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

Comments:

(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)