Poisson point process in matlab

Question

I am new with poisson point process. I did one simluation (matlab) as below. My intensity lambda = 50;

clear all;
lambda=50;
 npoints = poissrnd(lambda);
  pproc = rand(npoints, 2);
  plot(pproc(:, 1), pproc(:, 2), '.');

Then I have plot,

However, the link

http://connor-johnson.com/2014/02/25/spatial-point-processes/

showed me that when intensity lamuda = 0.2, smaller than 1 , he got

The link also showed the code in Python.Please check it.

Here is my question, why intensity is smaller than 1, he still can plot something here? If I let my code's lamda = 0.2, there will be no value to plot. I think I miss something about Poisson point process? or it's a programming problem? I want to simulate this lambda = 0.2 in matlab.

Thank you so much for your help.

total lambda = 0.4. I want to have 100 unit. This means unit intensity = 0.4 * 100 = 40. thank you so much.

 la=0.4;
 lala=0.4*100;
  npoints = poissrnd(lala);

  pproc = rand(npoints, 2);

  plot(pproc(:, 1).*100, pproc(:, 2).*100, '.');

Orran

Yes, there's a misunderstanding of the math. Look at the example more closely: the rate of events per unit time per unit area is 0.2, but lambda (the argument to the poisson function) is not 0.2. Your code only considers one unit of area, which will usually get zero events per unit time (but sometimes 1 or more). This isn't really a programming problem (thus I'm not leaving this as an Answer) – Glutamate

Note that you're just plotting a uniform distribution; the Poisson distribution is only used once to get a number of points. Edit: the author does point out, "the only thing Poisson about this is the number of points." – Glutamate

@Glutamate But a Poisson point process is just that: a Poisson number of points with uniform spatial distribution – Liederkranz

@LuisMendo What do you mean by "Poisson number"? Is there some "Poissonness" that sticks around with npoints after is assigned? ;-) – Glutamate

@Glutamate Thanks for your reply. Yes, you are right. I go zero events per unit time. that's why I can not have the figure. Could you please give me more detail about this? and How can I fix it for intensity smaller than 1? If there is a related link, it will be helpful. I am still learning PPP now. Thank you so much. – Orran

@Glutamate Dear Lack, I just edit my question. Could you please check the code for me very quick. It's short. Thank you so much. – Orran

@LuisMendo Hey, bro. I just re-edit. I think Lack is right. Please check it. – Orran

Answer 1

I think your solution should be following. (Note : Two different methods of drawing PPP spatial distribution)

clear all;
clc;
close all;

lambda=50;

%Method 1
pproc  = poissrnd(lambda, 100, 2);
size(pproc)
plot(pproc(:, 1), pproc(:, 2), '.');
title('Poisson with poissrnd')

%Method 2
pproc2 = random('Poisson', lambda, 100, 2);
size(pproc2)
figure;
plot(pproc2(:, 1), pproc2(:, 2), '.');
title('Poisson with Random statement')

Housemaster

Answer 2

total lambda = 0.4. I want to have 100 unit. This means unit intensity = 0.4 * 100 = 40. thank you so much.

 la=0.4;
 lala=0.4*100;
  npoints = poissrnd(lala);

  pproc = rand(npoints, 2);

  plot(pproc(:, 1).*100, pproc(:, 2).*100, '.');

Please check the figure in my question at the end.

Orran