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I'm reading a theorem at page $43$ of these notes, i.e.,


Proposition 7.12. Let $M$ be a continuous local martingale with respect to a filtration $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$ such that $$ M_0=\begin{array}{ll} 0 & \text { a.s } \end{array} \text { and } \lim _{t \rightarrow \infty}\langle M\rangle_t=\infty \quad \text { a.s. } $$

Let us also define $$ \tau(s)=\inf \left\{t>0:\langle M\rangle_t \geq s\right\}, \quad B_s=M_{\tau(s)}, \quad \mathcal{G}_s=\mathcal{F}_{\tau(s)}. $$

Then $B$ is a standard Brownian motion with respect to $\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$.

Proof. As already mentioned, the idea is to use Lévy's theorem, i.e., to show that

  • (i) $B$ has continuous trajectories.
  • (ii) $B$ is a local martingale with respect to $\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$.
  • (iii) $\langle B\rangle_s=s$, i.e., $\left(B_s^2-s, s \in \mathbb{R}_{+}\right)$is a local martingale with respect to $\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$.

Let us verify these three statements.

  • (i) As $M$ is continuous, $t \rightarrow\langle M\rangle_t$ is also continuous. Moreover, if $\langle M\rangle$ is constant on some interval, then $M$ also is, so the function $s \mapsto B_s=M_{\tau(s)}$ is continuous.

  • (ii) Let $\tau_n=\inf \left\{t>0:\left|M_t\right| \geq n\right\}, n \geq 1$. For each $n, M^{\tau_n}$ is a martingale such that $$ \mathbb{E}\left(\sup _{t \in[0, T]}\left|M_{t \wedge \tau_n}\right|^2\right)<\infty, \quad \forall T>0, $$ so by the optional stopping theorem (version 2), we have $$ \mathbb{E}\left(M_{\tau\left(s_2\right) \wedge \tau_n} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=M_{\tau\left(s_1\right) \wedge \tau_n} \quad \text { a.s., } \quad \forall s_2>s_1 \geq 0 . $$ By the dominated convergence theorem (and some details), this implies that $$ \mathbb{E}\left(M_{\tau\left(s_2\right)} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=M_{\tau\left(s_1\right)} \quad \text { a.s. } \quad \quad (\star) $$ i.e., $$ \mathbb{E}\left(B_{s_2} \mid \mathcal{G}_{s_1}\right)=B_{s_1} \quad \text { a.s. } $$ i.e., $B$ is a martingale with respect to $\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$.

  • (iii) Let $X_t=M_t^2-\langle M\rangle_t$. By assumption, $X^{\tau_n}$ is a martingale $\forall n$, so $$ \mathbb{E}\left(X_{\tau\left(s_2\right) \wedge \tau_n} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=X_{\tau\left(s_1\right) \wedge \tau_n} \quad \text { a.s., } \quad \forall s_2>s_1 \geq 0 $$ Then again by the dominated convergence theorem (and some details), we obtain that $$ \mathbb{E}\left(X_{\tau\left(s_2\right)} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=X_{\tau\left(s_1\right)} \quad \text { a.s. } $$ i.e., $$ \mathbb{E}\left(M_{\tau\left(s_2\right)}^2-\langle M\rangle_{\tau\left(s_2\right)} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=M_{\tau\left(s_1\right)}^2-\langle M\rangle_{\tau\left(s_1\right)} \quad \text { a.s. } $$ As $\langle M\rangle_{\tau(s)}=s$ by definition, we obtain: $$ \mathbb{E}\left(B_{s_2}^2-s_2 \mid \mathcal{G}_{s_1}\right)=B_{s_1}^2-s_1 \quad a . s ., \quad \forall s_2>s_1 \geq 0 $$ i.e., $\left(B_s^2-s, s \in \mathbb{R}_{+}\right)$is a martingale with respect to $\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$.


My understanding I have a problem understanding how $(\star)$ is obtained. To apply dominated convergence theorem, we need $$ |M_{\tau\left(s_2\right) \wedge \tau_n}| \le Z \quad \text{a.s.} \quad \forall n\in \mathbb N, $$ for some integrable random variable $Z$. I could not see how to have such $Z$.

Could you elaborate on how DCT is applied to get $(\star)$.

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  • $\begingroup$ $\vert M_t\vert\le n$ for all $t\le\tau_n$, a.s., by definition of $\tau_n$. So $Z=n$. $\endgroup$
    – Will
    Commented Mar 19, 2023 at 19:34
  • $\begingroup$ @Will but we take the limit $n \to \infty$... $\endgroup$
    – Akira
    Commented Mar 19, 2023 at 19:35
  • $\begingroup$ Oh you need an upper bound independent of $n$ ok sorry $\endgroup$
    – Will
    Commented Mar 19, 2023 at 19:38
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    $\begingroup$ $\tau_n$ goes to $+\infty$ with $n$. So $X_{\tau(s_1)\wedge\tau_n}$ converges almost surely to $X_{\tau(s_1)}$. And $X$ is a uniformly integrable martingale, so so is $X^{\tau(s_2)}$, hence $X_{\tau(s_2)\wedge\tau_n}$ converges in $L^1$ to $X_{\tau(s_2)}$ as $n\to+\infty$. This is a way to show $(\star)$. $\endgroup$
    – Will
    Commented Mar 19, 2023 at 19:43
  • $\begingroup$ Actually I meant $M$ instead of $X$, but it is true for both. $\endgroup$
    – Will
    Commented Mar 19, 2023 at 20:24

1 Answer 1

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By Vitali's convergence theorem, it suffices to show that $( M_{\tau (s_2) \wedge \tau_n}, n \in \mathbb N)$ is uniformly integrable. By OST, $( M_{t \wedge \tau (s_2) \wedge \tau_n}, t \ge 0)$ is a martingale. By Doob's maximal inequality, $$ \begin{align} \mathbb E \big [ \sup_{s \in [0, t]} | M_{s \wedge \tau (s_2) \wedge \tau_n} |^2 \big ] &\le 4 \mathbb E [ | M_{t \wedge \tau (s_2) \wedge \tau_n} |^2 ] \\ &= 4 \mathbb E [ \langle M \rangle_{t \wedge \tau (s_2) \wedge \tau_n} ] \\ &\le 4 \mathbb E [ \langle M \rangle_{\tau (s_2)} ] = 4 s_2. \end{align} $$

By monotone convergence thereom, $$ \mathbb E \big [ \sup_{s \in [0, \infty)} | M_{s \wedge \tau (s_2) \wedge \tau_n} |^2 \big ] \le 4 s_2. $$

Notice that $$ | M_{\tau (s_2) \wedge \tau_n} |^2 \le \sup_{s \in [0, \infty)} | M_{s \wedge \tau (s_2) \wedge \tau_n} |^2. $$

Hence $$ \mathbb E \big [ | M_{\tau (s_2) \wedge \tau_n} |^2 \big ] \le 4 s_2 \quad \forall n \in \mathbb N. $$

The claim then follows from below result (taken from these notes), i.e.,

Example 5.4.3. Let $(X_i)_{i \in I}$ be a family of real random variables. If $\sup _{i \in I} \mathbb E [ |X_i|^r ] < \infty$ for some $r>1$, then $(X_i)$ is uniformly integrable.

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    $\begingroup$ Looks correct to me $\endgroup$
    – Will
    Commented Mar 20, 2023 at 20:36
  • $\begingroup$ @Will Thank you so much for your verification! $\endgroup$
    – Akira
    Commented Mar 20, 2023 at 20:37

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