Past exam of the probability course of the University of Oxford / 2021 / 1 / c

Write $\Phi$ for the standard normal distribution function. The Poisson process in this part has rate one and starts at $N_0=0$.

Set $T_0=0$ and $E_j=T_j-T_{j-1}$. The Poisson interarrival times $E_j$ are independent and identically distributed random variables with the rate-one exponential distribution. In particular, $\mathbb{E}{E}_j=1$ and $\mathrm{Var}(E_j)=1$, so the finite, positive variance hypothesis of the central limit theorem holds. Since $T_n={\sum }_{j=1}^n{E}_j$,

The unit-time increments $D_j=N_j-N_{j-1}$ are independent and identically distributed random variables, by independent increments and stationary increments of the Poisson process. Each has the Poisson distribution of parameter one, hence expected value one and variance one. The central limit theorem therefore gives

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.