# Does $E[(X-a)(Y-b)] = E(X-a) E(Y-b)$ for all a, b $\in$ $\mathbb{R}$ imply X and Y are independent?

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

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

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

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

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.

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]

• 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. Apr 8, 2021 at 20:46
• @LucaMac Thanks. That was a "brain fart" on my part. Apr 8, 2021 at 20:52