2026-06-09 19:28:26 星期二
把老师上课随手写的小题目都汇总在这里，稍微写一下.
没时间整理格式了，让我复习完一遍就行了...下周就考了...
条件期望
随机积分
- 求
\[\lim_{||\Delta_n|| \to 0} \sum_{i=1}^{n} B\left( \frac{t_i + t_{i-1}}{2} \right) \left( B(t_i) - B(t_{i-1}) \right) = ?
\]
证. 记 \(\tau_i = \frac{t_{i-1} + t_i}{2}\).
\[\begin{aligned}
&\sum_{i=1}^{n} B\left(\frac{t_{i-1} + t_i}{2}\right) \left( B(t_i) - B(t_{i-1}) \right) \\
&= \sum_{i=1}^{n} B(t_{i-1}) \left( B(t_i) - B(t_{i-1}) \right) + \sum_{i=1}^{n} \left( B(\tau_i) - B(t_{i-1}) \right) \left( B(t_i) - B(t_{i-1}) \right) \\
&= L_n + K_n.
\end{aligned}
\]
\[L_n \xrightarrow{L^2} \int_a^b B(t) \, dB(t) = \frac{1}{2} \left( B(b)^2 - B(a)^2 \right) - \frac{1}{2}(b-a).
\]
\[\begin{aligned}
K_n &= \sum_{i=1}^{n} \left( B(\tau_i) - B(t_{i-1}) \right) \left[ \left( B(t_i) - B(\tau_i) \right) + \left( B(\tau_i) - B(t_{i-1}) \right) \right] \\
&= \sum_{i=1}^{n} \left( B(\tau_i) - B(t_{i-1}) \right) \left( B(t_i) - B(\tau_i) \right) + \sum_{i=1}^{n} \left( B(\tau_i) - B(t_{i-1}) \right)^2 \\
&= I_n + J_n.
\end{aligned}
\]
\[J_n \xrightarrow{L^2} \sum_{i=1}^{n} \frac{t_i - t_{i-1}}{2} = \frac{b-a}{2}.
\]
\[E(I_n^2) = \sum_{i=1}^{n} E \left[ \left( B(\tau_i) - B(t_{i-1}) \right)^2 \right] \cdot E \left[ \left( B(t_i) - B(\tau_i) \right)^2 \right]
= \sum_{i=1}^{n} \left( \frac{t_i - t_{i-1}}{2} \right)^2
\leq \| \Delta_n \| \frac{b-a}{4} \to 0,
\]
即 \(I_n \xrightarrow{L^2} 0\).
\(\therefore K_n \xrightarrow{L^2} \frac{b-a}{2}\).
\(\therefore\) 原极限 \(= \frac{1}{2} \left( B(b)^2 - B(a)^2 \right)\).
- 证明
\[E ( B ( t ) ^ { n } ) = \left\{ \begin{array} { l l } { 0 } & { n = 2 k + 1 } \\ { \dfrac { ( 2 k ) ! \, t ^ { k } } { 2 ^ { k } k ! } } & { n = 2 k } \end{array} \right. \quad k = 1 , 2 , \cdots
\]
Proof. 由于 \(B(t) \sim N(0, t)\)，
\[E(B(t)^{2k}) = \frac{1}{\sqrt{2\pi t}} \int_{-\infty}^{+\infty} x^{2k} e^{-\frac{x^2}{2t}} dx.
\]
\[\begin{aligned}
\int_{-\infty}^{+\infty} x^{2k} e^{-\frac{x^2}{2t}} dx
&= \left. -t \cdot x^{2k} e^{-\frac{x^2}{2t}} \right|_{-\infty}^{+\infty} + t(2k+1) \int_{-\infty}^{+\infty} x^{2k-2} e^{-\frac{x^2}{2t}} dx \\
&= t(2k+1) \int_{-\infty}^{+\infty} x^{2k-2} e^{-\frac{x^2}{2t}} dx.
\end{aligned}
\]
两边除以 \(\sqrt{2\pi t}\)，即
\[E(B(t)^{2k}) = (2k-1)t \, E(B(t)^{2k-2}).
\]
递推得
\[\begin{aligned}
E(B(t)^{2k}) &= t \cdot (2k-1) \cdot t \cdot (2k-3) \cdots t \cdot 1 \\
&= (2k-1)(2k-3)\cdots 3 \cdot 1 \cdot t^k \\
&= \frac{(2k-1) \cdot 2k \cdot (2k-3) \cdot (2k-2) \cdots 3 \cdot 4 \cdot 1 \cdot 2}{2 \cdot 4 \cdots (2k)} \, t^k \\
&= \frac{(2k)!}{2^k k!} \, t^k.
\end{aligned}
\]
- Find the variance of \(\displaystyle \int_0^1 |B(t)| \, dB(t)\).
Proof.
\[E \int_0^1 |B(t)|\, dB(t) = 0.
\]
\[\begin{aligned}
\operatorname{Var}\left( \int_0^1 |B(t)|\, dB(t) \right)
&= E \left( \int_0^1 |B(t)|\, dB(t) \right)^2 \\
&= \int_0^1 E(|B(t)|^2)\, dt \\
&= \int_0^1 E(B(t)^2)\, dt \\
&= \int_0^1 t \, dt = \frac{1}{2}.
\end{aligned}
\]
Exercise
-
For fixed \(t > 0\) and \(s > 0\), find the distribution of the random variable \(X = B(t) + B(s)\), where \(B(t)\) is a Brownian motion.
-
Let \(B(t)\) be a Brownian motion, and \(0 < s \leq t \leq u \leq v\). Show that the random variables \(aB(s) + bB(t)\) and \(\frac{1}{v}B(v) - \frac{1}{u}B(u)\) are independent for any \(a, b \in \mathbb{R}\) satisfying the condition \(as + bt = 0\).
-
Let \(B(t)\) be a Brownian motion. Show that \(\lim_{t \to 0^+} t B\left(\frac{1}{t}\right) = 0\) a.s.
-
Let \(B(t)\) be a Brownian motion. Define
\[W(t) = \begin{cases}
t B\left(\frac{1}{t}\right) & \text{for } t > 0, \\
0 & \text{for } t = 0.
\end{cases}
\]
Show that \(W(t)\) is a Brownian motion.
- Let \(B(t)\) be a Brownian motion. Show that
\[\mathbb{E}[(B(s) - B(t))^{2n}] = \frac{(2n)!}{2^n n!} |s - t|^n, \quad \mathbb{E}[(B(s) - B(t))^{2n+1}] = 0.
\]
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