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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