Proving the process $(Y_t)_{t\in[0,\infty)}$ is a standard Brownian motion

Question

Let $a\in(0,\infty)$. Let $(X_t)_{t\in[0,a]}$ be a standard Brownian motion and $(W_s)_{s\in[0,\infty)}$ be a standard Brownian motion such that $W_s$ is independent of $X_t$ for all $s \geqslant 0$ and $t \in [0, a]$. Define a new process $(Y_t)_{t∈[0,\infty)}$ as follows.

\begin{equation} Y_{t} = \begin{cases} X_t & \text{if}~0\leqslant t\leqslant a\\ W_{t-a}+X_a & \text{if}~t>a. \end{cases} \end{equation}

Justify whether the above process $(Y_t)_{t\in[0,\infty)}$ is a standard Brownian Motion.

We must check if the process $Y_t$ satisfies all the properties (independent increments, sample path continuity, etc.). To do so, you have to pick four-time points, let them be $t_j<t_{j+1}\leqslant t_i<t_{i+1}$, then vary where $a$ belongs in the in the inequality, and then take cases.

In the case where $t_j\leqslant a\leqslant t_{j+1}\leqslant t_i<t_{i+1}$ we have that $Y_{t_{j+1}}-Y_{t_{j}} = W_{t_{j+1}-a}+X_a-X_{t_{j}}$ and $Y_{t_{i+1}}-Y_{t_{i}}=W_{t_{i+1}-a}-W_{{t_i}-a}.$ How do I prove that $W_{t_{j+1}-a}$ is independent to $X_a-X_{t_{j}}$? I don't think that if $W$ is a process independent of $X$ and $Y$, then it's also necessarily independent of $X-Y$. Can someone direct me to proof of such a statement, if it exists? I would also appreciate any alternative approaches to the question.

Answer 1

The quadratic variation of $Y$ is $t\,.$

(The rest of the things you need to apply for Levy characterisation I leave to you.)

Proof. We only have to consider partitions $0<t_1<\dots<t_n$ of the interval $[0,t]$ where $t_n=t>a\,.$ WLOG we can also assume that there is an $t_k$ in every such partition such that $t_k=a\,.$
\begin{align} &\sum_{i=1}^n\big(Y_{t_{i+1}}-Y_{t_i}\big)^2=\sum_{i=k+1}^n \big(Y_{t_{i+1}}-Y_{t_i}\big)^2+\sum_{i=1}^k\big(Y_{t_{i+1}}-Y_{t_i}\big)^2\\[2mm] &=\sum_{i=k+1}^n \big(W_{t_{i+1}-a}-X_a-W_{t_i-a} +X_a\big)^2+\sum_{i=1}^k\big(X_{t_{i+1}}-X_{t_i}\big)^2\,. \end{align} The limit of the second term is $a\,.$ The first term is \begin{align} &\sum_{i=k+1}^n \big(W_{t_{i+1}-a}-W_{t_i-a}\big)^2\\ \end{align} whose limit is $t-a\,.$

Answer 2

First, lets clarify what independent processes means Independent stochastic processes and independent random vectors. The definition for the two processes to be independent is given by PlanetMath:

Two stochastic processes $\lbrace X(t)\mid t\in T \rbrace$ and $\lbrace Y(t)\mid t\in T \rbrace$ are said to > be independent, if for any positive > integer $n<\infty$, and any sequence > $t_1,\ldots,t_n\in T$, the random > vectors > $\boldsymbol{X}:=(X(t_1),\ldots,X(t_n))$ > and > $\boldsymbol{Y}:=(Y(t_1),\ldots,Y(t_n))$ > are independent.

So in this particular, we have independence of the vectors

$$(W_{t_{1}},...,W_{t_{n}})\text{ and }(X_{s_{1}},...,X_{s_{n}})$$

and so this also implies $W_{t_{2}}-W_{t_{1}}$ is independent of $X_{s_{2}}-X_{s_{1}}$.

The rest of the proof follows as the above answer mentioned https://math.stackexchange.com/a/4882724/383044.