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?


  • $\begingroup$ In the finite case: Suppose we have $500$ samplings of $x_1$ and $500$ samplings of $x_2$ then $x_{12}$ is the concatenation of the two ($1000$ samplings). $\endgroup$ – Covvar Nov 24 '16 at 11:59
  • $\begingroup$ so it's sort of like adding the distributions together I guess. $\endgroup$ – Covvar Nov 24 '16 at 12:00

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)

  • $\begingroup$ I assumed 'aggregate' meant all 200, not 50 from each for a 100 in all. I believe 50-50 'mixture' of 100 would mean to take from Gp1 and Gp2 each with probability 1/2 until you get a total of 100 values (0's or 1's). Vague question. $\endgroup$ – BruceET Nov 24 '16 at 22:18
  • $\begingroup$ @BruceET - I assumed each $X_i$ could take values in $\{0,1,2,3,\ldots ,99,100\}$ though I might be wrong $\endgroup$ – Henry Nov 24 '16 at 22:24
  • $\begingroup$ And we may both be wrong. I'm not making any claims. $\endgroup$ – BruceET Nov 24 '16 at 22:30
  • $\begingroup$ Yeah, that's totally what I meant, but could you explain the last row? How did you get it? $\endgroup$ – Covvar Nov 25 '16 at 11:23
  • 2
    $\begingroup$ @Covvar: $E[X_m^2]$ is the average of $E[X_1^2]$ and $E[X_2^2]$ for exactly the same reason and I used $\text{Var}[X]=E[X^2]-(E[X^2])$ $\endgroup$ – Henry Nov 25 '16 at 11:29

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.


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