Fix a deterministic horizon $T$ and $\varepsilon>0$. For every localization index $N$,
$$
\mathbb P\!\left(\sup_{t\leq T}|X_n(t)-X(t)|>\varepsilon\right)
\leq \mathbb P(T_N\leq T)
+\frac1\varepsilon\lVert X_n^{T_N}-X^{T_N}\rVert
$$
by <Markov inequality>. Because $T_N\uparrow\infty$ almost surely, the first term tends to zero as $N\to\infty$. For fixed $N$, the second tends to zero as $n\to\infty$. Taking first the limit superior in $n$ and then $N\to\infty$ proves
$$
\sup_{t\leq T}|X_n(t)-X(t)|\longrightarrow0
$$
in probability. This is precisely <uniform convergence on compacts in probability>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 / 2 / b / Solution

{title2=u.c.p.}
{wiki=Convergence_of_random_variables#Uniform_convergence_in_probability}

Processes $X_n$ converge uniformly on compacts in probability to $X$ when, for every $T<\infty$ and $\varepsilon>0$,
$$
\mathbb P\!\left(\sup_{t\leq T}|X_n(t)-X(t)|>\varepsilon\right)\to0.
$$

= UCP convergence
{c}
{synonym}

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 Incoming links: Uniform convergence on compacts in probability