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).


1 Answer 1


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, $$E^u[Y_\tau^4]=4X_u E^u[Y_\tau^3-3\tau Y_\tau]+6(X_u^2-u) E^u[Y_\tau^2-\tau] +3E^u[\tau^2]=3E^u[\tau^2]\,. \tag{4}$$ $\hspace{6.6in} \Box$

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

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