# Solved – Is a sum of two binomial distributions with different \$p\$ also binomial?

I have two independent random variables which follow binomial distributions
$$X sim B (n_1, p_1)$$ and $$Y sim B (n_2, p_2)$$.

Can we say that $$Z = X + Y$$ is also binomially distributed $$Z sim B (n_1+n_2, (p_1+p_2)/2)$$? Can you explain the answer to me please?

I have read many articles without finding the answer. I know that if $$p = p_1 = p_2$$ then we can say that $$Z sim B (n_1+n_2, p)$$. But what can we say when $$p_1neq p_2$$?

Contents

(a) Let $$X sim mathsf{Binom}(10,.2),$$ $$Y sim mathsf{Binom}(10,.8),$$ and $$W sim mathsf{Binom}(20,.5).$$ Then $$Var(X) = Var(Y) = 1.6$$ and $$Var(X+Y)=3.2.$$ But $$Var(W) = 5.$$ So $$X+Y$$ and $$W$$ can't have the same distribution.

(b) $$P(X+Y = 0) = P(X=0,Y=0) = P(X=0)P(Y=0) ne P(W = 0).$$

``pbinom(0,10,.2)*pbinom(0,10,.8) [1] 1.099512e-08 pbinom(0,20,.5) [1] 9.536743e-07 ``

(c) For $$X sim mathsf{Binom}(10,.2)$$ and $$Y sim mathsf{Binom}(10,.8),$$ here is R code to make a histogram of a large sample from $$Z = X + Y.$$ Then the red dots show the distribution $$mathsf{Binom}(20, .5).$$ The dots don't match the histogram.

``set.seed(1234) m = 10^6 x = rbinom(m, 10, .2) y = rbinom(m, 10, .8) z = x + y mean(x); var(x) [1] 2.000237    # aprx E(X) = 2 [1] 1.603947    # aprx Var(X) = 1.6 mean(z); var(z) [1] 10.00093    # aprx E(Z) = E(X) + E(Y) = 2 + 8 = 10 [1] 3.208881    # aprx Var(Z) = Var(X)+Var(Y) = 1.6+1.6 = 3.2  cutp = (-1:20)+.5 hist(z, prob=T, br=cutp, col="skyblue2")  k = 0:20;  pdf=dbinom(k,20,.5)  points(k,pdf, col="red", pch=19) ``

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