Negative quadrant dependence of a Poisson count and an arrival time ($\mathbb{P}(N_s\le k,T_n\le t)\le \mathbb{P}(N_s\le k)\mathbb{P}(T_n\le t)$)

For a Poisson process, the count and Poisson arrival time have the displayed inequality at every deterministic $s,t\ge 0$. A large count and a late arrival reflect opposite directions of the same fluctuation.

Here is a proof from independent increments. If $s\le t$, write $N_t=N_s+V$ with $V$ independent of $N_s$. Conditional on $N_s=u$, the event $T_n\le t$ has probability $g(u)=\mathbb{P}(V\ge n-u)$, a nondecreasing function of $u$. The event $N_s\le k$ has a nonincreasing indicator function $f(u)$. For any random variable $W$ and an independent copy $W^{\prime }$, oppositely monotone bounded functions satisfyTaking $W=N_s$ proves the claim. If $s>t$, condition instead on $N_t=u$; then $\mathbb{P}(N_s\le k\mid N_t=u)$ is nonincreasing in $u$ and $1_{\{u\ge n\}}$ is nondecreasing, giving the same inequality.

The inequality is preserved under joint convergence in distribution when the marginal distributions of the limit are continuous. Consequently, a limiting bivariate normal distribution with standard normal marginal distributions cannot have positive correlation coefficient. For $0<\rho <1$, conditioning on $Z_1=z<0$ makes $\mathbb{P}(Z_2\le 0\mid Z_1=z)>1/2$, so $\mathbb{P}(Z_1\le 0,Z_2\le 0)>1/4$, contradicting the inequality. For $\rho =1$, that quadrant has probability $1/2$, which also contradicts it.