 Solution

Fix $T>0$. The given normal distribution and part (d) implyOne can deduce determinism without any moment assumption on the bracket. Put $Z=e^{-⟨X{⟩}_T}$. Taking $\lambda =1,2$ gives $\mathbb{E}Z=e^{-T}$ and $\mathbb{E}{Z}^2=e^{-2T}$, so $\mathrm{Var}(Z)=0$. Therefore $⟨X{⟩}_T=T$ almost surely. Applying this at every rational time and using continuity of quadratic variation gives $⟨X{⟩}_t=t$ simultaneously for all $t\ge 0$ outside a single null set.

The Lévy characterization of Brownian motion states that a continuous local martingale starting at zero with this bracket is Brownian motion in its filtration. To see the independent-increment conclusion directly, the Itô formula shows that $e^{i\theta X_t+{\theta }^{2}t/2}$ is a martingale on any fixed bounded time interval: it is a local martingale with a deterministic bound on its modulus. ThusThe deterministic conditional characteristic function identifies an $N(0,t-s)$ increment independent of ${\mathcal{F}}_s$. Together with the given path continuity and $X_0=0$, this proves $X$ is Brownian motion.