= Solution
By the usual definition of a <Lévy process>, $X_0=0$ almost surely. The two <martingale> assumptions give
$$
\mathbb E X_t=0,\qquad \mathbb E X_t^2=t.
$$
For a continuous <Lévy process>, the jump measure in the <Lévy–Khintchine formula> is zero, and the process has the form of a Brownian Gaussian component plus a deterministic drift. Thus its <characteristic function> is
$$
\mathbb E e^{iuX_t}=\exp\left(iumt-\tfrac12\sigma^2tu^2\right).
$$
The first moment makes $m=0$, and the second moment makes $\sigma^2=1$. Consequently
$$
\boxed{X_1\sim N(0,1),\qquad (X_t)\text{ is standard Brownian motion}.}
$$
Equivalently, continuity and the martingale identity $X_t^2-t$ identify the <quadratic variation> as $t$, so the <Lévy characterization of Brownian motion> gives the same conclusion. A continuous nonzero drift is excluded by the mean-zero martingale assumption.
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