81 lines
4.3 KiB
TeX
81 lines
4.3 KiB
TeX
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\documentclass[全部作业]{subfiles}
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\input{mysubpreamble}
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\begin{document}
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\setcounter{chapter}{6}
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\section{反射原理与极值分布}
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\subsection{练习题\thesection}
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\begin{enumerate}
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\questionandanswerSolution[1]{
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求$B(t)=x$条件下$B^{*}(t)$的分布函数与概率密度函数。
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}{
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$(B^{*}(t),B(t))$的联合密度函数为$h(z,x) = \displaystyle \frac{2(2z-x)}{\sqrt{2 \pi t^{3}}} e^{- \frac{(2z - x)^{2}}{2t}}$($z\geqslant x, x\leqslant z$)，$B(t)$的边际密度函数为$p(x)=\frac{1}{\sqrt{2\pi t}} e^{- \frac{x^{2}}{2t}} (x \in \mathbb{R})$，所以在$B(t)=t$的条件下，$B^{*}(t)$的条件密度函数为
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$$
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p_{B^{*}(t)|B(t)} (z|x) = \frac{h(z,x)}{p(x)} = \frac{2(2z-x)}{t} e^{- \frac{2(z^{2}-xz)}{t}} (z\geqslant x \lor 0)
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$$
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所以$B^{*}(t)$的条件分布函数为
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$$
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F_{B^{*}(t)|B(t)}(z|x)=\begin{cases}
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\left. -e^{-\frac{2(u^{2}-xu)}{t}} \right|_{u=x \lor 0}^{z}=1-e^{- \frac{2(z^{2}-xz)}{t}},\quad & 当z\geqslant x\lor 0时 \\
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0,\quad & 当z < x \lor 0 时 \\
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\end{cases}
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$$
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}
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\questionandanswerSolution[2]{
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求$B^{*}(t)-B(t)$的概率密度函数。
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}{
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因为$B^{*}(t)-B(t) = \displaystyle \max_{0\leqslant u\leqslant t} [B(t-u)-B(t)]$，且$\{ B(t-u)-B(t)\ ;\ u \in [0,t] \}$为标准布朗运动，所以
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$$
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B^{*}(t) - B(t) \xlongequal{d} \max_{0\leqslant u\leqslant t} B(u) = B^{*}(t) \xlongequal{d} [B(t)]
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$$
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从而$B^{*}(t)-B(t)$的概率密度函数为$ \displaystyle \sqrt{\frac{2}{\pi t}}e^{- \frac{u^{2}}{2t}} (u\geqslant 0)$。
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}
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\questionandanswerSolution[3]{
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求$P(B(1)\leqslant x | B(u)\geqslant 0, 0\leqslant u\leqslant 1)$。
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}{
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根据对称性，
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$$
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\begin{aligned}
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P(B(1)\leqslant x|B(u)\geqslant 0, 0\leqslant u\leqslant 1) & = P(B(1)\geqslant x |B(u)\leqslant 0, 0\leqslant u\leqslant 1) \\
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& = P(B(1)\geqslant -x | B^{*}(1)=0) \\
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\end{aligned}
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$$ \\
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$(B^{*}(1),B(1))$的联合密度函数为
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$$
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h(z,x)=\frac{2(2z-x)}{\sqrt{2 \pi}} e^{- \frac{(2z-x)^{2}}{2}}\quad (z\geqslant x \lor 0)
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$$
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$B^{*}(1)$的边际密度函数为 $\displaystyle p(z) = \sqrt{\frac{2}{\pi}} e^{- \frac{x^{2}}{2}} (z\geqslant 0)$，所以
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$$
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\text{原式}=\begin{cases}
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- \int_{-x}^{0} u e^{- \frac{u^{2}}{2}} \mathrm{d}u = 1-e^{- \frac{x^{2}}{2}},\quad & 当 x\geqslant 0 时 \\
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0,\quad & 当 x<0 时 \\
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\end{cases}
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$$
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}
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\questionandanswerProof[4]{
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证明对任意$z\geqslant 0$，$\tau_z$与$\tau_{-z}$同分布。
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}{
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对任意$t\geqslant 0$，有
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$$
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P(\tau_{-z}> t) = P(B(u)>-z, 0\leqslant u\leqslant t) \xlongequal{对称性}P(B(u)<z, 0\leqslant u\leqslant t) = P(\tau_z>t)
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$$
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即$\tau_z$与$\tau_{-z}$同分布。
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}
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\questionandanswerProof[6]{
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证明当$0<z \to \infty$时，$\tau_z \xrightarrow{a.s.} \infty$。
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}{
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因为对任意$t>0$和$z>0$有
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$$
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0\leqslant P(\tau_z\leqslant t) = \int_{0}^{t} \frac{z}{\sqrt{2 \pi s^{3}}} e^{- \frac{z^{2}}{2s}} \mathrm{d}s \xlongequal{\text{令} s=\frac{z^{2}}{u^{2}}} \sqrt{\frac{2}{\pi}} \int_{\frac{z}{\sqrt{t}}}^{+\infty} e^{- \frac{u^{2}}{s}} \mathrm{d}u \xrightarrow{z \to +\infty} 0
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$$
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所以 $\displaystyle \lim_{z \to +\infty} P(\tau_z\leqslant t)=0 (\forall t>0)$，又因为$\tau_z$关于$z$单调增，所以事件$\{ \tau_z\leqslant t \}$关于$z$单调减，从而对任意$t >0$有
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$$
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0 = \lim_{n \to \infty} P(\tau_n\leqslant t) = P\left( \bigcap_{n = 1} ^{\infty} \{ \tau_n\leqslant t \} \right) =P\left( \lim_{n \to \infty} \tau_n\leqslant t \right) =P\left( \lim_{z \to +\infty} \tau_z\leqslant t \right)
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$$
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进而由
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$$
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0\leqslant P\left( \lim_{n \to +\infty} \tau_z<+\infty \right) =P\left( \bigcup_{m=1} ^{\infty} \left\{ \lim_{z \to +\infty} \tau_z\leqslant m \right\} \right) \leqslant \sum_{m=1}^{\infty} P \left( \lim_{z \to +\infty} \tau_z\leqslant m \right) =0
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$$
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知$\displaystyle P\left( \lim_{z \to +\infty} \tau_z = +\infty \right) =1$，于是当$z \to +\infty$时有$\tau_z \xrightarrow{a.s.} +\infty$。
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}
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\end{enumerate}
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\end{document}
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