2026-06-08 18:44:51 星期一
好困好累😩
上一节定义了随机积分，随机积分是否满足Newton-Leibniz公式？下面的例子告诉我们并不满足：
\[\int_a^b B(t) \, dB(t) = \lim_{\|\Delta_n\| \to 0} \sum B(t_{i-1}) (B(t_i) - B(t_{i-1}))
\]
\[= \frac{1}{2} \left( B(b)^2 - B(a)^2 - (b - a) \right)
\]
因为 \(E(B(s)B(t)) = \min(s,t)\).
下面先考虑\(f\) 是确定的函数，只与\(t\) 有关.
考虑 \(f\in C^2\)，由 Taylor 展开
\[f(x)-f(x_0)=f'(x_0)(x-x_0)+\frac{1}{2}f''(x+\theta(x-x_0))(x-x_0)^2,\quad \theta\in (0,1)
\]
给定划分 \(\Delta_n = \{a=t_0<t_1<\cdots<t_n=b\}\)
\[\begin{aligned}
f(B(b))-f(B(a))&=\sum_{i=1}^n\left(f(B(t_i))-f(B(t_{i-1}))\right)\\
&=\sum_{i=1}^n f'(B(t_{i-1}))(B(t_i)-B(t_{i-1}))\\
&\quad+\frac{1}{2}\sum_{i=1}^n f''(B(t_{i-1})+\theta_i(B(t_i)-B(t_{i-1})))(B(t_i)-B(t_{i-1}))^2\\
&=A+B
\end{aligned}
\]
\[A\to \int_a^b f'(B(t))dB(t)\quad \text{且}\quad B\to \frac{1}{2}\int_a^b f''(B(t))dt \quad (\|\Delta_n\|\to 0)
\]
综上，
\[f(B(b))-f(B(a))=\int_a^b f'(B(t))dB(t)+\frac{1}{2}\int_a^b f''(B(t))dt
\]
此为 Itô 公式。
例 \(f(x)=x^2\)，
\[B(b)^2-B(a)^2=2\int_a^b B(t)dB(t)+(b-a)
\]
\(\Rightarrow\) \(B(t)^2=2\int_a^t B(s)dB(s)+t\) 是一个 Doob 分解。
例： \(f(x)=x^4\)，\(f(x)=e^x\)，问 \(e^{B(b)}-e^{B(a)}=?\)
sol. \(e^{B(b)}-e^{B(a)}=4\int_a^b (B(t))^3 dB(t) + 6\int_a^b (B(t))^2 dt\)
再考虑 \(f(t,x)\)，有连续偏导数 \(f_t', f_x', f_{xx}''\)。
\[f(t,x) - f(s,x_0) = f(t,x) - f(s,x) + f(s,x) - f(s,x_0)
\]
\[= f_x'(s+\lambda(t-s), x)(t-s) + f_x'(s,x_0)(x-x_0) + \frac{1}{2} f_{xx}''(s, x_0 + \mu(x-x_0))(x-x_0)^2, \quad \lambda,\mu \in (0,1)
\]
\[\begin{aligned}
f(b,B(b)) - f(a,B(a)) &= \sum_{i=1}^n \left( f(t_i, B(t_i)) - f(t_{i-1}, B(t_{i-1})) \right) \\
&= \sum_{i=1}^n \frac{\partial f}{\partial t}(t_{i-1} + \lambda_i(t_i - t_{i-1})) (t_i - t_{i-1}) \\
&\quad + \sum_{i=1}^n \frac{\partial f}{\partial x}(t_{i-1}, B(t_{i-1})) (B(t_i) - B(t_{i-1})) \\
&\quad + \frac{1}{2} \sum_{i=1}^n \frac{\partial^2 f}{\partial x^2}(t_{i-1}, B(t_{i-1}) + \mu_i(B(t_i) - B(t_{i-1}))) (B(t_i) - B(t_{i-1}))^2
\end{aligned}
\]
\[\downarrow \text{ 令 } n \to \infty, \ \|\Delta_n\| \to 0
\]
\[\begin{aligned}
LHS= \int_a^b \frac{\partial f}{\partial t}(t, B(t)) dt + \int_a^b \frac{\partial f}{\partial x}(t, B(t)) dB(t) + \frac{1}{2} \int_a^b \frac{\partial^2 f}{\partial x^2}(t, B(t)) dt
\end{aligned}
\]
于是我们得到了 Itô 公式：
\[\begin{aligned}
f(b, B(b)) - f(a, B(a)) = \int_a^b \frac{\partial f}{\partial t}(t, B(t)) dt + \int_a^b \frac{\partial f}{\partial x}(t, B(t)) dB(t) + \frac{1}{2} \int_a^b \frac{\partial^2 f}{\partial x^2}(t, B(t)) dt
\end{aligned}
\]
例：
\[\begin{aligned}
b B(b)^2 - a B(a)^2 = \int_a^b B(t)^2 dt + 2 \int_a^b t B(t) dB(t) + \frac{1}{2} (b^2 - a^2)
\end{aligned}
\]
⭐定义 An Itô process is a stochastic process of the form
\[ X(t) = X(a) + \int_a^t f(s) dB(s) + \int_a^t g(s) ds, \quad t \in [a,b],
\]
where \(X(a)\) is \(\mathcal{F}_a\)-measurable, \(f \in L_{ad}^2([a,b] \times \Omega)\), \(g \in L_{ad}^1([a,b] \times \Omega)\).
为方便也可写为 \(dX_t = f(t) dB_t + g(t) dt\) 的“stochastic differential”形式记号，真实含义是
\[X(t) = X(a) + \int_a^t f(s) dB(s) + \int_a^t g(s) ds.
\]
Let \(X_t\) be an Itô process, \(\theta(t,x)\) be a continuous function with continuous
\(\frac{\partial\theta}{\partial t},\frac{\partial\theta}{\partial x},\frac{\partial^2\theta}{\partial x^2}\). Then \(\theta(t,X_t)\) is also an Itô process, and
\[d\theta(t,X_t) = \frac{\partial\theta}{\partial t}(t,X_t) dt + \frac{\partial\theta}{\partial x}(t,X_t) dX_t + \frac{1}{2} \frac{\partial^2\theta}{\partial x^2}(t,X_t) (dX_t)^2
\]
where \((dX_t)^2 = f(t)^2 dt\).
\[d\theta(t,X_t) = \left( \frac{\partial\theta}{\partial t}(t,X_t) + \frac{\partial\theta}{\partial x}(t,X_t) g(t) + \frac{1}{2} \frac{\partial^2\theta}{\partial x^2}(t,X_t) f(t)^2 \right) dt + \frac{\partial\theta}{\partial x}(t,X_t) f(t) dB(t)
\]
\[\theta(t,X_t) = \theta(a,X_a) + \int_a^t \left( \frac{\partial\theta}{\partial s}(s,X_s) + \frac{\partial\theta}{\partial x}(s,X_s) g(s) + \frac{1}{2} \frac{\partial^2\theta}{\partial x^2}(s,X_s) f(s)^2 \right) ds + \int_a^t \frac{\partial\theta}{\partial x}(s,X_s) f(s) dB(s)
\]
例: \(dX_t = f(t) dB(t) - \frac{1}{2} f(t)^2 dt\), \(\theta(x) = e^x\).
\[\begin{aligned}
d(e^{X_t}) &= e^{X_t} dX_t + \frac{1}{2} e^{X_t} (dX_t)^2 \\
&= e^{X_t} \left( f(t) dB(t) - \frac{1}{2} f(t)^2 dt \right) + \frac{1}{2} e^{X_t} f(t)^2 dt \\
&= f(t) e^{X_t} dB(t)
\end{aligned}
\]
\(X_0 = 0\), \(X_t = \int_0^t f(s) dB_s - \frac{1}{2} \int_0^t f(s)^2 ds\), \(Y_t = e^{X_t} \Rightarrow \begin{cases} dY_t = f(t) Y_t dB_t \\ Y_0 = 1 \end{cases}\)
解为 \(Y_t = e^{X_t}\), \(X_t = \int_0^t f(s) dB_s - \frac{1}{2} \int_0^t f(s)^2 ds\).
Multidimensional Itô formula
\(B_1(t), \dots, B_m(t)\) are independent Brownian motions.
\(X_i(t), i = 1, \dots, n\) are Itô processes
\[dX_i(t) = \sum_{j=1}^m f_{ij}(t) \, dB_j(t) + g_i(t) \, dt,
\]
where \(f_{ij} \in L_{ad}^2([a,b] \times \Omega)\), \(g_i \in L_{ad}^1([a,b] \times \Omega)\).
Let
\[B(t) = (B_1(t), \dots, B_m(t))^T, \quad g(t) = (g_1(t), \dots, g_n(t))^T,
\]
\[f(t) = (f_{ij}(t))_{n \times m}, \quad X(t) = (X_1(t), X_2(t), \dots, X_n(t))^T.
\]
Then
\[dX(t) = f(t) \, dB(t) + g(t) \, dt.
\]
下面是一个展开 要会这个 怎么把theta给写出来
Theorem. Let \(\theta(t, x_1, \dots, x_n)\) be continuous with continuous \(\frac{\partial \theta}{\partial t}, \frac{\partial \theta}{\partial x_i}, \frac{\partial^2 \theta}{\partial x_i \partial x_j}\). Then
\[\begin{aligned}
d\theta(t, X_t^{(1)}, X_t^{(2)}, \dots, X_t^{(n)}) &= \frac{\partial \theta}{\partial t}(t, X_t^{(1)}, \dots, X_t^{(n)}) dt \\
&\quad + \sum_{i=1}^n \frac{\partial \theta}{\partial x_i}(t, X_t^{(1)}, \dots, X_t^{(n)}) dX_t^{(i)} \\
&\quad + \frac{1}{2} \sum_{i,j=1}^n \frac{\partial^2 \theta}{\partial x_i \partial x_j}(t, X_t^{(1)}, \dots, X_t^{(n)}) dX_t^{(i)} dX_t^{(j)}
\end{aligned}
\]
例：
- \(\theta(X,Y)=XY\)
\[d(X_t Y_t)=Y_t dX_t + X_t dY_t + \frac{1}{2}(dX_t dY_t + dY_t dX_t)
\]
\[= Y_t dX_t + X_t dY_t + dX_t dY_t
\]
\[\begin{cases} dX_t = f(t) dB(t) + \xi(t) dt \\ dY_t = g(t) dB(t) + \eta(t) dt \end{cases}
\]
\[d(X_t Y_t) = \left( Y_t \xi(t) + X_t \eta(t) + f(t) g(t) \right) dt + \left( Y_t f(t) + X_t g(t) \right) dB(t)
\]
\[\begin{array}{c|cc} & dB_i(t) & dt \\ \hline dB_j(t) & \delta_{ij} dt & 0 \\ \hline dt & 0 & 0 \end{array}
\]
-
\(\theta(x)=x^2\)
\[\begin{aligned}
d(X_t^2) &= 2X_t dX_t + (dX_t)^2 \\
&= 2X_t \big( f(t) dB(t) + \xi(t) dt \big) + f(t)^2 dt \\
&= \big( 2X_t \xi(t) + f(t)^2 \big) dt + 2X_t f(t) dB(t)
\end{aligned}
\]
-
Langevin equation
\[dX_t = \alpha \, dB(t) - \beta X_t \, dt, \quad X_0 = x_0, \quad \alpha \in \mathbb{R}, \quad \beta > 0
\]
\(\theta(t,x)=xe^{\beta t}\),
\[\begin{aligned}
d( x_t e^{\beta t} ) &= \beta x_t e^{\beta t} dt + e^{\beta t} ( \alpha dB(t) - \beta x_t dt ) \\
&= \alpha e^{\beta t} dB(t)
\end{aligned}
\]
写成积分形式：
\[X_t=e^{-\beta t}x_0+\alpha\int_0^te^{-\beta(t-s)}dB(s)
\]
$\mathbb{E}[X_t] = e^{-\beta t} x_0 \to 0, \quad \operatorname{Var}(X_t) = \frac{\alpha^2}{2\beta}(1 - e^{-2\beta t}) \to \frac{\alpha^2}{2\beta}, t \to \infty $
-
\(X_t = \cos B(t)\), \(Y_t = \sin B(t)\), \(V_t = (X_t, Y_t)^T\)
\[dX_t = -\sin B(t) \, dB(t) - \frac{1}{2} \cos B(t) \, dt
\]
\[dY_t = \cos B(t) \, dB(t) - \frac{1}{2} \sin B(t) \, dt
\]
\[\Rightarrow \begin{cases}
dV_t = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} V_t \, dB(t) - \frac{1}{2} V_t \, dt \\[1em]
V_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
\end{cases}
\]
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