By <Brownian scaling>,
$$
|B_t|^{-1}\stackrel d=
t^{-1/2}|Z+t^{-1/2}B_0|^{-1},
$$
where $Z$ has the standard three-dimensional <multivariate normal distribution>. The right-hand side tends to zero in <convergence in probability>[probability], since $Z$ has no atom at the origin. Part (d) gives almost-sure convergence to $Y$, which also implies convergence in probability to $Y$. <Uniqueness of a limit in probability> therefore gives $Y=0$ almost surely. Hence $|B_t|\to\infty$ almost surely, proving the <transience of Brownian motion in dimension at least three> in dimension three.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 201 / 3 / e / Solution

If $X_n$ converges in probability to both $X$ and $Y$, then $X=Y$ almost surely. Indeed, the <triangle inequality> and a <union bound> show that $\mathbb P(|X-Y|>\varepsilon)=0$ for every $\varepsilon>0$.

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