m******2
发帖数: 564 | 1 无drift时候那个
Hitting Time Density = 2 pdf(M,W) M=W
这看起来很奇怪，但是它是事实。
Shreve用偏微分推导了半天的Joint Density for Maximum of Brownian Motion and
its terminal value 在w=m那点的密度恰为在t时刻打到m的概率密度的一半。
有没有高手给理解一下啊？
这可是考验功力的问题啊！ | r**a
发帖数: 536 | 2 Is this strange?
The key is as follows:
\begin{align}
P(\tau_m \leq t) &= P(\tau_m \leq t, W(t)\leq m)+P(\tau_m \leq t, W(t)\geq m
)\
&=P(W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
&=P(\tau_m\leq t, W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
&=2P(W(t)\geq m)
\end{align}
The naive understanding is that you have two possible paths letting $\tau_m<
t$ if $W(t)\geq m$ according to the reflection principle. Those two paths
correspond to the same $\tau_m$. So if the possibility getting one path is p
, then the possibility getting $\tau_m$ is double.
【在 m******2 的大作中提到】
: 无drift时候那个
: Hitting Time Density = 2 pdf(M,W) M=W
: 这看起来很奇怪，但是它是事实。
: Shreve用偏微分推导了半天的Joint Density for Maximum of Brownian Motion and
: its terminal value 在w=m那点的密度恰为在t时刻打到m的概率密度的一半。
: 有没有高手给理解一下啊？
: 这可是考验功力的问题啊！
| m******2
发帖数: 564 | 3 你说的这个我知道。
但是关键是能否严格的证明一下，毕竟variable一个是Wt Mt，一个是Wt和t | m******2
发帖数: 564 | | m******2
发帖数: 564 | 5 The key is as follows:
\begin{align}
P(\tau_m \leq t) &= P(\tau_m \leq t, W(t)\leq m)+P(\tau_m \leq t, W(t)\geq m
)\
&=P(W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
&=P(\tau_m\leq t, W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
&=2P(W(t)\geq m)
\end{align}
我知道你这个宏观的证明，可是这个证明和概率密度之间的关系有什么联系？ | r**a
发帖数: 536 | 6 You need to read Shreve's book more carefully! Go check his Eq.3.7.2 and the
formula just above Eq.3.7.4.
m
【在 m******2 的大作中提到】
: The key is as follows:
: \begin{align}
: P(\tau_m \leq t) &= P(\tau_m \leq t, W(t)\leq m)+P(\tau_m \leq t, W(t)\geq m
: )\
: &=P(W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
: &=P(\tau_m\leq t, W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
: &=2P(W(t)\geq m)
: \end{align}
: 我知道你这个宏观的证明，可是这个证明和概率密度之间的关系有什么联系？
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