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Itô's lemma

Itô's lemma

Statement

Let $B(t)$ be Brownian motion and let $f \in C^2(\mathbb{R})$. Then for $a < b$,

\[f(B(b)) - f(B(a)) = \int_a^b f'(B(t))\, dB(t) + \frac{1}{2}\int_a^b f''(B(t))\, dt \quad \text{a.s}.\]

Proof

Taylor expansion

Let \(\Pi_n = \{t_0,\, t_1,\, \cdots,\, t_n\}\) be a partition of $[a, b]$ with \(a = t_0 < t_1 < \cdots < t_n = b\).

  • Using Taylor’s theorem, we have

    \[\begin{align*} f(B(b)) - f(B(a)) &= \sum_{i=1}^n f(B(t_i)) - f(B(t_{i-1})) \\ &= \sum_{i=1}^n f'(B(t_{i-1}))(B(t_i) - B(t_{i-1})) + \frac{1}{2}\sum_{i=1}^n f''(\xi_{n,i})(B(t_i) - B(t_{i-1}))^2 \\ &= \sum_{i=1}^n f'(B(t_{i-1}))(B(t_i) - B(t_{i-1})) + \frac{1}{2} \underbrace{\sum_{i=1}^n (f''(\xi_{n,i}) - f''(B(t_{i-1})))(B(t_i) - B(t_{i-1}))^2}_{(a)} \\ &\quad + \frac{1}{2} \underbrace{\sum_{i=1}^n f''(B(t_{i-1}))((B(t_i) - B(t_{i-1}))^2 - (t_i - t_{i-1}))}_{(b)} \\ &\quad + \frac{1}{2}\sum_{i=1}^n f''(B(t_{i-1}))(t_i - t_{i-1}), \end{align*}\]

    where \(\xi_{n,i} = B(t_{i-1}) + \lambda_i (B(t_i) - B(t_{i-1}))\) for some \(\lambda_i \in (0, 1)\).

  • The first term will converge to the stochastic integral \(\int f'(B(t))\, dB(t)\), while the last term becomes the ordinary Riemann integral \(\frac{1}{2}\int f''(B(t))\, dt\).
  • The remaining two sums $(a)$ and $(b)$ will vanish in the limit.

Error term (a)

We show that a subsequence of \((a)\) converges to \(0\) almost surely as \(\| \Pi_n \| \to 0\).

  • Define

    \(\varphi_n = \max_{1 \leq i \leq n,\, 0 < \lambda < 1} \left\vert f''(B(t_{i-1}) + \lambda (B(t_i) - B(t_{i-1}))) - f''(B(t_{i-1})) \right\vert .\)

    • $\varphi_n$ converges to $0$ almost surely since $f’’$ is uniformly continuous on any compact set.
  • $\sum_{i=1}^n (B(t_i) - B(t_{i-1}))^2$ converges to $b - a$ in \(L^2(\Omega)\).

    • Therefore there is a subsequence that converges to $b - a$ almost surely.

Error term (b)

We show that \((b)\) converges to \(0\) in probability as \(\| \Pi_n \| \to 0\).

  • Fix \(L > 0\). Then we have

    \[\mathbb{P} \left[ |(b)| > \varepsilon \right] = \mathbb{P} \left[ (|(b)| > \varepsilon) \cap \max_{0 \leq i \leq n} |B(t_i)| \leq L \right] + \mathbb{P} \left[ (|(b)| > \varepsilon) \cap \max_{0 \leq i \leq n} |B(t_i)| > L \right].\]
  • If \(\max_{0 \leq i \leq n} \vert B(t_i) \vert \leq L\), then \((b)\) converges to \(0\) in $L^2(\Omega)$.
    • If we let \(M_L = \max_{\vert x \vert \leq L}(\vert f''(x)\vert^2)\), then we have \(\mathbb{E}[(b)] = 0\) and \(Var((b)) \leq 2 M_L \| \Pi_n \| (b-a)\).
    • Hence $(b)$ converges to $0$ in $L^2(\Omega)$, and therefore in probability.
  • The second term can be made arbitrarily small.
    • The Doob submartingale inequality implies that
    \[\begin{align*} \mathbb{P} \left[ (|(b)| > \varepsilon) \cap \max_{0 \leq i \leq n} |B(t_i)| > L \right] &\leq \mathbb{P} \left[ \max_{0 \leq i \leq n}|B(t_i)| > L \right]\\ &\leq \mathbb{P} \left[ \max_{a \leq t \leq b} |B(t)| > L \right]\\ &\leq \frac{1}{L} \mathbb{E} |B(b)|\\ &= \frac{1}{L} \sqrt{\frac{2b}{\pi}}. \end{align*}\]
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