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

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.