Sum of Poisson Distributions

The Sum of passion random variable has the number of success value ,that result comes from the passion experiments. The Sum of Probability distributions of the passion random variable is called passion distribution. Suppose we perform a Poisson experiment, in which the average number of successes within a given region is λ. Then, the sum of Poisson probability distribution is:

 P(x ; lambda) = (e - lambda) (lambda^x) / (x!)

where x is the actual number of the  Poisson experiment, and e is approximately equal to 2.71828. Now we see the sum of properties of passion distribution and then some example problems.


Properties of Poisson Distributions:

Mean and variance of  the  sum of poisson distributions is equal to λ.

Mean    E[X] = λ

Variance  Var[X]=λ

Probability mass Function:

Probability mass function of the probability distribution is

P[ X=k] = e^-lambda lambda^k/ (k!)

Management of  the sum of Poisson distribution is

 M(t) = exp ( lambda( e^t - 1) )

Generating function of the sum of  Poisson distribution is

 G(s) = e^lambda(s-1)

Large value of the parameter  - The Poisson distribution can be efficiently calculated by a normal distribution for large values of λ. Let

X1 ~ Poisson(λ1),    X2 ~ Poisson(λ2 ),   X1 and X2 independent.

Then,

 [X1+X2] ~~ Poisson (lambda1+lambda2).

Example Problems for Poisson Distribution:

Example problem 1: If a publisher of  technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing at least one such error is 0.004 and errors are independent from page to page

(i) what is the probability of its 500 page novels will contain exactly one page with error.

(ii) at-most three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].

Solution :

n = 500 , p = 0.004
np = 2 = λ
(i)  P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii)  P(at most 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x!    [limits 0 to 3]
= e−2 (2)x  /  x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569

Is this topic algebra problems hard for you? Watch out for my coming posts.

Exercises Problems for Poisson Distributions

Example problem 1: If a publisher of  technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing at least one such error is 0.002 and errors are independent from page to page

(i) what is the probability of its 500 page novels will contain exactly one page with error.

(ii) atmost three pages with errors.
[e−1 = 0.3678 ].

Answer:  0.9811