= Solution
Each completed visit to radius $r_2$ begins a new radial excursion. By part (a), the conditional probability that the following excursion reaches radius $R$ before radius $r_1$ is
$$
q_R=\frac{\log(r_2/r_1)}{\log(R/r_1)}.
$$
The <Strong Markov property> at the successive stopping times $\tau_{2k}$ makes these trials independent with the same success probability. Consequently $N(R)$ has a <geometric distribution> on $\{1,2,\ldots\}$ with parameter $q_R$:
$$
\mathbb P(N(R)>k)=(1-q_R)^k.
$$
As $R\to\infty$, $q_R\to0$ and
$$
q_RN(R)\xrightarrow d\operatorname{Exp}(1),
\qquad
q_R\log R\longrightarrow\log(r_2/r_1).
$$
<Slutsky theorem> now gives
$$
\frac{N(R)}{\log R}
\xrightarrow d\operatorname{Exp}\!\left(\log(r_2/r_1)\right).
$$
Back to article page
