Independent increments? The questions are simple:

Does the process $ X(t) = \int_0^t B(s)ds$ have independent increments?
What about $X(t) = \int_{t-r}^{t}B(s)ds$?

Here $B$ denotes the standard Brownian motion.
Thanks a lot!
 A: Where $B(s)$ is a Brownian motion?  No, the increments are certainly not independent for either of those, because the increment from $t$ to $t+\epsilon$ and the increment from 
$t+\epsilon$ to $t+2\epsilon$ are both close to $B(t) \epsilon$.
A: Well I am not sure to fully understand the heuristic argument of Robert Israel so I give my answer hoping that someone can spot my mistakes if any. 
So first let's remark that thanks to the representation of $X_t$ as the sum of two gaussian processes that are jointly gaussian.
Indeed, using integration by part formula (or Itô's lemma if you want to) we have :  
$X_t=t.B_t+\int_0^tr.dB_r$
(the fact that a Wiener integral is a gaussian process is considered known the jointy gaussian of the process $tB_t$ and $\int_0^tr.dB_r$ is also not difficult to derive but I can elaborate on this if asked for).
So now, are increments of this gaussian process independents ? To get to the conclusion it suffices to examine the covariation of the increments. I get for $0<s<t<u<v$ the following boring calculation :
$E[(X_t-X_s).(X_v-X_u)]=E[(t.(B_t-B_s)+B_s.(t-s)+\int_s^tr.dB_r).(v.(B_v-B_u)+B_u.(v-u)+\int_u^vr.dB_r)]$
$=E[t.(B_t-B_s).v.(B_v-B_u)]+E[t.(B_t-B_s).B_u.(v-u)]+E[t.(B_t-B_s).\int_u^vr.dB_r]+E[B_s.(t-s).v.(B_v-B_u)]+E[B_s.(t-s).B_u.(v-u)]+E[B_s.(t-s).\int_u^vr.dB_r]+E[\int_s^tr.dB_r.v.(B_v-B_u)]+E[\int_s^tr.dB_r.B_u.(v-u)]+E[\int_s^tr.dB_r.\int_u^vr.dB_r]$
Here all the terms are null (this can be deduced for terms from Itô's isometry) except the fifth which is equal to $(t-s).(v-u).min(u,s)=(t-s).(v-u).s$  
Whereas from the fact that $Y$ is centered :
$E[(X_t-X_s)].E[(X_v-X_u)]=0$
As both expressions are not equal this is enough to prove that independence of increments does not occur.
The proof is rather inelegant though,
Best regards
