For a given Process $K_{t}=\exp(B_{t}+\theta t)$ with $\theta\in\mathbb{R}$ and $B_{t}$ a Wiener process i want to show, that $K_{t}$ does have dependent increments. My idea is: $$ \begin{split} K_{t+s} &= \exp(B_{t+s}+\theta(t+s)) \\ &= \exp(B_{t}+\theta t)\exp(B_{t+s}-B_{t}+\theta s)\\ &= K_{t}\exp(B_{t+s}-B_{t}+\theta s) \end{split} $$ With $B_{0}=0$ We have $K_{0}=1$ and we define the two increments \begin{align} Z_{1}:=K_{1}-K_{0}=K_{1}-1\,,\,Z_{2}:=K_{2}-K_{1}=K_{1}\left[\exp(\theta)\exp(B_{2}-B_{1})-1\right] \end{align} Since $\exp(\theta)\exp(B_{2}-B_{1})$ is independent of $K_{1}$ and both increments depend on $K_{1}$ it follows, that $K_{t}$ doesn't have independent increments.
An argumentation i found is, that $K_{t}=\exp(B_{t}+\theta t)$ solves the SDG \begin{align} dK_{t}=(\theta+1/2 )K_{t}dt+K_{t}dB_{t} \end{align} with initial value $K_{0}=1$. It is said, that by the form of the SDG, the dependence of the increments follow. How can you argue this way?
We consider a diffusion process satisfying
$$dX_t = b(X_t) dt + \sigma(X_t) dB_t.$$
A process with independent increments with finite mean is a martingale and therefore has constant expectation. A process of the form above with a nontrivial drift term $b$ is "generically" not a martingale, because
$$\int_s^t b(X_u) du = E[X_t]-E[X_s]$$
provided $\sigma$ is sufficiently nice that the stochastic integral term forms a martingale. (For example this is true if $\sigma$ is bounded, as well as under much milder assumptions.) Now if $b$ is not identically zero then the integral on the left side "generically" can't be zero for all $s,t$.
Assuming your work is correct, you must be more careful in the case $\theta=-1/2$, for then the drift is in fact zero. In this case you basically need to argue that when $K$ is growing, the noise intensity is strengthening.
Edit: as @Did pointed out, there is a flaw here: a process with independent mean zero increments is a martingale. So one should use something like the second argument even when $\theta \neq -1/2$.