Gluing two Brownian paths together

Question

I need help with the following problem.

Let $B_1$ and $B_2$ be two independent random elements distributed as the Wiener measure. Define $$ B(t) = \begin{cases}B_1(t), \text{ if }t<1/2; \\ B_2(t), \text{ if }t\ge1/2; \end{cases}$$ Show that $B$ satisfies i), ii) and iii). In particular it has the same finite dimensional distributions as $B_i$. However it is not an element of $C$, almost surely.

Conditions 1), ii) and iii) are given as

i) $B(0) = 0$ almost surely;

ii) for each $t \in [0, 1]$, $B(t)$ is a $\mathcal{N}(0; t)$ random variable;

iii) if $0 \le t_0 < t_1 < \dots < t_k \le 1$ then the increments $B(t_1) - B(t_0)$; $B(t_2) - B(t_1)$; $\dots$; $B(t_k) - B(t_{k-1})$ are independent random variables.

Conditions i) and iii) are easy to check. I can also prove that $B$ is a.s. not continuous at $1/2$. However, I'm struggling to prove iii). My difficult lies to prove that $B_{i+1}-B_{1}$ will be independent of the other increments when $t_i<1/2\le t_{i+1}$. I've tried to write the difference as $[B_2(t_{1+1})-B_2(1/2)]+[B_2(1/2)-B_1(1/2)]+[B_1(1/2)-B_1(t_i)]$ but I couldn't advance more because I think the term $B_2(1/2)-B_1(1/2)$ will be not necessarily independent from the the others increments.

I will appreciate any help. Thanks!

Answer 1

If $B$ satisfied condition (iii) then it would have the same finite-dimensional distributions as a Brownian motion. As such $B$ would have a continuous modification. As $B$ is already cadlag, such a modification is precluded by the jump ($B(1/2-)\not= B(1/2)$ a.s.) at time $t=1/2$.

Answer 2

I also believe this not to be true. Let's look at the increments $B_2(\tfrac{1}{2}+\varepsilon)-B_1(\tfrac{1}{2}-\varepsilon)$ and $B_1(\tfrac{1}{2}-\varepsilon)-B_1(0)$.

Let $N >1$ and $\delta, \varepsilon > 0$. Then we have that

\begin{align} \mathbb{P}(B_2&(\tfrac{1}{2}+\varepsilon)-B_1(\tfrac{1}{2}-\varepsilon)\in [-N,N], B_1(\tfrac{1}{2}-\varepsilon)\in B(0,\delta))\\ &\geqslant \mathbb{P}(B_2(\tfrac{1}{2}+\varepsilon)\in [-N+\delta,N-\delta], B_1(\tfrac{1}{2}-\varepsilon) \in B(0,\delta))\\ &=\mathbb{P}(B_2(\tfrac{1}{2}+\varepsilon)\in [-N+\delta,N-\delta])\mathbb{P}(B_1(\tfrac{1}{2}-\varepsilon) \in B(0,\delta))\\ &> P(B_2(\tfrac{1}{2}+\varepsilon)-B_1(\tfrac{1}{2}-\varepsilon)\in [-N,N])\mathbb{P}(B_1(\tfrac{1}{2}-\varepsilon)\in B(0,\delta)) \end{align}

The last line holds true for $\varepsilon$ and $\delta$ small enough as $B_2(\tfrac{1}{2}+\varepsilon)-B_1(\tfrac{1}{2}-\varepsilon)$ is normally distributed with mean $0$ and variance $(\tfrac{1}{2}-\varepsilon)+(\tfrac{1}{2}+\varepsilon)=1$ (as it is the sum of two independent normally distributed random variables) and $B_2(\tfrac{1}{2}+\varepsilon)$ is normally distributed with mean $0$ and variance $\tfrac{1}{2}+\varepsilon$.