[proofplan] We verify that $X$ is a centred Gaussian process with the same covariance [function](/page/Function) as Brownian motion, namely $\operatorname{Cov}(X_s, X_t) = \min(s,t)$, which determines the finite-dimensional [distributions](/page/Distribution). [Continuity](/page/Continuity) at $t > 0$ is immediate from the definition; continuity at $t = 0$ requires an argument using the law of the iterated logarithm or a direct tail bound. [/proofplan] [step:Verify that $X$ has the correct finite-dimensional distributions] The process $X$ is defined by $X_t = t B_{1/t}$ for $t > 0$ and $X_0 = 0$. Since $B$ is a centred Gaussian process, and the map $(B_{1/t_1}, \ldots, B_{1/t_n}) \mapsto (t_1 B_{1/t_1}, \ldots, t_n B_{1/t_n})$ is a linear transformation, the vector $(X_{t_1}, \ldots, X_{t_n})$ is jointly centred Gaussian for any $0 < t_1 < \cdots < t_n$. A centred Gaussian process is determined by its covariance function. For $0 < s \leq t$, compute \begin{align*} \operatorname{Cov}(X_s, X_t) = \operatorname{Cov}(s B_{1/s}, \, t B_{1/t}) = st \operatorname{Cov}(B_{1/s}, B_{1/t}) = st \cdot \min(1/s, 1/t) = st \cdot \frac{1}{t} = s. \end{align*} Since $s = \min(s, t)$, this matches the covariance function of standard Brownian motion. Therefore $X$ and $B$ have identical finite-dimensional distributions. [/step] [step:Establish path continuity at $t > 0$] For $t > 0$, the map $t \mapsto X_t = t B_{1/t}$ is the composition of the continuous map $t \mapsto (t, 1/t)$ from $(0,\infty)$ to $(0,\infty)^2$ with the map $(t, s) \mapsto t B_s$. Since $s \mapsto B_s$ is continuous (by hypothesis) and multiplication is continuous, $t \mapsto X_t$ is continuous on $(0, \infty)$. [/step] [step:Establish path continuity at $t = 0$] We must show $X_t = t B_{1/t} \to 0$ as $t \downarrow 0$, which is equivalent to $B_s / s \to 0$ as $s \to \infty$ (under the substitution $s = 1/t$). For any $\varepsilon > 0$ and $s > 0$, the random variable $B_s / s$ has distribution $\mathcal{N}(0, 1/s)$, so \begin{align*} \mathbb{P}(|B_s / s| > \varepsilon) = \mathbb{P}(|N| > \varepsilon \sqrt{s}), \qquad N \sim \mathcal{N}(0,1). \end{align*} Taking $s = n \in \mathbb{N}$, the Gaussian tail bound gives $\mathbb{P}(|B_n / n| > \varepsilon) \leq 2 \exp(-\varepsilon^2 n / 2)$. Since $\sum_{n=1}^\infty \exp(-\varepsilon^2 n / 2) < \infty$, the [Borel--Cantelli Lemma](/theorems/507) yields $|B_n / n| \leq \varepsilon$ for all sufficiently large $n$, almost surely. Since $\varepsilon > 0$ was arbitrary, $B_n / n \to 0$ almost surely. To upgrade from integer times to continuous times: for $n \leq s \leq n+1$, \begin{align*} \left|\frac{B_s}{s}\right| \leq \frac{|B_n|}{n} + \frac{\sup_{n \leq u \leq n+1} |B_u - B_n|}{n}. \end{align*} By the [Scaling Property](/theorems/1175)(ii), $\sup_{0 \leq u \leq 1} |B_{n+u} - B_n|$ has the same distribution as $\sup_{0 \leq u \leq 1} |B_u|$, which is an almost surely finite random variable. Dividing by $n \to \infty$ shows the second term tends to zero almost surely (by the same Borel--Cantelli argument applied to each $\varepsilon > 0$). Therefore $B_s / s \to 0$ as $s \to \infty$, hence $X_t \to 0 = X_0$ as $t \downarrow 0$. [/step]