Let $W_t$ be a Brownian motion on $(\Omega, F, (F_t)_t, P)$. Find all values of $a$ and $b$ such that the stochastic integral $$X_t=\int_0^t a+\frac{bu}{t} \;dW_u$$ is a Brownian motion.
1)So I need to check that $X_0=0$ almost surely, and since here we have $t$ in the denominator I need to show that the limit of $X_t$ when $t$ approaches $0$ is $0$ a.s. If $b=0$ then it is fine. If $b\neq 0$ this leads to prove that $\lim_{t\to 0} \frac{1}{t}\int_0^tu \:dW_u =0$ since I can write $X_t=aW_t+\frac{b}{t}\int_0^tu \:dW_u$. Now since $u$ is smooth the stochastic integral reduces to the Riemann-Stiltjes integral $$ \frac{1}{t}\int_0^tu \:dW_u =W_t- \frac{1}{t}\int_0^t W_u du$$, therefore since almost every trajectory of $W_t$ is continuous the limit is zero almost everywhere.
2) Now the increments $X_{t+h}-X_t=\int_t^{t+h}(a-\frac{bu}{t+h})dW_u-b\int_0^t\frac{hu}{t(t+h)}dW_u$, so if I want this to be independent of $F_t$ then $b$ must be zero, otherwise the second integral is measurable w.r.t $F_t$. (I'm not sure if this is a valid/ correct explanation). If so, then $X_t=aW_t$, so it is a Brownian motion iff $a^2=1$ (because iff in this case the increments have the correct normal distribution). If the observation in point is correct I guess I wouldn't need my discussion of point 1, but I'm not sure and I would like some feedback on how i proved that point anyway.
Thank you in advance.
Levy's characterisation and the use of It$ô$'s lemma give the same answer. I think the independence argument in your approach is not quite enough to conclude that $b$ must be zero, unfortunately (please see my comment below).
Levy's Characterisation: The aim here is to show that there is a choice of $a$ and $b$ such that $X_t=\int_{0}^{t}(a+b\frac{u}{t})\text dW_u$ is a martingale with $X_0=0$ and with quadratic variation $\langle X\rangle_t=t$. Now, as you note, $$ X_t = a\int_{0}^{t}\text dW_u + b\left(W_t-\frac{1}{t}\int_0^{t}W_u~\text du\right) = (a+b)\int_{0}^{t}\text dW_u -\frac{b}{t}\int_0^{t}W_u~\text du \tag{1} $$ From $(1)$ it is immediate that the choice $b=0$ guarantees both $X_0=0$ and $X_t$ being a martingale. Furthermore, we now see that we require $(a+b)^2t = \langle X \rangle_t = t$, so that $a=\pm 1$.
It$\bf ô$'s Lemma: Identical to the previous argument, but just in terms of the SDE for $(1)$. $$ \text dX_t = (a + b)~\text dW_t +\frac{b}{t}\left(\frac{1}{t}\int_0^{t}W_u~\text du-W_t\right)\text dt \tag{2} $$
Your approach: Your argument in $(1)$ is good. If your argument using independence was true, we should be able to come to it in the following way. Given that $X_{t+h}-X_t$ and $X_t$ are jointly normal, thier independence follows if, and only if, we make a choice of $a$ and $b$ that guarantees $\mathbb Cov(X_{t+h}-X_t,X_t)=0$. The independent increments property of $W_t$ implies we want $$ \mathbb Cov(X_{t+h}-X_t,X_t) = -\frac{tbh}{t+h}\left(\frac{a}{2}+\frac{b}{3}\right) = 0 $$ from which we see $b= -\frac{3}{2}a, 0$ are both valid choices. Consequently, we need more to conclude.