traffic= poissrnd(lambda), lambda value is small

traffic= poissrnd(lambda), if I choose lambda between (0 and 1) (1> lambda >0) .what will be happened to the expected traffic ?

Answer 1

[David Goodmanson]

Hello Hassan,

Since lambda is the mean value of the distribution, for small lambda there will not be a lot of traffic. If lambda is .1 and you take 100 draws with poissrnd(.1,1,100), then roughly speaking you would expect about 100*.1 = 10 ones, 90 zeros and the occasional draw of value two or greater. More precisely, for values n = 0,1,2,3 the expected number of draws are

lambda = .1;
n = 0:3;
Ndraws = 100;
expected_draws = Ndraws*(lambda.^n./factorial(n))*exp(-lambda)

expected_draws = 90.4837 9.0484 0.4524 0.0151

[Hassan Al-Khateeb]

sorry, for this mistake the value of lambda is greater than 0 and less than 1 (e.g lambda=0.5) ? is it true !!!! what will be the expected traffic for lambda =0.5??

[David Goodmanson]

Hello Hassan,

Same answer as above, basically. For lambda = .5, the mean of the poisson distribution is now 1/2. Putting lambda = .5 in the code above,

expected_draws = 60.6531   30.3265    7.5816    1.2636

so poissrnd(.5,1,100) would give the number of draws of 0,1,2,3 somewhere in that vicinity.