I know the definition of a (standard) Brownian motion $\{B_t:t\geq0\}$. (i) $B_0=0$. (ii) the process has independent increments. (iii) for all $t,s\geq0$, $B_{t+s}-B_s$ are normally distributed with expectation zero and variance $s$. (iv) almost surely, $t\mapsto B_t$ is continuous.
I confused the following question.
Let $\{B_t:t\geq0\}$ be a Brownian motion. If $\{X_t:t\geq0\}$ have same finite-dimensional distribution as a Brownian motion, and $X(t)$ is almost surely continuous. Then, $\{X_t:t\geq0\}$ is a Brownian motion.
Why is this? Why finite-dimensional distribution equivalence derive the definion (i), (ii), (iii)?
Thank you for your cooperation.
Best Answer
(i) Since $X_0 = B_0$ in distribution, we have $$\mathbb{P}(X_0 \in A) = \mathbb{P}(B_0 \in A) = \delta_0(A)$$ for any measurable set $A$. Hence, $X_0=0$ almost surely.
(ii) Take $0=t_0 < \ldots < t_n$. By assumption, the random vectors $U:= (X_{t_1},\ldots,X_{t_n})$ and $V:=(B_{t_1},\ldots,B_{t_n})$ have the same distribution. Since $(B_t)_{t \geq 0}$ is a Brownian motion, $V$ is Gaussian, and so is $U$. If we define $$f(x_1,\ldots,x_n) := \begin{pmatrix} x_1 \\ x_2-x_1 \\ \vdots \\ x_n-x_{n-1} \end{pmatrix},$$ then $$f(U) = \begin{pmatrix} X_{t_1} \\ X_{t_2} -X_{t_1} \\ \vdots \\ X_{t_n}-X_{t_{n-1}} \end{pmatrix} \quad \text{and} \quad f(V)=\begin{pmatrix} B_{t_1} \\ B_{t_2}-B_{t_1} \\ \vdots \\ B_{t_n}-B_{t_{n-1}} \end{pmatrix}$$ have the same distribution and both are Gaussian. Now recall that the entries $G_j$ of a Gaussian random vector $G=(G_1,\ldots,G_n)$ are independent iff they are uncorrelated, i.e. if the covariance matrix of $G$ is a diagonal matrix. Since the Brownian motion $(B_t)_{t \geq 0}$ has independent increments, it follows that the covariance matrix of the Gaussian random vector $f(V)$ is a diagonal matrix, and so is the covariance matrix of $f(U)$, which means that the entries $X_{t_j}-X_{t_{j-1}}$ are independent.
(iii) By assumption, $(B_s,B_{t+s})$ and $(X_s,X_{t+s})$ have the same distribution. This implies that $X_{t+s}-X_s = B_{t+s}-B_s$ in distribution, and so $X_{t+s}-X_s \sim N(0,t)$.