Concatenation of truncated Brownian motion and increment from the stopped Brownian motion is again a Brownian motion

Question

Consider the concatenation mapping $\Phi: C_{(0)}\times [0, \infty) \times C_{(0)}$, where $C_{(0)}:=\{f \in C[0, \infty): f(0) = 0 \}$, $$ \Phi(f, t, g) := \begin{cases} f(s) &\text{ if } 0 \le s < t, \\ f(t) + g(s-t) &\text{ if } t \le s < \infty \end{cases} $$ ($\Phi$ "glues" $f$ and $g$ together at the point $t$). Take as facts the strong Markov property and that $\Phi$ is continuous and hence measurable.

I would like to show that given a Brownian motion $B$, the concatenation of the truncated Brownian motion $B(t \wedge \tau)$ and the increment $B(\cdot + \tau) - B(\tau)$ is a Brownian motion, that is, that $$ \Phi(B(\cdot \wedge \tau), \tau, B(\cdot + \tau) - B(\tau)) $$ is a Brownian motion.

My idea is to verify the axioms (for Brownian motion) directly.

The strong Markov property means that the probability of a measurable rectangle of the form $$ \left \{ ( B( \cdot \wedge \tau ), \tau ) \in A \right\} \times \left \{ B (\cdot + \tau) - B (\tau) \right \} $$ equals the product of the probabilities of the two rectangles.

I would need help with the axioms for independent increments and that the increments are normally distributed with zero mean and variance equal to the length of the increment. That is, 1) that given $0 \le t_0 < t_1 < t_2 < t_3$, $$ \Phi(B(\cdot \wedge \tau), \tau, B(\cdot + \tau) - B(\tau))(t_3) -\Phi(B(\cdot \wedge \tau), \tau, B(\cdot + \tau) - B(\tau))(t_2) $$ and $$ \Phi(B(\cdot \wedge \tau), \tau, B(\cdot + \tau) - B(\tau))(t_1) -\Phi(B(\cdot \wedge \tau), \tau, B(\cdot + \tau) - B(\tau))(t_0) $$ are independent and 2) that given any $0 \le s<t$, $$ \Phi(B(\cdot \wedge \tau), \tau, B(\cdot + \tau) - B(\tau)) -\Phi(B(\cdot \wedge \tau), \tau, B(\cdot + \tau) - B(\tau))(s) $$ is distributed according to $N(0, t-s)$.

Answer 1

Here is my own solution. Reading The answer from TheBridge, and not being able to completely follow the line of reasoning, I made a new attempt and - barring I'm in error - was able to solve it myself. (I think this just is the same argument TheBridge is making.)

In particular it does not seem that we need the Strong Markov Property.

First let us determine the distribution of $W_t := \Phi(B(t \wedge \tau), \tau, B(\cdot + \tau) - B(\tau))$: Given any $s >0$, consider the partition of $\Omega$ by the sets $$ A_1=\{\tau \le s \} $$ and $$ A_2 = \{ s < \tau \}. $$ For some arbitrary set $B \in \mathcal B (\mathbb R)$, Let us consider $$ \left\{ W_s \in B \right\} = \bigcup_{i=1}^2 \left \{ W_s \in B \cap A_i \right \}. $$ For $\omega \in A_1$ we have that $\tau (\omega) \le s$ and hence from the definition of $\mathrm \Phi$, \begin{align*} W_s(\omega) &= \underbrace{B (\tau(\omega)\wedge\tau(\omega), \omega )}_{f(\tau(\omega))} + \underbrace{B((s - \tau(\omega)) + \tau(\omega), \omega) - B(\tau(\omega), \omega)}_{g(s - \tau(\omega))} \\ &= B(s, \omega). \end{align*} Thus $$ \left \{ W_s \in B \cap A_1 \right \} = \left \{ B_s \in B \cap A_1 \right \}. $$ For $\omega \in A_2$ we have that $s < \tau(\omega)$ and hence from the definition of $\mathrm \Phi$, \begin{align*} W_s(\omega) &= B (s \wedge \tau(\omega), \omega ) \\ &= B (s , \omega ). \end{align*} Thus $$ \left \{ W_s \in B \cap A_2 \right \} = \left \{ B_s \in B \cap A_2 \right \}, $$ and we conclude \begin{align*} \left\{ W_s \in B \right\} &= \bigcup_{i=1}^2 \left \{ W_s \in B \cap A_i \right \} \\ &= \bigcup_{i=1}^2 \left \{ B_s \in B \cap A_i \right \} \\ &= \left \{ B_s \in B \right \}, \end{align*} and hence $$ \mathrm P \left\{ W_s \in B \right\} = \mathrm P \left \{ B_s \in B \right \} $$ Now, consider $s_1 < s_3$ and the increment $W_{s_2} - W_{s_1}$. By partitioning $\Omega $ into $$ \{ s_2 < \tau \} \bigcup \{ s_1 < \tau \le s_2 \} \bigcup \{ \tau \le s_1 \}, $$ it follows from an analogous argument that $$ W_{s_2} - W_{s_1} \sim B_{s_2} - B_{s_1}. $$

Further, the same kind of argument, and using that $B$ has independent increment, also works to show that for any $0 \le s_1 < s_2 < s_3 < s_4$, and any $A, B \in \mathcal{B}(\mathbb R)$ \begin{align*} \mathrm P \left \{ W_{s_2} - W_{s_1} \in A, W_{s_4} - W_{s_3} \in B \right \} &= \mathrm P \left \{ B_{s_2} - B_{s_1} \in A, B_{s_4} - B_{s_3} \in B \right \} \\ &= \mathrm P \left \{ B_{s_2} - B_{s_1} \in A\right\} \mathrm P\left \{ B_{s_4} - B_{s_3} \in B \right \} \\ &= \mathrm P \left \{ W_{s_2} - W_{s_1} \in A\right\} \mathrm P\left \{ W_{s_4} - W_{s_3} \in B \right \}. \end{align*} Hence $W$ has independent increments.

Answer 2

This kind of problem is always very intuitively clear but still a little hard to formalize, here are my two cents. You'll let me take the notation $W_t=\Phi(B(t \wedge \tau), \tau, B(t + \tau) - B(\tau))$ that will lighten my answer.

Let's take it the hard way and condition on the events $A_i=\{t_{i-1}\le\tau<t_i\}$ for $i=1,...,4$ (taking $t_0=0$) and $A_5=\{\tau\ge t_4\}$ to close the gap and cover all $\Omega$. For $A_1$ there is no problem as we know from strong Markov property that $W_t=B(t+\tau)-B(\tau)$ is a Brownian motion so axioms holds on the intervals of interest. In the same way for $A_5$, $B$ is a Brownian motion on $0,\tau$ which covers $t_4$ so that's ok.

We are left with 3 cases, let's take the simplest one $A_3$ that separates both intervals so that $\Delta(2,1)= W_{t_2}-W_{t_1}$ and $\Delta(4,3)=W_{t_4}-W_{t_3}$ are independent by the way you state the strong Markov property conditionally on $\tau$, both also follow a centered Gaussian of variance $t_i-t_{i-1}$ $i=2,4$ from the same line of argument as for $A_1$ (for $\Delta(4,3))$ and $A_5$ (for ($\Delta(2,1)$).

Then last case to examine is $A_2$ (as $A_4$ will follow symmetrically from the same reasoning just by switching indices). So let's break further interval $(t_1,t_2)$ in two pieces a.k.a. $(t_1,\tau)$ and $(\tau,t_2)$ and observe that we have in fact reduced our case to the case of $\Delta(2,1)=\Delta(2,\tau)+\Delta(\tau,1)$ and of course $\Delta(4,3)$.

Indeed $\Delta(2,\tau)$ and $\Delta(\tau,1)$ are from the same line of argument explained for the case $A_3$ two independent (conditionally on $\tau$) Gaussian random variables and further note that they are both independent from $\Delta(4,3)$ (which is Gaussian of course I hope you see why this is trivial here now). The sum of two independent Gaussian is Gaussian with variance the sum of the 2 marginal variances and as $\Delta(2,1)=\Delta(2,\tau)+\Delta(\tau,1)$ and $Var(\Delta(2,1))=Var(\Delta(2,\tau))+Var(\Delta(\tau,1))=t_2-\tau+\tau-t_1=t_2-t_1$ we are hopefully done here right ?

Regards

Edit : From the proof of MrFranzén, it appears that $\Phi(B(\cdot \wedge \tau), \tau, B(\cdot + \tau) - B(\tau))$ is simply $B_t$ so the transformation which appears complicated at first is nothing more than the identity. To make the matter a little more interesting, and linked to the idéea of patching together different path at a random time we could use another independent BM $B'$ and see that $\Phi(B(\cdot \wedge \tau), \tau, B'(\cdot + \tau) - B'(\tau))$ is still a BM, this time note equal to $B$ (but only up to time $\tau$). In this framework the proof above still works fine.