Prove two random variables are negatively correlated

Question

Given a raw estimator

\begin{equation} I=I(X>b)= \begin{cases}1, \text{ if } X > b\\ 0, \text{ otherwise } \end{cases} \end{equation}

for some random variable $X$ and some constant $b$ in the range of the distribution of $X$.

I was trying to prove that $I$ and $-X$ are negatively correlated. Here is what I tried so far, and then I am stuck.

\begin{align} & Cov(I,-X) = \mathbb E\{I(-X)\} - \mathbb E \{I\} \mathbb E \{-X\} \\ = & - \mathbb E\{IX\} + \mathbb E\{I\} \mathbb E\{X\} \\ = & - \big[ \mathbb E\{X|X>b\} \Pr (X > b) \big] + \mathbb E\{X\} \Pr(X > b) \\ = & \big[ \mathbb E\{X\} - \mathbb E\{X|X > b\} \big] \Pr(X > b). \end{align}

Any help will be greatly appreciated!

Answer 1

You're almost done. The next step is to write $$\operatorname{E}[X] = \operatorname{E}[X \mid X > b] \Pr[X > b] + \operatorname{E}[X \mid X \le b]\Pr[X \le b], \tag{1}$$ and then observe that $$\operatorname{E}[X \mid X \le b] \le b. \tag{2}$$ So if we let $a = \operatorname{E}[X \mid X > b]$ and $p = \Pr[X > b]$, $$\operatorname{Cov}[I, -X] \le (ap + b(1-p) - a)p = (-a(1-p) + b(1-p))p = (b-a)(1-p)p. \tag{3}$$ But for the same reason that $(2)$ is true, we also have $$a = \operatorname{E}[X \mid X > b] > b. \tag{4}$$ Therefore, since $0 < p < 1$, and $a > b$, then $$\operatorname{Cov}[I,-X] \le (b-a)(1-p)p < 0.$$