# Lévy's characterization of Brownian motion: right-continuous processes

Let $$(X_u)_{u}$$ be right-continuous martingale such that $$X_0=0$$ such that $$(X^2_u-u)_u,(X_u^3-3uX_u)_u,(X_u^4-6uX_u^2+3u^2)_u$$ are martingales (with respect to the canonical filtration).

Prove that $$X_u$$ has continuous paths and deduce that $$(X_u)_u$$ is a Brownian motion.

Supposing that we proved that $$X_u$$ has continuous paths then the Brownian motion property could be deduced from Lévy's characterization of BM.

How to prove the path-continuity of $$X_u$$? (It doesn't seem trivial).

Cool question.

Proposition Let $$(X_u)_{u}$$ be right-continuous martingale with $$X_0=0$$, such that $$(X^2_u-u)_u,(X_u^3-3uX_u)_u,(X_u^4-6uX_u^2+3u^2)_u$$ are martingales. Then for every integer $$M \ge 1$$, the path $$(X_u)_{u}$$ is a.s. continuous in $$[0,M]$$.

Proof: Let $$\{\mathcal F_t\}$$ be the canonical filtration of $$\{X_t\}$$. Fix $$u \ge 0$$ and write $$E^u[\,\cdot\,]:=E[\,\cdot \,| \mathcal F_u]$$. Denote $$\mathcal G_t:=\mathcal F_{u+t}$$ for $$t \ge 0$$. Observe that the process $$\{Y_t\}_{t \ge 0}$$ defined by $$Y_t:=X_{u+t}-X_u \tag{1}$$ is a $$\{\mathcal G_t\}$$-martingale.

Claim: For any bounded $$\{\mathcal G_t\}$$-stopping time $$\tau$$, we have $$E^u[Y_\tau^4]=3E^u[\tau^2]\,.$$

The claim is proved below. Now we will use it to complete the proof of the proposition.

Fix $$\epsilon>0$$ and $$\delta>0$$. Let $$P^u[\,\cdot\,]:=P[\,\cdot \, | \mathcal F_u]$$. Applying the claim to $$\tau:=\delta \wedge \min\{t\ge 0: |Y_t| \ge \epsilon\}$$ gives $$P^u[|Y_\tau| \ge \epsilon]\cdot\epsilon^4 \le E^u[Y_\tau^4] \le 3\delta^2\,,$$ so $$P[|Y_\tau| \ge \epsilon] \le 3\delta^2 \epsilon^{-4}\,. \tag{*}$$ For $$k \ge 1$$, we will use $$(*)$$ for $$\delta_k={32}^{-k}$$ and $$\epsilon_k=2^{-k}$$ to bound the probability of $$A_k:=\bigcup_{j=0}^{{32}^k M-1} \Big\{\max_{0 \le t \le \delta_k} |X_{j\delta_k+t}-X_{j\delta_k}| \ge \epsilon_k\Big\} \,.$$ We obtain $$P(A_k) \le 32^k M \cdot 3\delta_k^2 \epsilon_k^{-4}=3M\epsilon_k\,.$$ By the Borel-Cantelli lemma, almost surely only finitely many of the events $$A_k$$ occur. This implies that $$\{X_t\}$$ is a.s. continuous in $$[0,M]$$. $$\hspace{6.6in} \Box$$

Proof of Claim

$$X_u-u^2= E^u[(X_u+Y_\tau)^2-(u+\tau)]\,, \quad \text{so} \quad E^u[Y_\tau^2-\tau]=0 \,. \tag{2}$$ Similarly, $$X_u^3-3uX_u=E^u[(X_u+Y_\tau)^3-3(u+\tau)(X_u+Y_\tau)]\,, \quad \text{so by} \; (2),$$ $$E^u[Y_\tau^3-3\tau Y_\tau]=0 \,. \tag{3}$$ Also, $$X_u^4-6uX_u^2+3u^2=E^u[(X_u+Y_\tau)^4-6(u+\tau)(X_u+Y_\tau)^2+3(u+\tau)^2]\,,$$ so $$0=E^u[Y_\tau^4+4X_u Y_\tau^3+6X_u^2 Y_\tau^2-6u Y_\tau^2-12\tau X_u Y_\tau-6\tau(X_u^2+Y_\tau^2)+6u\tau+3\tau^2] \,.$$ Therefore, $$E^u[Y_\tau^4]=4X_u E^u[Y_\tau^3-3\tau Y_\tau]+6(X_u^2-u) E^u[Y_\tau^2-\tau] +3E^u[\tau^2]=3E^u[\tau^2]\,. \tag{4}$$ $$\hspace{6.6in} \Box$$

• A right-continuous local martingale? Jun 22 at 19:43
• Sorry if we considered a local martingale $X$ and $V_u=u$ the corresponding continuous finite variation. Jun 22 at 19:52
• I think so but did not check it carefully. Is the answer above clear enough for you to accept? Jun 22 at 20:47
• @mathex Choose $\epsilon_k<\epsilon/3$. Then for $u,u'$ as in your comment, $$|X_u-X_{u'}| \le |X_u-X_{q\delta_k}|+|X_{q\delta_k}- X_{(q+1)\delta_k}|+|X_{(q+1)\delta_k}-X_{u'}| \le 3\epsilon_k \le\epsilon$$ Jun 24 at 7:40
• What I wrote was for the case where $u'>(q+1)\delta_k$. In the remaining easier case, $|u'-q\delta_k| \le \delta_k$ so you can just use two terms in the triangle inequality Jun 24 at 16:47