= Negative quadrant dependence of a Poisson count and an arrival time
{title2=$\mathbb P(N_s\leq k,T_n\leq t)\leq\mathbb P(N_s\leq k)\mathbb P(T_n\leq t)$}
For a <Poisson process>, the count and <Poisson arrival time> have the displayed inequality at every deterministic $s,t\geq0$. A large count and a late arrival reflect opposite directions of the same fluctuation.
Here is a proof from <independent increments>. If $s\leq t$, write $N_t=N_s+V$ with $V$ independent of $N_s$. Conditional on $N_s=u$, the event $T_n\leq t$ has <probability> $g(u)=\mathbb P(V\geq n-u)$, a nondecreasing <function> of $u$. The event $N_s\leq k$ has a nonincreasing <indicator function> $f(u)$. For any <random variable> $W$ and an independent copy $W'$, oppositely monotone bounded <functions> satisfy
$$
2\operatorname{Cov}(f(W),g(W))=\mathbb E[(f(W)-f(W'))(g(W)-g(W'))]\leq0.
$$
Taking $W=N_s$ proves the claim. If $s>t$, condition instead on $N_t=u$; then $\mathbb P(N_s\leq k\mid N_t=u)$ is nonincreasing in $u$ and $\mathbf1_{\{u\geq n\}}$ 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 $\mathbb P(Z_2\leq0\mid Z_1=z)>1/2$, so $\mathbb P(Z_1\leq0,Z_2\leq0)>1/4$, contradicting the inequality. For $\rho=1$, that quadrant has <probability> $1/2$, which also contradicts it.
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