Is a sum of two binomial distributions with different $p$ also binomial? [duplicate]

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 $$p_1\neq p_2$$?

(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)


• Why does the variance of a Binom$(10,0.2)$ random variable equal $1.8$? Isn't the variance of a Binom$(n,p)$ random variable equal to $np(1-p)$ which in this case equals $10\times 0.2\times 0.8 = 1.6$? Jun 8 '20 at 14:15
• @DilipSarwate. Thanks much for correction note. Late night typo. Fixed it. Added correct values to simulation, which I might have done before. Jun 8 '20 at 18:30