Wiener process - proof of independent increments I have defined the Wiener process to be a stochastic process $X_t$ with values in $\mathbb{R}$ such that $X_0=0$, the paths $t \mapsto X_t$ are continuous, and for any times $0<t_1<\dots<t_n$ and Borel sets $A_1,\dots,A_n \subset \mathbb{R}$:
$$
\mathbb{P}(X_{t_1} \in A_1, \dots, X_{t_n} \in A_n) = \int_{A_1}\dots\int_{A_n}p_{t_1}(0,x_1)\dots p_{t_n-t_{n-1}}(x_{n-1},x_n) \; \textrm{d}x_1\dots \textrm{d}x_n
$$
where 
$$
p_t(x,y) = \frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}} 
$$
is the transition density. From this definition, how do I prove that for any $0=t_0 \leq t_1 \leq \dots \leq t_n$, the increments
$$
X_{t_1}-X_{t_0}, \dots, X_{t_n}-X_{t_{n-1}}
$$
are independent? In general, the only way I know how to show that two RVs are independent is to show that their joint density function factorises into the product of the marginals, but I can't see how to do that here. Thanks for any help.
 A: First consider the case $n=2$. By definition, the joint distribution of $(X_{t_1},X_{t_2})$ is given by
$$p_{t_1}(0,x_1) p_{t_2-t_1}(x_1,x_2) \, dx_1 \, dx_2 = q_{t_1}(x_1) q_{t_2-t_1}(x_2-x_1) \, dx_1 \, dx_2$$
where
$$q_t(x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right), \qquad x \in \mathbb{R}.$$
Consequently,
$$\begin{align*} \mathbb{E}\exp(\imath \xi X_{t_1} + \imath \, \eta (X_{t_2}-X_{t_1})) = \iint \exp(\imath \, \xi x_1+\imath \, \eta (x_2-x_1)) q_{t_1}(x_1) q_{t_2-t_1}(x_2-x_1) \, dx_2 \, dx_1 \end{align*}$$
For fixed $x_1$ we substitute, $y_2 := x_2 - x_1$ and obtain
$$ \begin{align*} \mathbb{E}\exp(\imath \xi X_{t_1} + \imath \, \eta (X_{t_2}-X_{t_1}) &= \iint \exp(\imath \, \xi x_1+\imath \, \eta y_2) q_{t_1}(x_1) q_{t_2-t_1}(y_2) \, dy_2 \, dx_1. \\ &= \left( \int e^{\imath \, \xi x_1} q_{t_1}(x_1) dx_1 \right) \left( \int e^{\imath \, \eta y_2} q_{t_2-t_1}(y_2) \, dy_2 \right) \\ &= \mathbb{E}e^{\imath \, \eta X_{t_1}} \cdot \mathbb{E}e^{\imath \eta (X_{t_2}-X_{t_1})}. \end{align*}$$
This finishes the proof for $n=2$. The general case is treated by induction.
