Let $B_t$, $t\in [0,T]$ be a $d$-dimensional standard Brownian motion. Let $\sigma:[0,T] \rightarrow \mathbb R^{d\times d}$ be a deterministic function such that $$\sigma(u) = diag( \sigma_1(u), \dots , \sigma_d(u))$$ i.e. it is a diagonal matrix with $$\int_0^T \|\sigma(u)\|^2du <\infty$$ then the integral w.r.t. $B$ is well-defined and let us define a new process:
$$W_t := \int_0^t \sigma(u) dB_u, \quad t\in [0,T].$$
It is clear that $W_t \sim N(0,\int_0^t \sigma^2(u) du)$ and that the components of $W=(W^{(1)},\dots, W^{(d)})$ are independent, but my question is: Does $W_t$ still satisfy the usual same properties as $B$, i.e.
1) Is $W$ stationary?
2) Does $W$ posess independent increments?
My feeling and computations say yes but the fact that one integrates over time might be a problem. I would like to double-confirm it with you! Thank you very much!
If the matrix $\sigma$ only depends on time (and the properties you defined) the process still has independent increments. For the stationarity the answer is no of course.
To see why increments stay independent let's look at $W_t-W_s$ and $W_s$ ($s<t$) (in the univariate case, the argument is the same in the multivariate case).
As $W$ is a Wiener integral, it is a Gaussian process as you noted, so we only need to look at $E[W_s.(W_t-W_s)]$ if its value is zero we are done, so :
$$E[W_s.(W_t-W_s)]=E[\int_0^s\sigma(u)dB_u.\int_s^t\sigma(u)dB_u]=E[\int_0^t\sigma(u)1[u<s]dB_u.(\int_0^t1[s<u<t]\sigma(u)dB_u)]$$
Now use Itô's isometry so that we get :
$$E[W_s.(W_t-W_s)]=\int_0^t\int_0^t1[s<u<t].1[u<s].\sigma^2(u)du=0$$
because the product of indicators functions is always null.
For the loss stationarity of increments let's look again at the second moment of the increments, you get :
$E[(W_t-W_s)^2]=\int_s^t \sigma(u)^2du$ this is not a function of $t-s$ in general so the increments cannot be stationary.
Best regards