/100 for x=1,...,100.
Let's assume you instead want to use, for example, a
triangular distribution, p(x) = constant * x for
x=1,...,100. Of course, you need to set the constant such
that the function sum to one, that is p(1)+...+p(100)=1.
In this case the constant is 1/5050, so p(1) = 1/5050 and p
(100) = 100/5050. Next step is to construct the cumulative
distribution, which in fact would correspond to stacking
the 100 probability bars on top of each other.
Now draw a uniform random number between zero and one.
Compare this number with your cumulative distribution and
make a note within which probability bar the drawn random
number lies. The x value corresponding to that probability
bar would then be a draw from your desired triangular
distribution. For example assume that, to make it more
easy for me, you drew the number 0.9851 from the uniform
distribution. Comparing this number with the cumulative
distribution you'll see that it lies within the last
probability bar, since the last bar is 0.0198 tall. You
can thus conclude that x=100 is a draw from your desired
triangular distribution. Continuing like this you'll note
that, in the long run, the drawn uniform number will "hit"
taller probability bars more often that shorter ones
implying more frequent draws from larger x numbers, just
as you wanted.
Best of luck

Anna Chen wrote:

> I would like to generate a population of numbers, from 1-
> 100 for example, distributed almost uniformly. However, I
> want the larger numbers to have more chance of popping up,
> so the population skews to the left.


As long as you mean, "the integers from 1 to 100", and as long as you're willing to actually specify all 100 probabilities, then RANDSAMPLE is just what you need. You might take a look at DISTTOOL, and the beta distribution. While this is a continuous dist'n defined on the unit interval, you can fiddle with the parameters to get roughly the shape you're looking for, a=3, b=1, for example. Then use BETAPDF to evaluate the beta density on (1:100)/101, normalize those values to sum to 1, and pass that to RANDSAMPLE, something like

   p = betapdf((1:100)/101,3,1);
   randsample(1:100,n,true,p/sum(p))

Hope this helps.