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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
    
Déjà vu. –  Did Jan 3 '13 at 0:26
    
@did not really. See the comments below your answer. –  user26698 Jan 6 '13 at 18:47
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The text you reproduce assumes from the start that the random variables are independent hence, as noted in the comments, the question seems moot. However:

The sum of non independent Bernoulli random variables can have a binomial distribution.

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, $$ \mathbb P(X=Y=1)=p^2\ne p^2(1-\theta^2q^2)=\mathbb P(X=1)\mathbb P(Y=1). $$

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In the "deja-vu" question of mine, I asked at 4), which was the most important to me, if it is trivial that these variables are independent or not - since it seemed odd to me that no sketch of a proof was in the text - and I mentioned I proved that they were independent. Your answer indicated, since you mentioned one had to use pen and paper to verify the independence, that a proof was indeed required. But now you're saying that a proof actually isn't required, since the variables are already assumed to be independent. What is true now ? –  user26698 Jan 6 '13 at 18:54
    
(Jonathan Christensen comment (not "comments") seems to imply that the independence of variables is not an assumption, but rather a conclusion one can draw from the definition of those variable, which, in my view, requires a proof.) –  user26698 Jan 6 '13 at 19:01
    
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 ? –  user26698 Jan 6 '13 at 23:09
    
No I do not. But what is difficult in the proof you delineate? Sum of products of nonnegative terms = product of the sums, et voilà! –  Did Jan 7 '13 at 9:17
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