Someone please help me, I have probability mass function as follwing, How can generate 100 random values of j from this probability mass function?

Note: λ>0 and 0≤α≤1/2 and 0≤P(X=j)≤1. (Take initial valus for λ, α and p(X=j) from thir domain as you like as example)

I want to generate randomly 100 numbers (values for j) with this probability mass function . But I really have no idea how and where to start.

Can somebody help me?

Thank you in advance

Answer 1

[Jeff Miller]

If you would like to generate 100 random j values, use the function KSSV gave you to generate the f values for the probability masses (they are not all 0's after the for loop, only before it). Then:

sump = cumsum(f);
r_unif = rand(100,1);
r_j = zeros(100,1);
for i=1:100
    r_j(i) = find(sump>r_unif(i),1);
end
figure;
histogram(r_j)

[mohammed elmenshawy]

Thank you very much for your cooperation.

Answer 2

[KSSV]

1. DEfine alpha and lambda values.
2. DEfine j value.
3. WRite your formula and substitute those values.

[mohammed elmenshawy]

sorry, I ask about j value ?

Meanning, i have λ, α and f(j) and ask about j value ?

If possible, I will change the value of f(j) only "but in its range" to get a new value for j and so on until I get to 100 values of j.

i write this (a means α and ld means λ )

syms j;
a=0.3; ld=0.5;
f=((1-a)^j-(-a)^j)*ld^j* exp(ld*a)/(factorial(j)*(exp(ld)-1));
J=solve(f,j);  

"i want solve it when f equal for example 0.8"

Hope you understood what I mean..thanks very much for your help

[KSSV]

a = 0.3 ; ld = 0.5;
% f = 0.8 ;
f = zeros(1,100)
for j = 1:100 
  f(j) = ((1-a)^j-(-a)^j)*ld^j* exp(ld*a)/(factorial(j)*(exp(ld)-1)) ;
end

[mohammed elmenshawy]

The result is all zeros, which is illogical here..

secondly, i want j value not f . "f is input and j which output"

[KSSV]

Firs few values are significant..rest are very small. Can you share the reference where you have this formula?

[mohammed elmenshawy]

https://link.springer.com/content/pdf/10.1007/s00184-009-0252-5.pdf

"equation 5 "

i want make simulation for j value

[mohammed elmenshawy]

Thank you very much for your cooperation.

Answer 3

[Bruno Luong]

alpha=0.01; lambda=0.99;

f=@(j) ((1-alpha).^j-(-alpha).^j).*lambda.^j*exp(lambda*alpha)./(factorial(j)*(exp(lambda)-1));

p=f(1:22); % f is practically 0 beyond 10  
c=cumsum(p); c=c/c(end);
n = 100; % 1e6;
[~,j] = histc(rand(1,n),[0 c]);