几何布朗运动：求 dX(t)

对 $X(t)=f(t,W(t))$ 用伊藤公式。

可得

$$\frac{\partial f}{\partial t}=\left(\mu-\frac{\sigma^2}{2}\right)X(t),\quad \frac{\partial f}{\partial W}=\sigma X(t),\quad \frac{\partial^2 f}{\partial W^2}=\sigma^2 X(t).$$

代入伊藤公式：

$$dX(t)=\left(\frac{\partial f}{\partial t}+\frac12\frac{\partial^2 f}{\partial W^2}\right)dt+\frac{\partial f}{\partial W}dW(t) =\mu X(t)dt+\sigma X(t)dW(t).$$

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Original Explanation

To apply Itô’s Lemma to $X(t)$, we treat $X$ as a function of $t$ and $W(t)$, i.e., $X(t) = f(t, W(t))$.

Let:
$$f(t, W(t)) = e^{\sigma W(t) + \left(\mu - \frac{\sigma^2}{2}\right)t}$$

First, compute the partial derivatives:

• Time derivative: $$\frac{\partial f}{\partial t} = \left( \mu - \frac{\sigma^2}{2} \right) e^{\sigma W(t) + \left(\mu - \frac{\sigma^2}{2}\right)t}$$

• First derivative with respect to $W$: $$\frac{\partial f}{\partial W} = \sigma e^{\sigma W(t) + \left(\mu - \frac{\sigma^2}{2}\right)t}$$

• Second derivative with respect to $W$: $$\frac{\partial^2 f}{\partial W^2} = \sigma^2 e^{\sigma W(t) + \left(\mu - \frac{\sigma^2}{2}\right)t}$$

Now, plug into Itô’s Lemma:

$$\begin{aligned} dX(t) &= \left( \frac{\partial f}{\partial t} + \frac{1}{2} \frac{\partial^2 f}{\partial W^2} \right) dt + \frac{\partial f}{\partial W} dW \\ &= \left( \left(\mu - \frac{\sigma^2}{2}\right) + \frac{1}{2} \sigma^2 \right) X(t) dt + \sigma X(t) dW(t) \\ &= \mu X(t) dt + \sigma X(t) dW(t) \end{aligned}$$

Final Answer:
$$dX(t) = \mu X(t) \, dt + \sigma X(t) \, dW(t)$$