 Solution

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.