Here is an inequality one finds in page 136 of Karatzas & Shreve - Brownian motion and Stochastic Calculus:

$$\begin{align}E\int_0^\infty Y_s^2\,ds&= E\int_0^\infty1_{\{T_s\leq T\}}X_{T_s}^2\,ds\\ &=E\int_0^{A_T+T}X_{T_s}^s\,ds\leq C^2(EA_T+T)<\infty,\end{align}$$

I don't see how $E[A_T] < \infty$.

Among the hypothesis, one does not find this condition:

2.7 Lemma. Let $\{A_t;0\leq t<\infty\}$ be a continuous, increasing (Definition 1.4.4) process adapted to the filtration of the martingale $M=\{M_t,\mathscr{F}_t;0\leq t<\infty\}$. If $X=\{X_t,\mathscr{F}_t;0\leq t<\infty\}$ is a progressively measurable process satisfying $$E\int_0^T X_t^2\,dA_t<\infty$$ for each $T>0$, then there exists a sequence $\{X^{(n)}\}_{n=1}^\infty$ of simple processes such that $$\sup_{T>0}\lim_{n\to\infty}E\int_0^T|X_t^{(n)}-X_t|^2\,dA_t=0.$$ Proof. We may assume without loss of generality that $X$ is bounded (cf. part $(b)$ in the proof of Proposition 2.6), i.e.,

I believe that usually we will consider $A_t = \langle M \rangle_t $ wich in the case when $M$ is a martingale implies that $E[A_T] < \infty$. Nevertheless this is not what is stated in thr Lemma.

Should we assume that $E[A_T] < \infty$? Isn't this a loss of generality?

Am I missing something here?


$\{A_{t}\}_{t\geq 0}$ is an increasing process so by definition $A_{t}$ is integrable for each $t\geq 0.$

| cite | improve this answer | |
  • $\begingroup$ Even for local martingales that are not martingales? $\endgroup$ – Conrado Costa Apr 5 '16 at 10:47
  • $\begingroup$ I do not think this is correct. $\endgroup$ – Potato Apr 16 '16 at 20:55
  • 1
    $\begingroup$ An adapted process $A$ is called increasing if $A_{t}(\omega)$ is right continuous and nondecreasing for a.s. $\mathbb{P}$ $\omega\in\Omega$ and $\mathbb{E}(A_{t})<\infty$ (see definition 1.4.4). $\endgroup$ – Fabio Andrés Gómez Apr 9 '17 at 22:03

It seems like a minor mistake to me, but it can be easily fixed: It suffices to prove the lemma for bounded $A$. To see this, define $A^m_t=A_{t\wedge\tau_m}$, where $\tau_m=\inf\{s:A_s\geq m\}$, and for any $k\in\mathbb{N}$ take $m(k)$ such that $$ E\int_{\tau_{m(k)}}^kX_t^2dA_t\leq 1/k. $$ By hypothesis, the lemma is true for $A^m$, so one can find a simple process $X^{(k)}$ such that $$ E\int_0^k|X_t^{(k)}-X_t|^2dA^{m(k)}_t\leq1/k. $$ One can choose $X^{(k)}_s$ to be $0$ for $s\geq \tau_{m(k)}$ (since $dA^{m(k)}=0$ on $[\tau_{m(k)},\infty)$) and by putting the two above together, $$ E\int_0^k|X_t^{(k)}-X_t|^2dA_t\leq2/k, $$ from which the claim follows.

| cite | improve this answer | |

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.