Expected value of two dependent variables is still a product of expectations

Question

For independent variables we have $E[XY]=E[X]E[Y]$. Now, since I could not find a statement that the converse is also true, I suspect that there are examples of dependent variables where this relation still holds. Could anybody here give me an example?- I am somewhat interested in how this could look like.

Answer 1

By definition, two random variables $X$ and $Y$ satisfy your equation if and only if their covariance is zero. Thus, any pair of dependent random variables with zero covariance give the counterexample you're looking for.

For instance, let $X$ be uniformly distributed between $-1$ and $1$, and let $Y = X^2$. Then $$\mathrm E[XY] = \mathrm E[X^3] = 0 = \mathrm E[X] = \mathrm E[X]\mathrm E[Y],$$ even though $X$ and $Y$ are obviously not independent.

Answer 2

Suppose that $(X,Y)$ is uniformly distributed on the unit disc. In calculating expectations $E[X]$, $E[Y]$, and $E[XY]$, it is best to convert the double integrals over the unit disc (usually set up in rectangular coordinates) to polar coordinates and note that $$\begin{align} \int_0^{2\pi} cos\, \theta\, \mathrm d\theta = 0 &\Rightarrow E[X] = 0,\\ \int_0^{2\pi} sin\, \theta\, \mathrm d\theta = 0 &\Rightarrow E[Y] = 0,\\ \int_0^{2\pi} cos\,\theta\, \sin\,\theta\, \mathrm d\theta = 0 &\Rightarrow E[XY] = 0. \end{align}$$ Thus, $E[XY] = E[X]E[Y]$ for these random variables. However, notice that although $Y$ can take on all values in $(-1,+1)$, if we know that $X = x, -1 < x < 1$, then $Y$ is necessarily restricted to $\left(-\sqrt{1-x^2}, +\sqrt{1-x^2}\right) \subset (-1,1)$ which shows that $X$ and $Y$ are dependent random variables.