# Why aren't all Normal and Rademacher variables independent?

In Wikipedia page: https://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent, we find a classical counter-example showing that two normally distributed and uncorrelated random variables may not be independent.

This implies $$X$$ having a normal distribution $$N(0,1)$$ and $$W$$ having the so-called Rademacher distribution.

This is OK, but what troubles me is the sentence "and assume $$W$$ is independent of $$X$$".

Do we really have to assume $$X$$ and $$W$$ are independent? Is it not obvious? Can we imagine such an $$X$$ so that $$X$$ and $$W$$ would not be independent?

• It depends on what you mean by "such an $X$:" are you asking whether there exists a bivariate random variable $(X,W)$ with Normal and Rademacher marginals for which $X$ and $W$ are not independent? (In that case consider any measurable subset $\mathcal{A}$ of the line for which $\Pr(X\in\mathcal A)=1/2$ and define $W$ in terms of its indicator function.) Or, as strongly suggested by your title, are you trying to ask something about the counterexample on the Wikipedia page? – whuber Oct 8 '19 at 17:06
• whuber, you are right; actually, the title is not as accurate as it should be...And yes, I was asking how it comes that X following N(0,1) and W following w=+1 and W=-1, each with probability 1/2, could be not independent. – Andrew Oct 8 '19 at 17:54

Imagine you have a box full of tickets on which are written various numbers in such a way that a blind draw of one ticket acts like observing $$X.$$ $$W$$ is a second number found on every ticket: half the tickets display a $$1$$ and the other half display a $$-1$$ (that's the definition of a Rademacher variable).

$$X$$ and $$W$$ are independent when observing $$X$$ gives you no clue about what $$W$$ is and vice versa. Here is the tickets-in-a-box translation of that intuitive description of independence: no matter what the value of $$X$$ may be, the proportions of tickets with any particular values of $$W$$ are always the same. In the terminology of conditional probability this says

$$\Pr(W\in\mathcal{A}\mid X\in\mathcal{B}) = \Pr(W\in\mathcal{A})$$

for any events $$\mathcal{A}$$ and $$\mathcal B.$$ Equivalently, multiplying both sides of this equation by $$\Pr(X),$$ we obtain

$$\Pr(X\in\mathcal{A}\text{ and }W\in\mathcal{B}) = \Pr(X\in\mathcal{A})\Pr(W\in\mathcal{B}).$$

That's the mathematical criterion of independence.

Thus, one way to create dependence is to write $$W=1$$ on some special half of the tickets. In fact, almost any half will do, because the other half--unless you chose these halves very carefully--will exhibit a different distribution of $$X$$ values.

As a concrete example, let $$X$$ have a standard Normal distribution and set $$W=1$$ when $$X$$ is positive and $$W=-1$$ otherwise. $$X$$ tells us exactly what's on $$W,$$ so $$X$$ and $$W$$ cannot be independent.

• whuber, this is exactly the kind of answer I was looking for: detailed and precisely explained in all details. It's now clear why we have to "assume" independence. Thanks for having taken time to write. – Andrew Oct 8 '19 at 18:50