Variance of aggregated distributions of binomial random variables Let two random variables: $$x_1 \sim Bin(100, 0.5) \\ x_2 \sim Bin(100, 0.6)$$
Now, we define a third random variable, $x_{12}$ which it's distribution is the aggregated distributions of $x_1$ and $x_2$, so it's not quite like $x_1 + x_2$ even though empirically the variance seems like the sum of the two variances. Is that the case? How can I show it?
Thanks. 
 A: It seems you may be describing a mixture of distributions rather than a sum
For the individual distributions you have 


*

*$E[X_1] =50$, $\text{Var}[X_1]=25$, $E[X_1^2] = 2525$   

*$E[X_2] = 60$, $\text{Var}[X_2]=24$, $E[X_2^2] = 3624$


For the sum you would have


*

*$E[X_1+X_2] = 50+60=110$, 

*$\text{Var}[X_1+X_2]=25+24=49$, 

*$E[(X_1+X_2)^2] =110^2+49= 12149$  


But for a mixture (half the observations from the first distribution and half from the second) you would have


*

*$E[X_m] = \frac{50+60}{2}=55$, 

*$E[X_m^2] = \frac{2525+3624}{2}=3074.5$

*$\text{Var}[X_m]=3074.5-55^2=49.5$, 


which is not quite the sameas the sum, though the variances are close (as you have noticed)
A: From a theoretical point of view it is exactly the distribution of $X_1 + X_2.$ But it is not binomial because the two Success probabilities are not the same.
$$E(X_1 + X_2) = n_1p_1 + n_2p_2 = 50 + 60.$$
Provided $X_1$ and $X_2$ are independent, you also have
$$Var(X_1 + X_2) = n_1p_1(1-p_1) + n_2p_2(1-p_2).$$
[In particular, this is not the same thing as $(n_1 + n_2)p_a(1-p_a),$ where
$p_a = (p_1 + p_2)/2.$] 
If this were an applied situation and the 100 observations for $X_1$ were
taken contemporaneously with the 100 observations for $X_2$ and in the same place, then I'd investigate circumstances to make sure about the independence.
For example, if the subjects for $X_1$ are 100 randomly chosen men from a population in which 50% are Democrats, and the subjects for $X_2$ are 100 randomly chosen women from a population in which 60% are Democrats, then independence seems reasonable. But I'd want to make sure that 'for
convenience' the researchers didn't select 100 married M/F couples and
use the 100 men and 100 women from those.
