From fa2577376e8feb851db05ace4e58bf629c4dbb76 Mon Sep 17 00:00:00 2001 From: "Keith A. Lewis" Date: Wed, 9 Oct 2024 09:14:02 -0400 Subject: [PATCH 1/2] sync --- py.md | 79 +++++++++++++++++++++++++++++++++++++++++++++++------------ 1 file changed, 64 insertions(+), 15 deletions(-) diff --git a/py.md b/py.md index cf41f31..de1cc4f 100644 --- a/py.md +++ b/py.md @@ -9,23 +9,38 @@ abstract: \newcommand\o[1]{\overline{#1}} \newcommand\u[1]{\underline{#1}} +\newcommand{\Cov}{\operatorname{Cov}} + +If $B_t$ is Brownian motion let $\o{B}_t = \max_{0\le s\le t}B_s$ be the running max. +Let $\tau_a = \inf\{t > 0\mid \o{B}_t > a\}$ be the first time $B_t$ hits level $a$. +Note $\tau_a < t$ if and only if $\o{B}_t > a$. -If $B_t$ is Brownian motion and $\o{B}_t = \max_{0\le s\le t}B_s$ is the running max, then $$ E[f(B_t) 1(\o{B}_t > a)] = E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(B_t > a)] $$ +__Proof__. Define the reflected Brownian motions $B_t^a = B_t$ for $t < \tau_a$ and $B_t^a = 2a - B_t$ for $t > \tau_a$. +We have +$$ +\begin{aligned} +E[f(B_t) 1(\o{B}_t > a)] &= E[f(B_t) 1(\o{B}_t > a,B_t > a)] + E[f(B_t) 1(\o{B}_t > a,B_t < a)]\\ +&= E[f(B_t) 1(B_t > a)] + E[f(B_t) 1(\o{B}_t > a,B_t < a)] \\ +&= E[f(B_t) 1(B_t > a)] + E[f(B^a_t) 1(\o{B^a}_t > a,B^a_t < a)] \\ +&= E[f(B_t) 1(B_t > a)] + E[f(2a - B) 1(\o{B^a}_t > a,2a - B_t < a)] \\ +&= E[f(B_t) 1(B_t > a)] + E[f(2a - B) 1(\o{B^a}_t > a, B_t > a)] \\ +&= E[f(B_t) 1(B_t > a)] + E[f(2a - B) 1(B_t > a)] \\ +\end{aligned} +$$ + Taking $f(x) = 1$ gives $P(\o{B}_t > a) = 2P(B_t > a)$. $1 - G(x) = 2(1 - F(x))$, $G(x) = 2F(x) - 1$, $x > 0$. $g(x) = 2f(x)$, $x > 0$. -Let $\tau_a = \inf\{t > 0\mid \o{B}_t > a\}$ be the first time $B$ -Note $\tau_a < t$ if and only if $\o{B}_t > a$. -Recall $E[X] = \int_0^\infty P(X > x)\,dx$ if $X$ is positive. +Recall $E[T] = \int_0^\infty P(T > t)\,dx$ if $T\ge0$. $P(\min\{\tau_a, T\} > t) = P(\tau_a > t, T > t) = P(\tau_a > t)1(T > t)$ @@ -33,27 +48,61 @@ $E[\min\{\tau_a, T\}] = \int_0^T P(\tau_a > t)\,dt$. Let $X_t = \mu t + \sigma B_t$. -$B_t - \alpha t$ is BM under $P_\alpha$ where $dP_\alpha/dP|_t = \exp(\alpha B_t - \alpha^2t/2)$. +Girsanov's theorem states $B_t - \alpha t$ is BM under $P_\alpha$ +where ${dP_\alpha/dP|_t = \exp(\alpha B_t - \alpha^2t/2)}$. +Let $g(x) = f(\mu t + \sigma x)$ and $\alpha = \mu/\sigma$. $$ \begin{aligned} -&E[g(X_t) 1(\o{X}_t > a)] \\ -\qquad &= E[f(B_t) 1(\max_{0\le s\le t}\mu s + \sigma B_s > a)] \\ -\qquad &= E[\exp(\alpha B_t - \alpha^2t/2) f(B_t - \alpha t) +E[f(X_t) 1(\o{X}_t > a)] &= E[g(B_t) 1(\max_{0\le s\le t}\mu s + \sigma B_s > a)] \\ +&= E[\exp(\alpha B_t - \alpha^2t/2) g(B_t - \alpha t) 1(\max_{0\le s\le t}\mu s + \sigma (B_s - \alpha s) > a)] \\ -\qquad &= E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) f(B_t - (\mu/\sigma) t) +&= E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) g(B_t - (\mu/\sigma) t) 1(\max_{0\le s\le t}\sigma B_s > a)] \\ -\qquad &= E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) f(B_t - (\mu/\sigma) t) +&= E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) g(B_t - (\mu/\sigma) t) 1(\max_{0\le s\le t}B_s > a/\sigma)] \\ -\qquad &= E[h(B_t) 1(\o{B}_t > a/\sigma)] \\ -\qquad &= E[(h(B_t) + h(2a - B_t)) 1(B_t > a/\sigma)], \\ +&= E[h(B_t) 1(\o{B}_t > a/\sigma)] \\ +&= E[(h(B_t) + h(2a - B_t)) 1(B_t > a/\sigma)], \\ \end{aligned} $$ -where $f(x) = g(\mu t + \sigma x)$ and $h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2) f(x - (\mu/\sigma) t)$. +where $h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2) g(x - (\mu/\sigma) t)$. -If $g(x) = 1$ then $f(x) = 1$ and $h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2)$. +If $f(x) = 1$ then $g(x) = 1$ and $h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2)$. $$ P(\o{X}_t > a) = E[(\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) - + \exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2))1(B_t > a)]. + + \exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2))1(B_t > a/\sigma)]. +$$ + +$E[e^{\alpha B_t - \alpha^2t/2} 1(B_t > a)] += E[1(B_t + \Cov(\alpha B_t,B_t) > a)] += P(B_t > a - \alpha t)$. + +$E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) 1(B_t > a/\sigma)] += P(B_t > a/\sigma - (\mu/\sigma)t)$. + +$E[\exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2) 1(B_t > a/\sigma)] +=\exp(2a\mu/\sigma) P(B_t > a/\sigma + (\mu/\sigma)t)$. + +$$ +E[\min\{\tau_a,T\}] = \int_0^T P(B_t > a/\sigma - (\mu/\sigma)t) + + \exp(2a\mu/\sigma) P(B_t > a/\sigma + (\mu/\sigma)t)\,dt \\ +$$ + +Let $Z$ be standard normal. + +$\int_0^T P(Z > a + bt)\,dt = +P(Z > a + bt)t|_0^T + \int_0^T t \phi(a + bt)b\,dt$ + +$u = P(Z > a + bt) = 1 - \Phi(a + bt)$, $dv = dt$; $du = -\phi(a + bt)b\,dt$, $v = t$. + +$s = a + bt$, $t = (s - a)/b$ + +$$ +\begin{aligned} +\int_0^T t \phi(a + bt)b\,dt &= \int_a^{a + bT} (s - a)/b \phi(s)b\,ds/b \\ +&= \int_a^{a + bT} (s - a) \phi(s)\,ds/b \\ +&= ((-\exp(s^2/2) - a \Phi(s))/b|_a^{a + bT} \\ +\end{aligned} $$ + From 4bf35fc77bc3f53101c0af81ca8d31dbf4e6f039 Mon Sep 17 00:00:00 2001 From: "Keith A. Lewis" Date: Thu, 10 Oct 2024 17:49:38 -0400 Subject: [PATCH 2/2] sync --- py.md | 72 ++++++++++++++++++++++++++++++++++------------------------- 1 file changed, 41 insertions(+), 31 deletions(-) diff --git a/py.md b/py.md index de1cc4f..727aedc 100644 --- a/py.md +++ b/py.md @@ -7,13 +7,11 @@ classoption: fleqn abstract: ... -\newcommand\o[1]{\overline{#1}} -\newcommand\u[1]{\underline{#1}} +\renewcommand\o[1]{\overline{#1}} +\renewcommand\u[1]{\underline{#1}} \newcommand{\Cov}{\operatorname{Cov}} If $B_t$ is Brownian motion let $\o{B}_t = \max_{0\le s\le t}B_s$ be the running max. -Let $\tau_a = \inf\{t > 0\mid \o{B}_t > a\}$ be the first time $B_t$ hits level $a$. -Note $\tau_a < t$ if and only if $\o{B}_t > a$. $$ E[f(B_t) 1(\o{B}_t > a)] = E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(B_t > a)] @@ -26,21 +24,35 @@ $$ E[f(B_t) 1(\o{B}_t > a)] &= E[f(B_t) 1(\o{B}_t > a,B_t > a)] + E[f(B_t) 1(\o{B}_t > a,B_t < a)]\\ &= E[f(B_t) 1(B_t > a)] + E[f(B_t) 1(\o{B}_t > a,B_t < a)] \\ &= E[f(B_t) 1(B_t > a)] + E[f(B^a_t) 1(\o{B^a}_t > a,B^a_t < a)] \\ -&= E[f(B_t) 1(B_t > a)] + E[f(2a - B) 1(\o{B^a}_t > a,2a - B_t < a)] \\ -&= E[f(B_t) 1(B_t > a)] + E[f(2a - B) 1(\o{B^a}_t > a, B_t > a)] \\ -&= E[f(B_t) 1(B_t > a)] + E[f(2a - B) 1(B_t > a)] \\ +&= E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(\o{B^a}_t > a,2a - B_t < a)] \\ +&= E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(\o{B^a}_t > a, B_t > a)] \\ +&= E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(B_t > a)] \\ +&= E[\bigl(f(B_t) + f(2a - B_t)\bigr) 1(B_t > a)] \\ \end{aligned} $$ Taking $f(x) = 1$ gives $P(\o{B}_t > a) = 2P(B_t > a)$. -$1 - G(x) = 2(1 - F(x))$, $G(x) = 2F(x) - 1$, $x > 0$. +Let $\u{B}_t = \min_{0 \le s \le t} B_s$. $(\u{-B})_t = -\o{B}_t$. -$g(x) = 2f(x)$, $x > 0$. +$$ +\begin{aligned} + E[f(B_t) 1(\u{B}_t < a)] &= E[f(-B_t) 1(\u{-B}_t < a)] \\ + &= E[f(-B_t) 1(-\o{B}_t < a)] \\ + &= E[f(-B_t) 1(\o{B}_t > -a)] \\ + &= E[\bigl(f(-B_t) + f(-(-2a - B_t)\bigr) 1(B_t > -a)] \\ + &= E[\bigl(f(-B_t) + f(2a + B_t)\bigr) 1(B_t > -a)] \\ + &= E[\bigl(f(B_t) + f(2a - B_t)\bigr) 1(B_t < a)] \\ +\end{aligned} +$$ + +Taking $f(x) = 1$ gives $P(\u{B}_t < a) = 2P(B_t < a)$. +Let $\tau_a = \inf\{t > 0\mid \u{B}_t < a\}$ be the first time $B_t$ hits level $a < 0$. +Note $\tau_a > t$ if and only if $\u{B}_t < a$. -Recall $E[T] = \int_0^\infty P(T > t)\,dx$ if $T\ge0$. +Recall $E[\tau] = \int_0^\infty P(\tau > t)\,dt$ if $\tau\ge0$. $P(\min\{\tau_a, T\} > t) = P(\tau_a > t, T > t) = P(\tau_a > t)1(T > t)$ @@ -54,15 +66,15 @@ Let $g(x) = f(\mu t + \sigma x)$ and $\alpha = \mu/\sigma$. $$ \begin{aligned} -E[f(X_t) 1(\o{X}_t > a)] &= E[g(B_t) 1(\max_{0\le s\le t}\mu s + \sigma B_s > a)] \\ +E[f(X_t) 1(\u{X}_t < a)] &= E[g(B_t) 1(\min_{0\le s\le t}\mu s + \sigma B_s < a)] \\ &= E[\exp(\alpha B_t - \alpha^2t/2) g(B_t - \alpha t) - 1(\max_{0\le s\le t}\mu s + \sigma (B_s - \alpha s) > a)] \\ + 1(\min_{0\le s\le t}\mu s + \sigma (B_s - \alpha s) < a)] \\ &= E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) g(B_t - (\mu/\sigma) t) - 1(\max_{0\le s\le t}\sigma B_s > a)] \\ + 1(\min_{0\le s\le t}\sigma B_s < a)] \\ &= E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) g(B_t - (\mu/\sigma) t) - 1(\max_{0\le s\le t}B_s > a/\sigma)] \\ -&= E[h(B_t) 1(\o{B}_t > a/\sigma)] \\ -&= E[(h(B_t) + h(2a - B_t)) 1(B_t > a/\sigma)], \\ + 1(\min_{0\le s\le t}B_s < a/\sigma)] \\ +&= E[h(B_t) 1(\u{B}_t < a/\sigma)] \\ +&= E[(h(B_t) + h(2a - B_t)) 1(B_t < a/\sigma)], \\ \end{aligned} $$ where $h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2) g(x - (\mu/\sigma) t)$. @@ -70,31 +82,30 @@ where $h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2) g(x - (\mu/\sigma) t)$. If $f(x) = 1$ then $g(x) = 1$ and $h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2)$. $$ -P(\o{X}_t > a) = E[(\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) - + \exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2))1(B_t > a/\sigma)]. +P(\u{X}_t < a) = E[\bigl(\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) + + \exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2)\bigr)1(B_t < a/\sigma)]. $$ -$E[e^{\alpha B_t - \alpha^2t/2} 1(B_t > a)] -= E[1(B_t + \Cov(\alpha B_t,B_t) > a)] -= P(B_t > a - \alpha t)$. +$E[\exp(\alpha B_t - \alpha^2t/2) 1(B_t < a)] += E[1(B_t + \Cov(\alpha B_t,B_t) < a)] += P(B_t < a - \alpha t)$. -$E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) 1(B_t > a/\sigma)] -= P(B_t > a/\sigma - (\mu/\sigma)t)$. +$E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) 1(B_t < a/\sigma)] += P(B_t < a/\sigma - (\mu/\sigma)t)$. -$E[\exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2) 1(B_t > a/\sigma)] -=\exp(2a\mu/\sigma) P(B_t > a/\sigma + (\mu/\sigma)t)$. +$E[\exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2) 1(B_t < a/\sigma)] +=\exp(2a\mu/\sigma) P(B_t < a/\sigma + (\mu/\sigma)t)$. $$ -E[\min\{\tau_a,T\}] = \int_0^T P(B_t > a/\sigma - (\mu/\sigma)t) - + \exp(2a\mu/\sigma) P(B_t > a/\sigma + (\mu/\sigma)t)\,dt \\ +E[\min\{\tau_a,T\}] = \int_0^T P(B_t < a/\sigma - (\mu/\sigma)t) + + \exp(2a\mu/\sigma) P(B_t < a/\sigma + (\mu/\sigma)t)\,dt \\ $$ Let $Z$ be standard normal. -$\int_0^T P(Z > a + bt)\,dt = -P(Z > a + bt)t|_0^T + \int_0^T t \phi(a + bt)b\,dt$ +$\int_0^T P(Z < a/\sqrt{t} + b\sqrt{t})\,dt$ -$u = P(Z > a + bt) = 1 - \Phi(a + bt)$, $dv = dt$; $du = -\phi(a + bt)b\,dt$, $v = t$. +$u = P(Z < a\sqrt{t} + b\sqrt{t}) = \Phi(a + bt)$, $dv = dt$; $du = \phi(a + bt)b\,dt$, $v = t$. $s = a + bt$, $t = (s - a)/b$ @@ -105,4 +116,3 @@ $$ &= ((-\exp(s^2/2) - a \Phi(s))/b|_a^{a + bT} \\ \end{aligned} $$ -