Itô's lemma
A Brownian path has no slope. What takes the place of the chain rule for \(f(W_t)\)?
Split the time interval into short steps, expand \(f\) in a Taylor series over each step, and add the terms up along the path. For a smooth path only the first-order terms count once the steps are short. A Brownian increment over a step of length \(\Delta t\) is of size \(\sqrt{\Delta t}\), and Itô's lemma says which Taylor terms then count; stage 1 lets you find them on one path before the page states the rule.
Before you start · 3 quick checks on Taylor polynomials, quadratic variation and the Itô integral, about 3 min
- Brownian motion. \(W_t\), \(t \ge 0\), is a standard Brownian motion: \(W_0 = 0\), its paths are continuous, and for \(s \lt t\) the increment \(W_t - W_s\) is normal with mean 0 and variance \(t - s\), independent of the path up to time \(s\) (step 9). \(\mathcal F_t\) is the information in the path up to time \(t\); a process is adapted if its value at each \(t\) is determined by \(\mathcal F_t\).
- Partition. For \(T \gt 0\) and \(n \ge 1\), \(t_k = kT/n\) for \(k = 0, \dots, n\), \(\Delta t = T/n\), and \(\Delta W_k = W_{t_k} - W_{t_{k-1}}\) (step 10).
- The Itô integral (step 12). If \(H\) is adapted, has continuous paths and \(\mathbb E\int_0^T H_t^2\,dt \lt \infty\), the left sums \(\sum_{k=1}^n H_{t_{k-1}}\Delta W_k\) converge in probability to \(\int_0^T H_t\,dW_t\) as \(n\) grows. This integral has mean 0.
- Taylor expansion. Over step \(k\) write \(a = W_{t_{k-1}}\) and \(d = \Delta W_k\). For a polynomial \(f\) of degree at most 3, exactly, \[ f(a + d) - f(a) = f'(a)\,d + \tfrac12 f''(a)\,d^2 + \tfrac16 f'''(a)\,d^3. \] For other smooth \(f\) more terms follow, or a remainder.
Fine print: how the paths are made
Paths come from a seeded random number generator, so every visit shows the same paths. Each path is sampled at 4097 equally spaced times on \([0, 1]\): its 4096 increments are \(Z/64\) with \(Z\) a standard normal draw, since \(\sqrt{1/4096} = 1/64\). A partition with \(n = 2^j\) steps uses every \((4096/n)\)th point of the same path, so doubling \(n\) splits every step in two.
Along these doubling partitions, the limits quoted in stage 1 hold for almost every path, not only in probability, because the mean square errors computed in stage 3 add up to a finite total over \(n = 2^j\).
Keep or drop Taylor terms along one path
Fix \(T = 1\). On each step, the change \(f(W_{t_k}) - f(W_{t_{k-1}})\) is split into term cards. Turn cards on, and the page adds the chosen cards over the steps, starting from \(f(W_0)\), and compares this running sum with \(f(W_t)\) on the same path.
| Card | General \(f\) | \(f(x) = x^2\) | \(f(x) = x^3\) |
|---|---|---|---|
| 1, first order | \(f'(a)\,d\) | \(2a\,d\) | \(3a^2 d\) |
| 2, second order | \(\tfrac12 f''(a)\,d^2\) | \(d^2\) | \(3a\,d^2\) |
| 3, third order | \(\tfrac16 f'''(a)\,d^3\) | \(0\) | \(d^3\) |
| 4, second order with \(\Delta t\) for \(d^2\) | \(\tfrac12 f''(a)\,\Delta t\) | \(\Delta t\) | \(3a\,\Delta t\) |
Card 4 is card 2 with the random square \(d^2\) replaced by its mean \(\Delta t\); it is known at the start of the step. Make your prediction first; the explorer opens after it.
Function
Cards in the running sum
| Quantity | At n |
|---|---|
| Card 1 total | |
| Card 2 total | |
| Card 3 total | |
| Card 4 total | |
| \(f(W_T) - f(W_0)\) | |
| Gap: chosen cards minus \(f(W_T) - f(W_0)\) |
Goals
Means, martingales and processes with drift
Let \(f\) have a continuous second derivative, and suppose \(\mathbb E\int_0^T f'(W_t)^2\,dt \lt \infty\). Then, with probability 1,
\[ f(W_T) = f(W_0) + \int_0^T f'(W_t)\,dW_t + \tfrac12\int_0^T f''(W_t)\,dt. \]In differential form, \(df(W_t) = f'(W_t)\,dW_t + \tfrac12 f''(W_t)\,dt\). The result is also called Itô's formula or the Itô–Doeblin formula. Stage 3 proves it for \(f(x) = x^3\). If also \(\mathbb E\int_0^T \lvert f''(W_t)\rvert\,dt \lt \infty\), taking means removes the Itô integral, which has mean 0:
\[ \mathbb E[f(W_T)] = f(W_0) + \tfrac12\int_0^T \mathbb E[f''(W_t)]\,dt. \]Cancel the dt term
Each process \(Y\) below is a function of \(W_t\) plus or minus \(c\) times a time integral, and starts at \(Y_0 = f(W_0)\). Choose \(c\). The page averages \(Y_t\) over 2000 paths on \([0, 1]\) with 128 steps, with the time integrals by the trapezoid rule on those steps, and draws the average against the exact mean.
A process with drift
Let \(X_t = \mu t + \sigma W_t\), so that \(\Delta X_k = \mu\,\Delta t + \sigma\,\Delta W_k\). The question uses path 1 of stage 1.
Let \(X_t = X_0 + \int_0^t \mu_s\,ds + \int_0^t \sigma_s\,dW_s\), where the starting value \(X_0\) is a fixed number, \(\mu\) and \(\sigma\) are adapted with continuous paths and \(\mathbb E\int_0^T \sigma_s^2\,ds \lt \infty\). Let \(f(t, x)\) have continuous partial derivatives \(f_t\), \(f_x\) and \(f_{xx}\). Then, with probability 1, for \(0 \le t \le T\),
\[ \begin{aligned} f(t, X_t) = f(0, X_0) &+ \int_0^t \big(f_t + \mu_s f_x + \tfrac12\sigma_s^2 f_{xx}\big)\,ds \\ &+ \int_0^t \sigma_s f_x\,dW_s, \end{aligned} \]with \(f\) and its derivatives evaluated at \((s, X_s)\). In differential form, \(df = f_t\,dt + f_x\,dX + \tfrac12 f_{xx}\,(dX)^2\), where \((dX)^2 = \sigma^2\,dt\) comes from the table \(dt \cdot dt = 0\), \(dt \cdot dW = 0\), \(dW \cdot dW = dt\) of step 10. When \(\mathbb E\int_0^t (\sigma_s f_x)^2\,ds \lt \infty\), the last integral is the Itô integral of step 12 and has mean 0; otherwise it is defined by stopping the process before it gets large, which this path does not cover.
Itô's lemma for the cube
Assumptions.
- \(W\) is a standard Brownian motion with \(W_0 = 0\). \(T \gt 0\), \(n \ge 1\), \(t_k = kT/n\), \(\Delta t = T/n\) and \(\Delta W_k = W_{t_k} - W_{t_{k-1}}\).
- Increments: \(\Delta W_k\) is independent of the path up to \(t_{k-1}\), and \(\Delta W_k = \sqrt{\Delta t}\,Z_k\) with \(Z_1, \dots, Z_n\) independent standard normal, so \(\mathbb E[Z_k] = \mathbb E[Z_k^3] = 0\), \(\mathbb E[Z_k^2] = 1\), \(\mathbb E[Z_k^4] = 3\) and \(\mathbb E[Z_k^6] = 15\).
- From step 12: \(\sum_{k=1}^n W_{t_{k-1}}^2\Delta W_k \to \int_0^T W_t^2\,dW_t\) in probability, since \(W_t^2\) is adapted, has continuous paths and \(\mathbb E\int_0^T W_t^4\,dt \lt \infty\).
- Riemann sums: \(\sum_{k=1}^n W_{t_{k-1}}\Delta t \to \int_0^T W_t\,dt\) on every path, since each path is continuous.
- Notation: the four parts of the expansion are \[ \begin{aligned} A_n &= 3\sum_{k=1}^{n} W_{t_{k-1}}^2\Delta W_k, \\ B_n &= 3\sum_{k=1}^{n} W_{t_{k-1}}\Delta t, \\ C_n &= 3\sum_{k=1}^{n} W_{t_{k-1}}\big((\Delta W_k)^2 - \Delta t\big), \\ D_n &= \sum_{k=1}^{n} (\Delta W_k)^3. \end{aligned} \]
Conclusion. With probability 1,
\[ W_T^3 = 3\int_0^T W_t^2\,dW_t + 3\int_0^T W_t\,dt, \]that is, \(d(W_t^3) = 3W_t^2\,dW_t + 3W_t\,dt\).
The same argument for a general f
Let \(f\) have a continuous second derivative. Taylor's theorem with the Lagrange remainder gives, on each step, \(f(W_{t_k}) - f(W_{t_{k-1}}) = f'(a)\,d + \tfrac12 f''(\xi_k)\,d^2\) for some \(\xi_k\) between \(a = W_{t_{k-1}}\) and \(W_{t_k}\). Add over \(k\) and split the second-order part into three sums:
\[ \begin{aligned} f(W_T) - f(W_0) = &\textstyle\sum_k f'(a)\,d + \tfrac12\sum_k f''(a)\,\Delta t \\ &+ \tfrac12\textstyle\sum_k f''(a)(d^2 - \Delta t) \\ &+ \tfrac12\textstyle\sum_k \big(f''(\xi_k) - f''(a)\big)d^2. \end{aligned} \]The first sum tends to \(\int_0^T f'(W_t)\,dW_t\) under the condition of stage 2, and the second to \(\tfrac12\int_0^T f''(W_t)\,dt\). If \(\lvert f''\rvert \le M\), the third has mean square \(\tfrac14\sum_k \mathbb E[f''(a)^2]\cdot 2(\Delta t)^2 \le \tfrac12 M^2 T\,\Delta t\), as in line 3. The last is at most \(\tfrac12 \max_k \lvert f''(\xi_k) - f''(a)\rvert \cdot \sum_k d^2\) in size: the maximum tends to 0 on every path, because \(f''\) is uniformly continuous on the bounded range of the path, and \(\sum_k d^2 \to T\) in probability. For \(f''\) that is not bounded, the same argument is applied up to the first time \(\lvert W\rvert\) reaches a level \(m\), and then \(m \to \infty\). The third-order terms of the cube are inside the last sum here; for \(x^3\) they were kept separately as \(D_n\).
Problems
\(W\) is a standard Brownian motion. Sampled increments in the problems are given numbers, not draws from the page.
More problems
Further reading
- K. Itô, "On a formula concerning stochastic differentials," Nagoya Mathematical Journal 3 (1951), 55–65.
- S. E. Shreve, Stochastic Calculus for Finance II: Continuous-Time Models (Springer, 2004), ch. 4, "Stochastic Calculus," section on the Itô–Doeblin formula.
- B. Øksendal, Stochastic Differential Equations, 6th ed. (Springer, 2003), ch. 4, "The Itô Formula and the Martingale Representation Theorem."
- I. Karatzas and S. E. Shreve, Brownian Motion and Stochastic Calculus, 2nd ed. (Springer, 1991), ch. 3, "Stochastic Integration," section on the change-of-variable formula.