How to use this probability distribution?

I have this probability distribution: Pi=1/(1+(i/31)^2,6) (Pi is the probability of i happens). But, how can I use this probability (Pi) for generate a random value? Which function I should use?

[Torsten]

And i ranges between which integer values ? And the Pi sum to 1 ?

Best wishes

Torsten.

[Rafael Pires]

i ranges between 1 and 100, Pi don't sum to 1. May you help me?

[Torsten]

If the Pi don't sum to 1, you don't have a probability distribution.

Or maybe the Pi are defined as

Pi=1/(1+(i/31)^2.6)/sum_{j=1}^{j=100}1/(1+(j/31)^2.6)

?

In any case, you can use

http://www.mathworks.com/matlabcentral/fileexchange/34101-random-numbers-from-a-discrete-distribution

Best wishes

Torsten.

[Rafael Pires]

sorry, Pi sum to 1.

[Torsten]

Use the MATLAB program from the file exchange above.

Best wishes

Torsten.

Answer 1

[John BG]

1.- it's good practice to avoid giving names to variables that are reserved for constants, like pi, so let me call the probability simply p, ok?

2.- when trying your function:

i=[0:10];p=1/(1+(i/31)^2,6)

errors happen,

so again let me modify the function you want to turn into a probability function, as follows:

   i=[0:10];p=1./(1+(i/31).^2.6)

3.- with

i=[1:100]

format long
sum(p)
=
  36.661222264870894

so a straight forward way to turn this into a probability density function is to divide all possibilities by

    i=[1:100];
    p=1./(1+(i/31).^2.6);
    P0=sum(p);
    p1=1/P0*(1./(1+(i/31).^2.6));

now

    sum(p1)
    =
       0.999999999999999

sum(p1) should be exactly 1.

4.- Because the exponent of i is larger than 2, the sum of the series converges as i gets away heading to +Inf.

One way to include all possible positive values of i [1,Inf] is

syms k 
f=1./(1+(k/31).^2.6)
int(f,k)
 =
(155*log((244140625*31^(4/5)*k^(1/5))/31 + 244140625))/13 + (155*exp((pi*2i)/13)*log(1650390625*exp((pi*3i)/13) - (1650390625*31^(4/5)*k^(1/5))/31))/13 - (155*exp((pi*5i)/13)*log(1650390625*exp((pi*1i)/13) - (1650390625*31^(4/5)*k^(1/5))/31))/13 - (155*exp((pi*1i)/13)*log(1650390625*exp((pi*8i)/13) + (1650390625*31^(4/5)*k^(1/5))/31))/13 + (155*exp((pi*4i)/13)*log(1650390625*exp((pi*6i)/13) + (1650390625*31^(4/5)*k^(1/5))/31))/13 - (155*exp((pi*7i)/13)*log(1650390625*exp((pi*4i)/13) + (1650390625*31^(4/5)*k^(1/5))/31))/13 + (155*exp((pi*10i)/13)*log(1650390625*exp((pi*2i)/13) + (1650390625*31^(4/5)*k^(1/5))/31))/13 - (155*exp((pi*3i)/13)*log(1650390625*exp((pi*11i)/13) - (1650390625*31^(4/5)*k^(1/5))/31))/13 + (155*exp((pi*6i)/13)*log(1650390625*exp((pi*9i)/13) - (1650390625*31^(4/5)*k^(1/5))/31))/13 - (155*exp((pi*9i)/13)*log(1650390625*exp((pi*7i)/13) - (1650390625*31^(4/5)*k^(1/5))/31))/13 + (155*exp((pi*12i)/13)*log(1650390625*exp((pi*5i)/13) - (1650390625*31^(4/5)*k^(1/5))/31))/13 + (155*exp((pi*8i)/13)*log(1650390625*exp((pi*12i)/13) + (1650390625*31^(4/5)*k^(1/5))/31))/13 - (155*exp((pi*11i)/13)*log(1650390625*exp((pi*10i)/13) + (1650390625*31^(4/5)*k^(1/5))/31))/13

your probability density function has as primitive, so let's integrate:

   double(int(1/((k/31)^(13/5) + 1),1,Inf))
   =
   39.060785674272687 - 0.000000000000000i

5.- one candidate to the probability density function you are after would be

P0=39.060785674272687
p=1/P0;1./(1+(i/31).^2.6)

for any positive integer >0, the range being

[1,Inf]

If you find this answer of any help solving your question,

please click on the thumbs-up vote link,

thanks in advance

John

[e-mail]

[Rafael Pires]

Thank you. That's ok, but how can I generate a random valou obeying this probability?

Answer 2

[Image Analyst]

You need to use inverse transform sampling. https://en.wikipedia.org/wiki/Inverse_transform_sampling

I've attached an example where I use that method for the Rayleigh distribution. If you don't know the analytical formula for the CDF, you could always construct the digital PDF and then use cumsum() to get the CDF.