Let $f \in C^2_C(\mathbb{R})$ and $$X_t = X_0 + \int_0^t \sigma(s) \, dB_s + \int_0^t b(s) \, ds$$ be an (one-dimensional) Itô process where $\sigma,b: [0,\infty) \times \Omega \to \mathbb{R}$ progressively measurable and pathwise bounded. Let $(\tau_n)_n$ be a common localization sequence of $b$, $\sigma$. Then there exist sequences of simple processes $(\sigma^{\Pi})_{\Pi}$, $(b^{\Pi})_{\Pi}$ such that $$\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \stackrel{L^2(\lambda_T \times \mathbb{P})}{\to} \sigma \cdot 1_{[0,\tau_n)} \qquad (|\Pi| \to 0) \\ b^{\Pi} \cdot 1_{[0,\tau_n)} \stackrel{L^2(\lambda_T \times \mathbb{P})}{\to}b \cdot 1_{[0,\tau_n)} \qquad (|\Pi| \to 0)$$ for all $n \in \mathbb{N}$. Denote by $X^{\Pi}$ the corresponding Itô process. Then one can actually show that $$\int_0^{t} \sigma^{\Pi}(s) \, dB_s \to \int_0^t \sigma(s) \, dB_s \qquad \quad \int_0^t b^{\Pi}(s) \, ds \to \int_0^t b(s) \, ds \tag{1}$$ as $|\Pi| \to 0$ where the limits are uniform in probability. Thus in particular, $X^{\Pi} \to X$ uniform in probability as $|\Pi| \to 0$

In a proof of Itô's formula for Itô processes,

$$f(X_t)-f(X_0)= \int_0^t f'(X_s) \sigma(s) \, dB_s + \int_0^t f'(X_s) b(s) \, ds + \frac{1}{2} \int_0^t f''(X_s) \sigma^2(s) \, ds \tag{2}$$

the author claims that it suffices to prove the formula for Itô processes where $b$, $\sigma$ are simple processes, because one can approximate (uniform in probability) an arbitrary Itô process by such processes, as already mentioned.

I don't see why the right-hand side in $(2)$ converges also uniform in probability, i.e. why $$\mathbb{P} \left( \sup_{0 \leq t \leq T} \left| \int_0^t f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) \, dB_s - \int_0^t f'(X_s) \cdot \sigma(s) \, dB_s \right| > \varepsilon \right) \to 0 \tag{3}$$ as $|\Pi| \to 0$. It looks similar to the statement in $(1)$, but there I can apply the maximum inequality and Itô isometry since I know that $\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \to \sigma \cdot 1_{[0,\tau_n)}$ in $L^2(\lambda_T \times \mathbb{P})$. Whereas in this case, I have $f'(X_s^{\Pi}) \to f'(X_s)$ uniform in probability, but as far as I can see no convergence in $L^2(\lambda_T \times \mathbb{P})$. How to conclude $(3)$...?

Any hints would be appreciated.


Actually, the proof is indeed similar to the proof of $(1)$. It's based on the fact that convergence in probability implies almost sure convergence of a subsequence:

By Doob's inequality, Itô's isometry and Tschbysheff inequality we have

$$\begin{align} & \quad \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^t (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon \right) \\ &\leq \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^{t \wedge \tau_n} (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon, \tau_n > T \right) + \mathbb{P}(\tau_n \leq T) \\ &\leq \frac{4}{\varepsilon^2} \cdot \underbrace{\mathbb{E} \left( \int_0^{T \wedge \tau_n} |f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)|^2 \, ds \right)}_{=:I} + \mathbb{P}(\tau_n \leq T) \end{align} $$

(Note that the boundedness of $f'$ implies the existence of the integrals.) Since $$f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s) = f'(X_s^{\Pi}) \cdot \big(\sigma^{\Pi}(s)-\sigma(s)\big) + \sigma(s) \cdot \big(f'(X_s^{\Pi})-f'(X_s) \big)$$ and $(a+b)^2 \leq 2a^2+2b^2$ we obtain $$\begin{align*} I &\leq 2 \mathbb{E} \bigg( \int_0^{T \wedge \tau_n} \underbrace{|f'(X_s^{\Pi})|^2}_{\leq \|f'\|^2_{\infty}} \cdot |\sigma^{\Pi}(s)-\sigma(s)|^2 \, ds \bigg) + 2 \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \end{align*}$$

The first addend converges to $0$ as $|\Pi| \downarrow 0$ since $\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \to \sigma \cdot 1_{[0,\tau_n)}$ in $L^2(\lambda_T \times \mathbb{P})$ by assumption. For the second one, we note that $X^{\Pi} \to X$ uniformly in probability implies

$$\sup_{s \leq t} |\sigma^2(s) (f'(X_s^{\Pi})-f'(X_s))|^2 \stackrel{\mathbb{P}}{\to} 0$$

as $|\Pi| \to 0$ since $f'$ is continuous. From Vitali's convergence theorem, we find

$$ \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \to 0$$

Similarily, one can prove the convergence of the other addends in the right-hand side of $(2)$.

  • $\begingroup$ Thanks for giving such a detailed answer! 2 Question: 1) Where is the subsequence coming into play? 2) From what follows uniform integrability required by Vitali's convergence theorem? $\endgroup$ – Johannes Gerer Feb 28 '15 at 11:40
  • $\begingroup$ @user4514 1. Ouh, I guess I decided to do it without subsequences (right now, I believe that we do not even need Vitali if we use the subsequence principle, but I have to check this). 2. We know that$$Z_{\Pi}:= \int_0^{T \wedge \tau_n} \sigma^2 |f'(X_s^{\Pi}-f'(X_s)|^2 \, ds \stackrel{\mathbb{P}}{\to} 0$$Moreover, since $\tau_n$ is a localizing sequence and $f'$ is bounded, we have$$|Z_{\Pi}| \leq T n 4 \|f'\|_{\infty}^2, $$ so we even have $$\sup_{\Pi} \|Z_{\Pi}\|_{L^{p}}<\infty$$ for all $p>1$. (If you find the answer/question helpful, you can upvoite it by clicking on the arrow next to it.) $\endgroup$ – saz Feb 28 '15 at 12:39
  • $\begingroup$ So it even follows from dominated convergence? $\endgroup$ – Johannes Gerer Feb 28 '15 at 13:13
  • $\begingroup$ @user4514 No, but my last comment shows that uniform integrability is satisfied. (Note that we only know $Z_{\Pi} \to 0$ in probability; not $Z_{\Pi} \to 0$ almost surely - therefore we cannot apply dominated convergence theorem.) [However, if we use subsequences, then it should follow from the dominated convergence theorem; yes.] $\endgroup$ – saz Feb 28 '15 at 13:29
  • $\begingroup$ Ok, but doesn't Vitali's convergence also require almost sure convergence? $\endgroup$ – Johannes Gerer Feb 28 '15 at 14:03

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.