通過 Fokker-Planck 獲得幾何 BM 的密度
嘗試使用 Fokker Planck 方程長期推導 GBM 的密度（我們知道這是對數正態）。無法弄清楚我哪裡出錯了 - 會欣賞幾組額外的眼睛！
大紫荊勳賢： $ dS_t = \mu S_t dt + \sigma S_t dW_t $
讓 $ p(t,T,x,y) = \mathbb{P}(S(T) = y|S(t) = x) $ 是過渡密度 $ S_t $ , 我們知道
$$ \begin{align} p(0,T,x,y) = \frac{1}{y\sigma \sqrt{2 \pi T}} \exp\left(-\frac{1}{2}\left(\frac{\log(y) - \left(\log(x) + (\mu - \frac{\sigma^2}{2})T\right)}{\sigma \sqrt{T}}\right)^2 \right). \end{align} $$ 我想使用以下事實得出這個結果 $ p $ （視為 $ p(T,y) $ ) 滿足 Fokker Planck PDE
$$ \begin{align*} \frac{\partial p}{\partial T} & = -\frac{\partial}{\partial y}(\mu yp) + \frac{\partial^2}{\partial y^2}\left(\frac{1}{2}\sigma^2 y^2 p\right) \ & = -\mu \left(p + y\frac{\partial p}{\partial y}\right) + \frac{1}{2}\sigma^2 \left(2p + 4y\frac{\partial p}{\partial y} + y^2 \frac{\partial^2 p}{\partial y^2}\right) \ & = \frac{\sigma^2 y^2}{2} \frac{\partial^2 p}{\partial y^2} + (2 \sigma^2 - \mu)y \frac{\partial p}{\partial y} + (\sigma^2 - \mu)p. \end{align*} $$ 為了解決這個偏微分方程，我首先使用變數的變化將其轉換為常數係數問題 $ w = \log y $ ， 所以 $ \frac{\partial p}{\partial y} = \frac{\partial p}{\partial w}\frac{1}{y} $ 和 $ \frac{\partial^2 p}{\partial y^2} = \frac{\partial^2 p}{\partial w^2}\frac{1}{y^2} - \frac{\partial p}{\partial w} \frac{1}{y^2} $ , 給
$$ \begin{align*} \frac{\partial p}{\partial T} & = \frac{\sigma^2}{2} \left(\frac{\partial^2 p}{\partial w^2} - \frac{\partial p}{\partial w}\right) + (2 \sigma^2 - \mu) \frac{\partial p}{\partial w}+ (\sigma^2 - \mu)p \ & = \frac{\sigma^2}{2} \frac{\partial^2 p}{\partial w^2} + \left(\frac{3\sigma^2}{2} - \mu\right) \frac{\partial p}{\partial w} + (\sigma^2 - \mu)p. \end{align*} $$ 採用傅里葉變換將其轉換為 ODE：
$$ \begin{align*} \frac{\partial \hat{p}}{\partial T} & = -\frac{\sigma^2 \omega^2}{2} \hat{p} + i\omega\left(\frac{3\sigma^2}{2} - \mu\right) \hat{p} + (\sigma^2 - \mu)\hat{p} \ \hat{p} & = \hat{p}0\exp\left((\sigma^2 - \mu)T\right)\exp\left(-\frac{\sigma^2 \omega^2}{2}T + i\omega\left(\frac{3\sigma^2}{2} - \mu\right)T \right) \end{align*} $$ 我正在使用的轉換在哪裡 $$ \mathcal{F}f = \int{-\infty}^\infty e^{-i \omega x} f(x) dx. $$ 自從 $$ \mathcal{F}\left\frac{1}{s \sqrt{2\pi}} \exp\left(\frac{1}{2}\left(\frac{w - m}{s}\right)^2\right)\right = \exp(-i\omega m - \omega^2s^2/2), $$ 和 $ m = -\left(\frac{3\sigma^2}{2} - \mu\right)T $ 和 $ s = \sigma \sqrt{T} $ 我們看到 $$ \begin{align*} \hat{p} & = \hat{p}_0\exp\left((\sigma^2 - \mu)T\right)\mathcal{F}\left[\frac{1}{\sigma \sqrt{2\pi T}}\exp\left(-\frac{1}{2}\left(\frac{w -\left(\mu - \frac{3\sigma^2}{2}\right)T}{\sigma \sqrt{T}}\right)^2 \right)\right]. \end{align*} $$ 現在，自從 $ \mathcal{F}[f \ast g] = \mathcal{F}[f] \mathcal{F}[g] $ 和 $ p_0 = p(0,w) = \delta(w - w_0) $ , $$ \begin{align*} p(T,w) & = \exp\left((\sigma^2 - \mu)T\right)\frac{1}{\sigma \sqrt{2\pi T}}\exp\left(-\frac{1}{2}\left(\frac{w - w_0 -\left(\mu - \frac{3\sigma^2}{2}\right)T}{\sigma \sqrt{T}}\right)^2 \right). \end{align*} $$ 終於換回來了 $ w \to \log y $ 我們得到了假設的解決方案 $$ \begin{align*} p(T,y) & = \exp\left((\sigma^2 - \mu)T\right)\frac{1}{\sigma \sqrt{2\pi T}}\exp\left(-\frac{1}{2}\left(\frac{\log(y) - \left(\log(y_0) + \left(\mu - \frac{3\sigma^2}{2}\right)T\right)}{\sigma \sqrt{T}}\right)^2 \right), \end{align*} $$ 這不是上面的對數正態密度。
嗨，bcf：這是一個很好的問題。正如你在下面指出的，
$$ \begin{align*} p_0 &= \delta(y-y_0)\ &=\delta(e^w-y_0). \end{align*} $$ 然後， $$ \begin{align*} p_0 * g &= \int_{-\infty}^{\infty}\delta(e^z-y_0) g(w-z) dz\ &=\int_{0}^{\infty}\delta(u-y_0) g(w-\ln u) \frac{1}{u}du\ &= \frac{1}{y_0}g(w-\ln y_0). \end{align*} $$ 因此，你的最後一個平等變成 $$ \begin{align*} p(T,y) &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{\ln y - \big[\ln y_0 + \big(\mu - \frac{3\sigma^2}{2}\big)T\big]}{\sigma \sqrt{T}}\bigg)^2 \bigg) \end{align*} $$ 讓 $ Z= \ln y - \big[\ln y_0 + \big(\mu - \frac{\sigma^2}{2}\big)T\big] $ . 然後， $$ \begin{align*} p(T,y) &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{\ln y - \big[\ln y_0 + \big(\mu - \frac{3\sigma^2}{2}\big)T\big]}{\sigma \sqrt{T}}\bigg)^2 \bigg)\ &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{\ln y - \big[\ln y_0 + \big(\mu - \frac{\sigma^2}{2}\big)T\big]}{\sigma \sqrt{T}} + \sigma \sqrt{T}\bigg)^2 \bigg)\ &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{Z}{\sigma \sqrt{T}} + \sigma \sqrt{T}\bigg)^2 \bigg)\ &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{Z}{\sigma \sqrt{T}}\bigg)^2 - Z - \frac{\sigma ^2}{2}T \bigg)\ &= \frac{1}{y_0}\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{Z}{\sigma \sqrt{T}}\bigg)^2 + \ln y_0 -\ln y \bigg)\ &= \frac{1}{y \sigma \sqrt{2\pi T}} \exp\bigg(-\frac{1}{2}\bigg(\frac{\ln y - \big[\ln y_0 + \big(\mu - \frac{\sigma^2}{2}\big)T\big]}{\sigma \sqrt{T}}\bigg)^2 \bigg). \end{align*} $$ 這是您在上面提供的過渡密度。