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a problem while using an anonymous function

Asked by Eliraz Nahum on 10 Oct 2018
Latest activity Commented on by Eliraz Nahum on 11 Oct 2018
hello,
I can't understand what's wrong. I will appreciate any help. I have a problem while trying to using the integral function with the handle of an anonymous function. I am adding a photo of the output error.
clear all
close all
clc
lamda=5;
x=[1:1:30];
p=zeros(1,length(x));
poiss_PDF=zeros(1,length(x));
poiss_int=zeros(1,length(x));
step=0;
poiss_pdf=@(x) (1/factorial(x))*(lamda^x)*exp(-lamda);
for x_i=x
step=step+1;
p(step)=poiss_pdf(x_i);
poiss_PDF(step)=poisspdf(x_i,lamda);
poiss_int(step)=integral(poiss_pdf,x(1),x(end));
end
plot(x,p,'b')
title (['Poisson PDF (calculated) with the parameter lamda= ',num2str(lamda)])
xlabel('x')
ylabel ('Distribution')
xlim([x(1) x(end)])
hold on
plot (lamda,poiss_pdf(lamda),'dr')
figure
plot(x,poiss_PDF,'g')
title (['Poisson PDF (from Matlab) with the parameter lamda= ',num2str(lamda)])
xlabel('x')
ylabel ('Distribution')
xlim([x(1) x(end)])
hold on
plot (lamda,poisspdf(lamda,lamda),'dr')
figure
plot(x,poiss_int,'k')
title (['Poisson integral (calculated) with the parameter lamda= ',num2str(lamda)])
xlabel('x')
ylabel ('Probability')
xlim([x(1) x(end)])
hold on
plot (lamda,poiss_pdf(lamda),'dr')

  2 Comments

jonas
on 10 Oct 2018
The problem is that factorial only works for integers, so your function is only defined in those points. This is not my field of expertise, but wikipedia tells me you can use a gamma-function instead.
Why you are using integral to handle discrete PDF like Poisson? For discrete you need to do a sum.
Or transform the discrete PDF to a sum of weighted dirac, not sur integral can handle though.

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2 Answers

Answer by the cyclist
on 10 Oct 2018
Edited by the cyclist
on 10 Oct 2018
 Accepted Answer

Replace
poiss_pdf=@(x) (1/factorial(x))*(lamda^x)*exp(-lamda);
with
poiss_pdf=@(x) (1./gamma(x+1)).*(lamda.^x).*exp(-lamda);
and your code will execute to completion. I did not try to figure out if the output is sensible.
Note that I did two things here:
  • Replaced your factorial with the appropriate gamma function
  • Used element-wise operations in the function, which I'm pretty sure is what you intended

  3 Comments

hey...why do I have to use the element-wise operations if I give the function each time a single value and not a vector. I thought that as long as the function deals with a single number (1X1) I don't have to use it.
Because internally, the integral function creates a vector from x(1) to x(end), and evaluates your function using that vector.
Ok, understood now. Thanks

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Answer by Jos (10584)
on 10 Oct 2018

You do not show the whole error message!
But I also assume that the problem is indeed in the use of factorial inside the function poiss_pdf that is being passed to integral. The function factorial(x) is only defined for integer values of x. You can use gamma(x+1) instead. See the documentation of gamma for more details.

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