# Proving the independence of random indicator variables when deriving the binomial distribution

Short question: Is independence of random indicator variables a necessary assumption to derive the Binomial distribution ?

But far as I can see, using the Definition (that is at the same time a theorem) from below, taken from Snells probability book, page 144, we can do without this assumption and prove that the random indicator variables are independent! (as opposed to, for example, the geometric distribution, were we need independence to derive it).

EDIT: A better explanation: Do we in the text below assume the variables are independent, or to we deduce it from their definition ?
To me it seems the latter is the case, since we concretely define the $X_j$ on a given sample space, so we can use the definition of independence of variables to test if they are indeed independent or not - so independence is proven, not assumed.

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Yes, you can prove that the $X_j$ are independent from the definition. – Jonathan Christensen Jan 2 '13 at 17:50
– Did Jan 3 '13 at 0:26

For an example, let $p$ in $(0,1)$, $q=1-p$, and $(X,Y)$ with values in $\{0,1\}^2$ with $$\mathbb P(X=Y=0)=q^2,\qquad\mathbb P(X=Y=1)=p^2,$$ and $$\mathbb P(X=0,Y=1)=(1+\theta)pq,\qquad\mathbb P(X=1,Y=0)=(1-\theta)pq,$$ for some $|\theta|\leqslant1$. Then $X+Y$ is binomial $(2,p)$ but $X$ and $Y$ are not independent except when $\theta=0$. To wit, the distribution of $X$ is $$\mathbb P(X=0)=q(1+\theta p),\qquad\mathbb P(X=1)=p(1-\theta q),$$ and the distribution of $Y$ is $$\mathbb P(X=0)=q(1-\theta p),\qquad\mathbb P(Y=1)=p(1+\theta q),$$ hence, for example, $$\mathbb P(X=Y=1)=p^2\ne p^2(1-\theta^2q^2)=\mathbb P(X=1)\mathbb P(Y=1).$$
Let me try once again: the book defines the distribution as a product measure. Hence, by definition of independence, the random variables are independent. Nothing to prove (except, if you insist, that $m(A_1\times\cdots\times A_n)=m(A_1)\cdots m(A_n)$ as soon as $m(\omega)=m(\omega_1)\cdots m(\omega_n)$--hence my (probably too generous) mention of pen and paper). – Did Jan 6 '13 at 19:11
exactly! Only that proving $m(A_1 \times \cdots A_n)=m(A_1)\cdots m(A_n)$ (at least to me) doesn't seem to be that trivial, since the proof goes $$m(A_1 \times \cdots \times A_n)=m(\bigcup_{(\omega_1\ldots,\omega_n) } \{(\omega_1,\ldots,\omega_n)\}=$$ $$=\sum_{(\omega_1\ldots,\omega_n)} m(\{(\omega_1,\ldots,\omega_n)\}) =\sum_{(\omega_1\ldots,\omega_n)} m(\omega_1)\cdots m(\omega_n)=m(A_1)\cdots m(A_n).$$ Or do you know an easier proof ? – MyCatsHat Jan 6 '13 at 23:09