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?

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    $\begingroup$ 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? $\endgroup$ – whuber Oct 8 '19 at 17:06
  • $\begingroup$ 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. $\endgroup$ – Andrew Oct 8 '19 at 17:54

The tickets-in-a-box model of random variables described at https://stats.stackexchange.com/a/54894/919 provides a helpful way to think about this.

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.

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    $\begingroup$ 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. $\endgroup$ – Andrew Oct 8 '19 at 18:50

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