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I suspect I'll post my own answer to this question shortly, but it may be of interest to see what answers others post.

A theorem found in Feller's famous book and elsewhere says that if $X,Y$ are independent random variables and $X+Y$ is normally distributed, then $X$ and $Y$ are normally distributed.

Is there a similar result for binomial distributions? I.e. can we show that $$ \begin{align} & \text{if }X_1,X_2\text{ are independent and } X_1+X_2\sim\operatorname{Binomial}(n,p) \\ & \text{then for some }n_1,n_2,\quad X_i\sim\operatorname{Binomial}(n_i,p)\text{ for }i=1,2\text{ ?} \end{align} $$

PS ADDED LATER: Could we assume throughout that random variables considered here take values in $\{0,1,2,3,\ldots\}$.

In fact, I suspect we can get a stronger statement: $$ \begin{align} & \text{if } X_1+X_2\sim\operatorname{Binomial}(n,p)\text{ then $X_1,X_2$ are independent} \\ & \text{and then for some }n_1,n_2,\quad X_i\sim\operatorname{Binomial}(n_i,p)\text{ for }i=1,2. \end{align} $$ Is that also true?

Both of these are converses of a proposition found in every textbook.

Michael Hardy
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    You cannot conclude independence of components of a sum, if you only know properties of the sum. For example given $X=X_1+X_2$, you cannot rule out $X_1=X_2=\frac{1}{2}X$ unless you assume independence of $X_1$ and $X_2$ a priori. – ccorn Jan 23 '14 at 17:39
  • Better example perhaps: $X_1=X_2+\epsilon$ with $|\epsilon|\leq1$. – ccorn Jan 23 '14 at 17:47
  • For your weaker statement, consider $X_1 = -1$ always gives this constant negative value (hence not a binomial distribution), and $X_2=1\sim B(1,1)$. The result $X_1+X_2\sim B(n,0)\sim B(0,p)$ is binomial, but its component may not be binomial ($X_1$), or may not share the same $p$ (X_2). – peterwhy Jan 23 '14 at 18:00
  • Consider I. $X$ and $Y$ are independent, II. $X$ and $Y$ are normal, and III. $X+Y$ is normal. The standard proposition is that if I and II hold, then so does III, while Cramer's theorem which you quote (cited in Feller) says that if I and III hold, then so does II. But clearly when III holds, II might hold but I need not hold (even if II does) since $X$ and $Y$ could be jointly normal correlated random variables too. Now replace normal by binomial. You are asking whether III implies both I and II. Could you provide some intuition why such a result might be true? – Dilip Sarwate Jan 27 '14 at 20:04
  • @DilipSarwate : I think I can provide a reason why someone might mistakenly think that's true. But a reason why it might be true is another matter. – Michael Hardy Jan 28 '14 at 04:10

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Counterexample:

Consider $X_1$ uniform on $\{0,2\}$. Assume that, if $X_1=0$ then $X_2$ is uniform on $\{0,1\}$, and that, if $X_1=2$ then $X_2$ is uniform on $\{0,-1\}$.

(In other words, $(X_1,X_2)$ is uniform on the set $\{(0,0),(0,1),(2,-1),(2,0)\}$.)

Then $X_1+X_2$ is Bin$(2,\frac12)$ but $(X_1,X_2)$ is not independent.

Did
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  • A bit rushed but I'll look at this. If the propositions are wrong, then are there some fairly weak additional hypotheses that make them right? – Michael Hardy Jan 23 '14 at 19:50
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    Since the conditional distribution of $X_2$ given $X_1$ is not the same for both values of $X_1$, can we still claim that $X_1$ and $X_2$ are independent? – Dilip Sarwate Jan 27 '14 at 03:18
  • @DilipSarwate No we cannot (claim this), and we did not since this is a counterexample to the so-called "stronger statement". – Did Jan 27 '14 at 06:33
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    Thanks for the clarification. Any thoughts on the weaker statement? Can the characteristic function $(1-p+p e^{it})^n$ of a binomial $(n,p)$ random variable be factored into the characteristic functions of two random variables in any way other than $(1-p+p e^{it})^m\cdot (1-p+p e^{it})^{n-m}$? – Dilip Sarwate Jan 27 '14 at 12:43
  • @DilipSarwate The support of the terms of the sum should be, up to opposite translations, subsets of the nonnegative integers, hence we are asking if the polynomial $(px+1−p)^n$ can be the product of two polynomials that are not suitable powers of $px+1−p$. It cannot, qed. – Did Sep 09 '16 at 13:40