速寫 Day32

PSG打T1
前面夾很久
但兩波失誤就爆了
哀哀只能說可惜啊

$$dX(t)=adt+bdB_t$$
$$(dX(t))^2=(adt+bdB_t)^2=b^2dt$$ 是由於$dt^2=0\space (Finite\space Variation)$，且 $dtdB_t=0 \space (continuous\space process)$
另外，$(dB_t)^2=dt$ due to Quadratic Variation of a Wiener process.

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$f(x)-f(x_0)$可用泰勒展開式展開為：

$$f(x)-f(x_0)=f’(x_0)(x-x_0)+\dfrac{1}{2}f’’(x_0)(x-x_0)^2$$
因此假定上述$X(t)$唯一隨機過程，得：
$$f(X(t))-f(X(0))=f’(X(0))(X(t)-X(0))+\dfrac{1}{2}f’’(X(0))(X(t)-X(0))^2$$
又，若$\Delta t\rightarrow 0$，且$\space X(t+\Delta t)-X(t)=dX(t)$, 則：
$$df(X(t))=f’(X(t))dX(t)+\dfrac{1}{2}f’’(X(t))(dX(t))^2$$

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$$dS_t=\mu S_tdt+\sigma S_tdB_t$$
令 $f(S_t)=ln(S_t)$，則：
$df=f’(S_t)dS_t+\dfrac{1}{2}f’’(S_t)(dS_t)^2$
$=\dfrac{1}{S_t}dS_t+\dfrac{1}{2}(-\dfrac{1}{S_t^2})(S_t^2\sigma^2dt)$
$=\dfrac{1}{S_t}(\mu S_tdt+\sigma S_tdB_t)-\dfrac{1}{2}\sigma^2dt$
$$=(\mu-\dfrac{\sigma^2}{2})dt+\sigma dB_t $$
$dX_t=\mu_tdt+\sigma_tdB_t$, where $B_t$ is a Wiener process.
若$f$為$t$與$X(t)$的函數，則：
$$df(t,X(t))=\dfrac{\partial f}{\partial t}dt+\dfrac{\partial f(X(t))}{\partial X(t)}dX(t)+\dfrac{1}{2}\dfrac{\partial^2 f(X(t))}{\partial X(t)^2}(dX(t))^2$$
$$df(t,X(t))=(\dfrac{\partial f}{\partial t}+\mu_t\dfrac{\partial f(X(t))}{\partial X(t)}+\dfrac{\sigma_t^2}{2}\dfrac{\partial^2 f(X(t))}{\partial X(t)^2})dt+\sigma_t\dfrac{\partial f(X(t))}{\partial X(t)}dB_t$$

我會繼續努力的。