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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.

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1 Answer 1

up vote 2 down vote accepted

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 $(2)$.

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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? –  JSG Feb 28 at 11:40
    
@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.) –  saz Feb 28 at 12:39
    
So it even follows from dominated convergence? –  JSG Feb 28 at 13:13
    
@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.] –  saz Feb 28 at 13:29
    
Ok, but doesn't Vitali's convergence also require almost sure convergence? –  JSG Feb 28 at 14:03

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