[proofplan] The proof exploits the [Strong Markov Property](/theorems/1180) to decompose $B$ at the stopping time $T$ into a pre-$T$ part and an independent post-$T$ Brownian motion. Since $-B^{(T)}$ has the same [distribution](/page/Distribution) as $B^{(T)}$, replacing the post-$T$ increment by its negative — which is precisely reflection about $B_T$ — produces a process with the same law as $B$. [/proofplan] [step:Decompose $B$ at the stopping time $T$ using the strong Markov property] By the [Strong Markov Property](/theorems/1180), the post-$T$ process \begin{align*} B^{(T)}_t = B_{T+t} - B_T, \qquad t \geq 0, \end{align*} is a standard Brownian motion independent of $\mathcal{F}_T^+$. In particular, $B^{(T)}$ is independent of the stopped path $(B_{t \wedge T})_{t \geq 0}$, which is $\mathcal{F}_T^+$-measurable. [/step] [step:Observe that $-B^{(T)}$ has the same distribution as $B^{(T)}$ and is equally independent] Since $B^{(T)}$ is a standard Brownian motion, the process $-B^{(T)}$ is also a standard Brownian motion (by [Invariance Properties](/theorems/1175)(i), taking the orthogonal matrix $U = -I_d$ with $d=1$). Moreover, $-B^{(T)}$ is a measurable [function](/page/Function) of $B^{(T)}$, and $B^{(T)}$ is independent of $(B_{t \wedge T})_{t \geq 0}$, so $-B^{(T)}$ is also independent of $(B_{t \wedge T})_{t \geq 0}$. Therefore the pairs \begin{align*} \bigl((B_{t \wedge T})_{t \geq 0}, \, B^{(T)}\bigr) \quad \text{and} \quad \bigl((B_{t \wedge T})_{t \geq 0}, \, -B^{(T)}\bigr) \end{align*} have the same joint distribution. [guided] Why does replacing $B^{(T)}$ by $-B^{(T)}$ preserve the joint distribution? Two ingredients are needed: (i) $B^{(T)} \overset{d}{=} -B^{(T)}$ (distributional symmetry of Brownian motion), and (ii) both $B^{(T)}$ and $-B^{(T)}$ are independent of the pre-$T$ path. Given (i) and (ii), for any bounded [measurable functions](/page/Measurable%20Functions) $\Phi$ and $\Psi$, \begin{align*} \mathbb{E}[\Phi((B_{t \wedge T})_t) \, \Psi(-B^{(T)})] &= \mathbb{E}[\Phi((B_{t \wedge T})_t)] \cdot \mathbb{E}[\Psi(-B^{(T)})] \\ &= \mathbb{E}[\Phi((B_{t \wedge T})_t)] \cdot \mathbb{E}[\Psi(B^{(T)})] \\ &= \mathbb{E}[\Phi((B_{t \wedge T})_t) \, \Psi(B^{(T)})], \end{align*} where the first and third equalities use independence and the second uses $B^{(T)} \overset{d}{=} -B^{(T)}$. This confirms the joint distributional equality. [/guided] [/step] [step:Identify the concatenation with $B^{(T)}$ as $B$ and the concatenation with $-B^{(T)}$ as $\widetilde{B}$] The original process $B$ can be reconstructed from the pair $((B_{t \wedge T})_t, B^{(T)})$ by concatenation: for $t \leq T$, $B_t = B_{t \wedge T}$, and for $t > T$, $B_t = B_T + B^{(T)}_{t-T}$. Similarly, the reflected process $\widetilde{B}$ is obtained by concatenating $(B_{t \wedge T})_t$ with $-B^{(T)}$: for $t \leq T$, $\widetilde{B}_t = B_t$, and for $t > T$, \begin{align*} \widetilde{B}_t = B_T + (-B^{(T)}_{t-T}) = B_T - (B_t - B_T) = 2B_T - B_t, \end{align*} which matches the definition. Since the two pairs have the same joint distribution and both $B$ and $\widetilde{B}$ are measurable functions of their respective pairs via the same concatenation map, $\widetilde{B}$ and $B$ have the same distribution on $C([0,\infty); \mathbb{R})$. Path [continuity](/page/Continuity) of $\widetilde{B}$ follows from the continuity of $B$ and the matching condition $\widetilde{B}_T = B_T$. Therefore $\widetilde{B}$ is a standard Brownian motion. [/step]