0
$\begingroup$

Let X and Y be two random variables such that $\forall$ a, b $\in$ $\mathbb{R}$, \begin{equation} E[(X-a)(Y-b)] = E(X-a) E(Y-b) \end{equation}

Does it imply X and Y are independent r.v.s?

$\endgroup$
0

2 Answers 2

3
$\begingroup$

$E((X-a)(Y-b))=E(XY)-aE(Y)-bE(X)+ab$

$E(X-a)E(Y-b)=E(X)E(Y)-aE(Y)-bE(X)+ab$

These are equal, independent of $a$ and $b$ iff $E(XY)=E(X)E(Y)$. In this case, independence of $X$ and $Y$ cannot be inferred.

$\endgroup$
2
$\begingroup$

Whenever $X,Y$ are integrable, we always have $$E[(X-a)(Y-b)] = E[XY - aY - bX + ab] = E[XY] - aE[Y] - bE[X] + ab \\ E(X-a)E(Y-b) = (E[X] - a)(E[Y] - b) = E[X]E[Y] - aE[Y] - bE[X] + ab$$

Compare the two lines to see that this just means $E[XY] = E[X]E[Y]$. That is, if we assume further that $X,Y$ are square-integrable, then all of this is just equivalent to the statement $X,Y$ are uncorrelated.

For some square integrable random variables, this can imply independence (for instance, if $X,Y$ are jointly normal, then they'd be independent) but in general, no, this wouldn't necessarily mean they're independent. [see for instance, this question]

$\endgroup$
2
  • $\begingroup$ More than normal is needed (joint normal). Indeed, $Y = XZ$, where $X \sim N(0,1)$ and $\mathbb P(Z=-1) = \mathbb P(Z=1) = \frac12$, is a counterexample. $\endgroup$
    – LucaMac
    Apr 8, 2021 at 20:46
  • $\begingroup$ @LucaMac Thanks. That was a "brain fart" on my part. $\endgroup$ Apr 8, 2021 at 20:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .