2
$\begingroup$

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).

$\endgroup$

1 Answer 1

3
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.