Reduction of independence to uncorrelatedness Consider two random variables $Y, Z \in \mathbb R$. Someone claims that $Y$ is independent of $Z$ if and only if $1\{Y \ge t\}$ is uncorrelated with $Z$ for all $t \in \mathbb R$. Is this true?
The statement seems suspect to me. I can see that they are independent iff $1\{Y \ge t\}$ is uncorrelated with $1\{Z \ge s\}$ for all $s,t \in \mathbb R$. But not the above.
 A: Since $1\{Y\ge t\}$ is uncorrelated with $Z$ for all $t\in \Bbb R$, we can show that $Y$ itself is uncorrelated with $Z$. To do this, first assume $Y\ge 0$ for simplicity. I use this trick to relate $Y$ to $1\{Y\ge t\}$:
$$
Y=\int_0^\infty 1\{Y\ge t\}\,dt \tag 1
$$
Therefore,
$$
\begin{align}
E[YZ]
&=E\left[Z\int_0^\infty 1\{Y\ge t\}\,dt\right]
\\&=\int_0^\infty E[Z1\{Y\ge t\}]\,dt
\\&=\int_0^\infty E[Z]E[1\{Y\ge t\}]\,dt
\\&=E[Z]\int_0^\infty P(Y\ge t)\,dt
\\&=E[Z]E[Y]
\end{align}
$$
To exchange the order of $\int$ and $E[]$, you can use Fubini's theorem. When $Y$ can be negative, replace $(1)$ with $Y = \int_0^\infty [1\{Y\ge t\}-1\{Y\le -t\}]\,dt$.
Therefore, it makes sense to look for an example of $Y$ and $Z$ which are uncorrelated but dependent, and hope that we additionally have $1_{Y\ge t}$ being uncorrelated with $Z$.
One common example of uncorrelated and dependent random variables is when $Y$ and $Z$ are given by the following joint pmf:





$Z=-1$
$Z=0$
$Z=1$




$Y=-1$
0
$\frac14$
0


$Y=0$
$\frac14$
0
$\frac14$


$Y=1$
0
$\frac14$
0




You can check that $1_{Y\ge t}$ is uncorrelated with $Z$ for all $t$, so this is a counterexample which solves your problem.
