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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.

There is a stronger conclusion than the requested limit. For $x>0$, the event $N_n-n\ge x\sqrt{n}$ forces $N_n\ge n+1$, because $N_n$ is an integer. In particular, at least $n$ arrivals have occurred by time $n$, so the $n$-th Poisson arrival time satisfies $T_n\le n$. This contradicts $T_n-n\ge y\sqrt{n}$ for $y>0$. Thus the two events are disjoint:

Separate a long future count from the early Poisson arrival time. Let ${\mathcal{F}}_t=\sigma (N_s:0\le s\le t)$ be the natural filtration, and defineThe truncated Poisson arrival time in $B_n$ is ${\mathcal{F}}_{n^2}$-measurable. The independent increments property makes $A_n$ independent of ${\mathcal{F}}_{n^2}$, and hence of $B_n$.

The numerator of $A_n$ is a centered Poisson random variable of parameter $n^4-n^2$. Apply the central limit theorem to the $n^4-n^2$ unit-time increments, then use $\sqrt{n^4-n^2}/n^2\to 1$ and Slutsky's theorem. This proves $A_n\stackrel{\mathrm{d}}{\to }N(0,1)$. Also, Markov's inequality and $\mathbb{E}{T}_n=n$ giveThus $B_n$ equals $(T_n-n)/\sqrt{n}$ outside an event of vanishing probability, so part (i) and Slutsky's theorem give $B_n\stackrel{\mathrm{d}}{\to }N(0,1)$. Exact independence of random variables for each $n$ now gives the joint limit $(A_n,B_n)\stackrel{\mathrm{d}}{\to }(Z_1,Z_2)$ with independent standard normal random variables.

Finally, $\mathbb{E}{C}_n=0$ and $\mathrm{Var}(C_n)=n^{-2}$, so Chebyshev's inequality gives $C_n\to 0$ in probability. The first coordinate in the question is $A_n+C_n$. The second differs from $B_n$ only on $\{T_n>n^2\}$, and so its difference tends to zero in probability. The vector version of Slutsky's theorem proves the required joint convergence in distribution. The boundary of $(-\infty ,x]\times [y,\infty )$ has zero limiting probability, since both marginal distributions have continuous cumulative distribution functions. The Portmanteau theorem therefore yieldsThis is $\mathbb{P}(Z_1\le x,Z_2\ge y)$ by independence of random variables.

The possible limiting correlation coefficients areThe direction of the dependence comes from inversion of the same counting process: extra arrivals produce a larger count but an earlier $n$-th Poisson arrival time. Shared fluctuations therefore enter the two coordinates with opposite signs. Positive limiting correlation cannot arise; this can be formalized using negative quadrant dependence of a Poisson count and an arrival time.

At the same time scale, $f(n)=n$, part (iii) rules out every positive-positive quadrant. A bivariate normal distribution with standard normal marginal distributions and $\rho >-1$ assigns positive probability to such a quadrant. Thus, under the assumed joint limit, this choice forces $\rho =-1$. At very different time scales, $f(n)=n^4$, part (iv) gives $\rho =0$.

Intermediate time scales retain an intermediate fraction of the same arrival fluctuations. In particular, the Poisson count-arrival joint central limit theorem gives $\rho =-1/\sqrt{\lambda }$ for $f(n)=\lambda n$ with $1\le \lambda <\infty$. These choices realize every $\rho \in [-1,0)$, and $f(n)=n^4$ supplies zero. This identifies the whole interval without requiring an additional limit calculation in this part.

The premise about a bivariate normal distribution is understood to apply when a joint limit exists. Merely requiring $f(n)\to \infty$ cannot guarantee one: taking $f(n)=n$ for even $n$ and $f(n)=n^4$ for odd $n$ gives different limits along the two subsequences just identified.

For a rate-one Poisson process with Poisson arrival times $T_n$, suppose $t_n\to \infty$ and $t_n/n\to \lambda \in [0,\infty ]$. Thenwhere $1/\sqrt{\infty }=0$ and $\mathrm{BN}(\rho )$ denotes the bivariate normal distribution with standard normal marginal distributions, including the singular cases. In particular, equal scales give $\rho =-1$ and separated scales give $\rho =0$.

To prove this, fix real $x,y$ and put $s_n=n+y\sqrt{n}>0$ for large $n$. The event $(T_n-n)/\sqrt{n}\le y$ is exactly $N_{s_n}\ge n$. Counts at the two deterministic times $t_n,s_n$ are sums of a shared count on $[0,\min (t_n,s_n)]$ and an independent count on the remaining interval. The central limit theorem for a centered Poisson random variable of diverging parameter follows by writing a count at a real time $u$ as its count at $\lfloor u\rfloor$, a sum of independent unit increments, plus an independent remainder with expected value at most one. Markov's inequality makes that remainder negligible after division by $\sqrt{u}$.

Apply this central limit theorem to the two independent count blocks whenever their lengths diverge. A block with bounded length, or length negligible compared with the scale of its coordinate, contributes a term tending to zero in probability. The standardized pair of counts therefore has a joint normal limit whose correlation coefficient isThis includes $\lambda =1$: the difference block then has negligible variance after normalization. In the arrival event the standardized count at $s_n$ must be at least $(n-s_n)/\sqrt{s_n}\to -y$. Reversing this inequality changes the sign of the second limiting normal random variable, giving the asserted negative correlation coefficient. Continuity of the marginal distributions permits the moving thresholds and the weak inequalities, including at the singular limits.

Every $\rho \in [-1,0)$ is realized by $t_n=\lambda n$ with $\lambda ={\rho }^{-2}\ge 1$, and zero by $t_n=n^4$. Conversely, even when $t_n/n$ has no limit, it has a subsequence converging in the extended interval $[0,\infty ]$. If the original pair has a joint bivariate normal distribution limit, that subsequence has the same limit, forcing its correlation coefficient into $[-1,0]$.