Differentiable approximation of Brownian diffusion with bounded volatility

Let $$\{W_t\}_{t\in[0;T]}$$ be a one-dimensional Brownian motion and let $$\{\mathcal F_t\}_{t\in[0;T]}$$ be the augmented filtration generated by this Brownian motion. Let $$\{\sigma_t\}_{t\in[0;T]}$$ be real-valued, progressively measurable and bounded. Is there a sequence of continuous and adapted processes $$\{\tilde\sigma^n_t\}_{t\in[0;T]}$$ such that $$\lim_{n\to \infty} \mathbb E\Bigg( \int^T_0 \bigg( \int^t_0 \tilde\sigma^n_s \mathrm ds - \int^t_0 \sigma_s \mathrm dW_s \bigg)^2 \mathrm dt \Bigg) = 0?$$

1 Answer

Yes. Let $$X_t := \int^t_0 \sigma_s \mathrm dW_s$$. Due to Theorem V.6 from the book Stochastic Integration and Differential Equations (second edition) by P.E. Protter, there is a continuous and adapted process $$\{\tilde X^n_t\}_{t\in[0;T]}$$ such that $$\tilde X^n_t = \int^t_0 n \cdot \big( \tilde X^n_s - X_s \big) \mathrm ds.$$ Hence we define $$\tilde\sigma^n_s := n \cdot ( \tilde X^n_s - X_s )$$, which is adapted and even continuous.

To prove the limit property, we first prove the following:

Lemma 1. For all $$\beta,\delta \in (0;\infty)$$, there exists $$\nu \in (0;\infty)$$ such that $$\mathbb P \bigg[ \sup_{0 \le s \le t \le \min(s+\nu,T)} \bigg| \int^t_s \sigma_u \mathrm dW_u \bigg| \le \beta \bigg] \ge 1 - \delta.$$ Proof: We discretize the interval $$[0;T]$$ and consider events that its increments stay bounded. For all integer $$0 \le k < N$$ and all $$\alpha \in (0;\infty)$$, we define $$A^\alpha_{k,N} := \bigg\{ \max_{\frac{T}{N} k \le t \le \frac{T}{N} (k+1)} \bigg| \int_{\frac{T}{N} k}^t \sigma_u \mathrm dW_u \bigg| \ge \alpha \bigg\}.$$ Due to the Burkholder–Davis–Gundy inequality (since $$\sigma$$ is bounded, say $$\vert\sigma\vert \le \overline\sigma$$), $$\mathbb E\bigg( \max_{T/N\cdot k \le t \le T/N\cdot (k+1)} \bigg| \int_{T/N\cdot k}^t \sigma_u \mathrm dW_u \bigg|^4 \bigg) \le C_4 \mathbb E\bigg( \bigg\langle \int_{T/N\cdot k}^\cdot \sigma_u \mathrm dW_u \bigg\rangle_{T/N\cdot (k+1)}^2 \bigg) \\= C_4 \mathbb E\bigg( \bigg( \int_{T/N\cdot k}^{T/N\cdot (k+1)} \big(\sigma_u\big)^2 \mathrm du \bigg)^2 \bigg) \le C_4 \bigg( \frac{T} N \cdot \overline\sigma^2 \bigg)^2 = N^{-2} C$$ and due to the Markov inequality, we obtain $$\mathbb P \big( A^\alpha_{k,N} \big) \le \alpha^{-4} \mathbb E\bigg( \max_{T/N\cdot k \le t \le T/N\cdot (k+1)} \bigg| \int_{T/N\cdot k}^t \sigma_u \mathrm dW_u \bigg|^4 \bigg) \le \alpha^{-4} N^{-2} C$$ Now we assume that $$\omega \in \Omega \backslash \bigcup_{k=0}^{N-1} A^\alpha_{k,N}$$ and assume $$s, t \in [0;T]$$ with $$s \le t \le s + T/N$$. Then we can find a $$k \in \{0,\ldots,N-1\}$$ such that $$T/N\cdot k \le s \le T/N\cdot (k+1) \le t \le T/N\cdot (k+2)$$ or $$T/N\cdot k \le s \le t \le T/N\cdot (k+1)$$. In the first case, we obtain $$\bigg| \bigg(\int^t_s \sigma_u \mathrm dW_u\bigg)(\omega) \bigg| \le \bigg| \bigg(\int_{T/N\cdot (k+1)}^t \sigma_u \mathrm dW_u\bigg)(\omega) \bigg| + \bigg| \bigg(\int_{T/N\cdot k}^{T/N\cdot (k+1)} \sigma_u \mathrm dW_u\bigg)(\omega) \bigg| \\ \quad + \bigg| \bigg(\int_{T/N\cdot k}^s \sigma_u \mathrm dW_u\bigg)(\omega) \bigg| \le 3 \alpha.$$ In the second case, we get the same result analogously.

Let $$\omega \in \Omega \backslash \bigcup_{k=0}^{N-1} A^\alpha_{k,N}$$ and $$s, t \in [0;T]$$ with $$|s - t| \le \frac{T}{N}$$. Then, $$\bigg| \bigg(\int^t_s \sigma_u \mathrm dW_u\bigg)(\omega) \bigg| \leq 3 \alpha$$ and so $$\Omega \backslash \bigcup_{k=0}^{N-1} A^\alpha_{k,N} \subseteq \bigg\{ \max_{s,t\in [0;T], |s-t| \le \frac T N} \bigg| \int^t_s \sigma_u \mathrm dW_u \bigg| \le 3 \alpha \bigg\}.$$ As a result, if $$N$$ is large enough, $$\mathbb P \bigg[ \max_{s,t\in [0;T], |s-t| \le \frac T N} \bigg| \int^t_s \tilde\sigma_u \mathrm dW_u \bigg| \le 3 \alpha \bigg] \\\ge 1 - \sum_{k=0}^{N-1} \mathbb P \big( A^\alpha_{k,N} \big) \ge 1 - \frac{C}{\alpha^{4} N^{1}} \ge 1 - \delta,$$ which proves the statement.

Since $$\tilde X^n$$ always moves into the direction of $$X$$, we also have the following:

Lemma 2. $$\sup_{t \in [0;T]} \vert \tilde X^n_t \vert \le \sup_{t \in [0;T]} \vert X_t \vert$$

Now since the increments of $$X$$ are bounded on an event of large probability due to Lemma 1, it is also straightforward to prove this:

Lemma 3. Let $$M^{\beta,\nu}:=\bigg\{\sup_{0 \le s \le t \le \min(s+\nu,T)} \bigg| \int^t_s \sigma_u \mathrm dW_u \bigg| \le \beta\bigg\}.$$ Then for all $$\omega\in M$$, we have $$\sup_{t\in [0;T]} \big\vert \tilde X^{\beta/\nu}_t - X_t \big\vert \le 3 \beta.$$

Now we prove the main statement. Let $$n:=\beta/\nu$$. Due to the Minkovski inequality, $$\sqrt{ \mathbb E\bigg( \int^T_0 \big( \tilde X^n_s - X_s \big)^2 \mathrm dt \bigg) } \\\le \sqrt{ \mathbb E\bigg( \mathbb 1_{M^{\beta,\nu}} \int^T_0 \big( \tilde X^n_s - X_s \big)^2 \mathrm dt \bigg) } + \sqrt{ \mathbb E\bigg( \mathbb 1_{\Omega\backslash M^{\beta,\nu}} \int^T_0 \big( \tilde X^n_s - X_s \big)^2 \mathrm dt \bigg) }$$ The first summand can be bound directly by $$3 \beta \sqrt T$$ using Lemma 2. The second summand can be bound using Hölder inequality by $$\mathbb E\bigg( \int^T_0 \mathbb 1_{\Omega\backslash M^{\beta,\nu}} \big( \tilde X^n_s - X_s \big)^2 \mathrm dt \bigg) \\\le \sqrt{ \mathbb E\bigg( \int^T_0 \mathbb 1_{\Omega\backslash M^{\beta,\nu}} \mathrm dt \bigg) } \sqrt{ \mathbb E\bigg( \int^T_0 \big( \tilde X^n_s - X_s \big)^4 \mathrm dt \bigg) } = \sqrt T \sqrt{1 - \mathbb P\big(M^{\beta,\nu}\big) } \sqrt{ \mathbb E\bigg( \int^T_0 \big( \tilde X^n_s - X_s \big)^4 \mathrm dt \bigg) }$$ The first factor can be made arbitrarily small if choosing $$\nu$$ small enough depending on $$\beta$$ due to Lemma 1, and the second factor is bounded due to the boundedness of $$\sigma$$ and Lemma 2.