Answer to Question #149551 in Statistics and Probability for ajay

Question

prove that binomial distribution does not have the additive or reproductive property also find the condition under which the additive property holds.

Answer 1

Let X1  and X2  be two independent Binomial variate with parameters (n1,p1) and (n2,p2) respectively.  Let Y=X1+X2. Then the MGF (moment generating function) of Y is

$M_Y(t) = M_{X_1}(t)*M_{X_2}(t) = (q_1 +p_1 e^t)^{n_1} * (q_2 +p_2 e^t)^{n_2}$

It is not MGF of Binomial distribution.

But it is MGF of Binomial distribution iif p1 = p2=p:

$(q_1 +p_1 e^t)^{n_1} * (q_2 +p_2 e^t)^{n_2} = (q +p e^t)^{n_1} * (q +p e^t)^{n_2}=(q +p e^t)^{n_1+n_2}$

Now it is MGF of Binomial distribution with parameters (n1+n2, p)

Hence, condition is p1 = p2 then Y=X1+X2∼B(n1+n2, p)

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