# About simluation of the poisson point process. Thank you

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sun on 14 Jan 2015
Commented: KalMandy on 20 Jan 2017
Dear friends,
I am new with poisson point process. I did one simluation 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, http://connor-johnson.com/2014/02/25/spatial-point-processes/
showed me that when intensity lamuda = 0.2, smaller than 1 , he got 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 poiion point process? or it's a programming problem?
Thank you so much for your help.

sun on 14 Jan 2015
Guys, I think I got it. I was plotting the unit figure. but the 2nd figure is not plotting the unit but 0 - 20. Thank you.
la=0.4;
lala=0.4*100;
npoints = poissrnd(lala);
pproc = rand(npoints, 2);
plot(pproc(:, 1).*100, pproc(:, 2).*100, '.');

Syed on 10 Oct 2016
Don't you think it should be something like below.
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')
KalMandy on 20 Jan 2017
hi, do we need any special toolbox to run these programs with PPP (Poison Point Process)?