1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.