GBM擊中障礙的機率
我嘗試使用布朗橋方法來確定機率
$$ P(S_t<\beta,t\in [0,T]|S_0,S_T) $$在哪裡 $ S_t $ 是通常 Black Scholes 設置中的 GBM。我們知道對於 BM $ W_t $ , $$ P(W_t<\beta,t\in [0,T]|W_T=x)=1-\exp\bigg(-\frac{2}{T}\beta(\beta-x)\bigg) $$ 然後我嘗試了以下方法：因為$$ S_t=S_0e^{(\mu-1/2\sigma^2)t-\sigma W_t} $$我們有 $$ \begin{aligned}&P(S_t<\beta,t\in [0,T]|S_0,S_T=s)\&=P\bigg(W_t<\frac{\ln(\beta/S_0)-(\mu-1/2\sigma^2)t}{\sigma}\bigg|S_0,W_T=\frac{\ln(s/S_0)-(\mu-1/2\sigma^2)T}{\sigma}\bigg)\&=1-\exp\bigg(-\frac{2}{T}\bigg(\frac{\ln(\beta/S_0)-(\mu-1/2\sigma^2)t}{\sigma}\bigg)\bigg(\frac{\ln(\beta/S_0)-\ln(s/S_0)}{\sigma}\bigg)\bigg)\&=1-\exp\bigg(-\frac{2}{\sigma^2T}\bigg(\ln(\beta/S_0)-(\mu-1/2\sigma^2)t\bigg)\ln(\beta/s)\bigg)\end{aligned} $$ 顯然，正確答案只是$$ 1-\exp\bigg(-\frac{2}{\sigma^2T}\ln(\beta/S_0)\ln(\beta/s)\bigg) $$ 所以我相信我快到了。我錯過了什麼？
在不平等
$$ \begin{align*} W_t<\frac{\ln(\beta/S_0)-(\mu-1/2\sigma^2)t}{\sigma}, \end{align*} $$ 右手邊取決於 $ t $ ，你不能直接使用上面的結果。在這種情況下，通常採用 Girsanov 變換。 讓 $ a= \frac{\mu-1/2\sigma^2}{\sigma} $ , $ b= \frac{\ln(\beta/S_0)}{\sigma} $ ， 和 $ c= \frac{\ln(s/S_0)}{\sigma} $ . 我們定義機率測度 $ \tilde{P} $ 這樣
$$ \begin{align*} \frac{d\tilde{P}}{dP}\big|t = e^{-\frac{1}{2}a^2 t - aW_t}, \end{align*} $$ 在哪裡 $ P $ 是原始機率測度。然後 $ \tilde{W}t = W_t + at $ 是一個標準的布朗運動 $ \tilde{P} $ . 讓 $ E $ 和 $ \tilde{E} $ 是對措施的期望 $ P $ 和 $ \tilde{P} $ . 那麼，對於任何 Borel 集 $ A $ , $$ \begin{align*} &\ E\left(\pmb{1}{{W_t+at < b, t \in [0, T]}} \pmb{1}{{W_T\in A}} \right)\ =&\ \tilde{E}\left(\left(\frac{d\tilde{P}}{dP}\big|T \right)^{-1}\pmb{1}{{W_t+at < b, t \in [0, T]}} \pmb{1}{{W_T\in A}} \right)\ =&\ \tilde{E}\left(e^{\frac{1}{2}a^2 T + aW_T}\pmb{1}{{W_t+at < b, t \in [0, T]}} \pmb{1}_{{W_T\in A}} \right)\ =&\ \tilde{E}\left(e^{-\frac{1}{2}a^2 T + a\tilde{W}T}\pmb{1}{{\tilde{W}t < b, t \in [0, T]}} \pmb{1}{{\tilde{W}_T-aT\in A}} \right)\ =&\ \tilde{E}\left(e^{-\frac{1}{2}a^2 T + a\tilde{W}T}\pmb{1}{{\tilde{W}T-aT\in A}}\tilde{E}\left(\pmb{1}{{\tilde{W}_t < b, t \in [0, T]}} ,|, \tilde{W}_T \right) \right)\ =&\ \tilde{E}\left(e^{-\frac{1}{2}a^2 T + a\tilde{W}T}\pmb{1}{{\tilde{W}_T-aT\in A}}\left[1-\exp\Big(-\frac{2}{T}b(b-\tilde{W}T)\Big) \right] \right)\ =&\ E\left(\frac{d\tilde{P}}{dP}\big|Te^{-\frac{1}{2}a^2 T + a\tilde{W}T}\pmb{1}{{\tilde{W}T-aT\in A}}\left[1-\exp\Big(-\frac{2}{T}b(b-\tilde{W}T)\Big) \right] \right)\ =&\ E\left(\pmb{1}{{W_T\in A}}\left[1-\exp\Big(-\frac{2}{T}b(b-W_T-aT)\Big) \right] \right). \end{align*} $$ 那是， $$ \begin{align*} E\left(\pmb{1}{{W_t+at < b, t \in [0, T]}} ,|, W_T\right)&=1-\exp\Big(-\frac{2}{T}b(b-W_T-aT)\Big), \end{align*} $$ 或者，等效地， $$ \begin{align*} E\left(\pmb{1}{{W_t+at < b, t \in [0, T]}} ,|, W_T=x\right)&=1-\exp\Big(-\frac{2}{T}b(b-x-aT)\Big). \end{align*} $$ 所以， $$ \begin{align*} &\ P\bigg(W_t<\frac{\ln(\beta/S_0)-(\mu-1/2\sigma^2)t}{\sigma}, t \in[0, T],\Big|,S_0,W_T=\frac{\ln(s/S_0)-(\mu-1/2\sigma^2)T}{\sigma}\bigg)\ =&\ E\left(\pmb{1}{{W_t+at < b, t \in [0, T]}} ,|, W_T=c-aT\right)\ =&\ 1-\exp\Big(-\frac{2}{T}b(b-c)\Big)\ =&\ 1-\exp\Big(-\frac{2}{\sigma^2T}\ln(\beta/S_0)\ln(\beta/s)\Big). \end{align*} $$