$X$ and $Y$ are non-negative rv, and $X = \mathbb E(Y|\mathcal F)$. Is it possible to show that $\operatorname{Cov(X, \min(Y, c))} \ge 0$?

68 Views Asked by At

Given that

  • (1) $X$ is a non-negative random variable
  • (2) The distribution of $Y | \mathcal F$ is determined by $X$. For example, $Y|\mathcal F = \text{Poisson}(X)$.
  • (3) $\mathbb E(Y|\mathcal F) = X$, where $\mathcal F$ is a sigma field

For example, $\mathcal F = \sigma(X)$.

is it possible to show that $\operatorname{Cov(X, \min(Y, c))} \ge 0$, where $c$ is a known constant?

Denoting $\mu = \mathbb E X$, the covariance can be written as $$ \mathbb E \left\{(X - \mu) \mathbb E[(Y -\mu)1_{\{Y \le c\}} | \mathcal F] \right\} + (c -\mu)\mathbb E \left\{(X - \mu) \mathbb E[1_{\{Y > c\}} | \mathcal F] \right\} $$