$pW_1(t) + \sqrt{1-p^2}W_2(t)$ Wiener process Let's define two independent wiener processes: $W_1, W_2$ and take $p \in (0,1)$.
I want to prove that $W(t) = pW_1(t)+\sqrt{1-p^2}W_2(t)$
I want to check four statements:
(i) $W(0) = 0$
(ii) W has independent increasments
(iii) W has Gaussian incresments ($W_t - W_s \sim \mathbb{N}(0, t-s))$
(iv) W has continuous paths: $W_{t}$ is continuous in $t$
My work so far
(i) W(0) = $0 \cdot p + \sqrt{1-p^2}\cdot 0  = 0$ $\\$
(iii) Let's fix $t \ge s$
$W_t(t) - W_s(t) = pW_1(t) + \sqrt{1-p^2}W_2(t) - pW_1(s) - \sqrt{1 - p^2}W_2(s) = $
$=p(W_1(t) - W_1(s)) + \sqrt{1-p^2}(W_2(t) - W_2(s))$
First expression has distrubution $\mathbb{N}(0, p^2(t-s))$ and second one $\mathbb{N(0, \sqrt{1-p^2}(t-s))}$ . So the sum of those expressions is distributed $\mathbb{N}(0, t-s)$.
(iv) $W_t$ is continous because it's a sum of continous functions
I have two questions: Is my justification correct ? And also - ow can I prove second condition i.e. how to prove that $W$ hasindependent increasments ?
 A: Fix $0 \leq t_0 < t_1<t_2<...<t_n$, and consider the increments $W(t_i) - W(t_{i-1})$ for $i=1,2,...,n$. We need to show that these are independent.
Note that $W(t_i) - W(t_{i-1}) = p(W_1(t_i) - W_1(t_{i-1})) + \sqrt{1-p^2}(W_2(t_i) - W_2(t_{i-1}))$.
We use a theorem called the disjoint blocks theorem for independent random variables.

Let $X_1,...,X_n$ be independent random variables. Let $S_1 \cup S_2 \cup ... \cup S_m = \{1,2,...,n\}$ be a disjoint decomposition of $1,2,...,n$. Let $f_i: \mathbb R^{|S_i|} \to \mathbb R$ be measurable functions. Then $f_i((X_i : i \in S))$ are independent random variables.

For example, if $X_1,X_2,X_3,X_4$ are independent, then using the disjoint blocks theorem we get that $e^{X_4}+\cos X_2$ and $X_3X_1^2 + \sin X_3$ are independent (see what the $S_i$ and $f_i$ are in this case).
Now, we have the random variables $W_j(t_i) - W_j(t_{i-1})$ for $i=1,2,...,n$ and $j=1,2$. By independence of increments for $W_1$ and $W_2$, and by independence of $W_1$ from $W_2$, these random variables are independent.
Now form the blocks $S_i = \{W_1(t_i) - W_1(t_{i-1}), W_2(t_i) - W_{2}(t_{i-1})\}$ which are a disjoint decomposition of those random variables. Then the functions $f_i$ are the same i.e. $f_i(x,y) = px + \sqrt{1-p^2}y$. Applying the disjoint blocks theorem tells you instantly that the increments of $W(t)$ are independent.
Hence, we are done.

ADDED FOR FURTHER READING
Note that $p^2 + (\sqrt{1-p^2})^2 = 1$ and both $p,\sqrt{1-p^2}$ are non-negative. $W$ can be seen as the first component of an orthogonal transformation (a rotation) of the two-dimensional Brownian motion $(W_1(t),W_2(t))$.
That is, $(p,\sqrt{1-p^2})$ is the first row of an orthogonal matrix $O$, such that $W(t) = [O(W_1(t),W_2(t))]_1$ where $[]_1$ means the first column.

*

*It is well known that combining independent Brownian motions as components provides a Brownian motion of higher dimensions.


*Orthogonal transformations preserve the standard Brownian motion


*and that components of Brownian motion are lower-dimension Brownian motions.
Combining these facts, $W$ is a Brownian motion.
