# Tag Info

1

If we are going to have the form \begin{align*} dr = A dt + BdW_t, \end{align*} Then both A and B are functions of $t$ and $r_t$, otherwise, $r_t$ is normal. However, note that \begin{align*} r_t = \exp\Bigg(\frac{1}{\sigma(t)}\bigg(\int_0^t \theta(s)\sigma(s) ds +\sigma(0)\ln r_0 + \int_0^t\sigma^2(s) dW_s\bigg)\Bigg). \end{align*} That is, $r_t$ is ...

1

$$d(S^p) = pS^p (\alpha +\sigma dW) + \frac{1}{2}p(p-1)S^p\sigma^2 dt$$ $$= pS^p \left[ \left(\alpha +\frac{1}{2}\sigma^2(p-1)\right)dt + \sigma dW \right]$$

1

Shreve's answer is the correct one: The drift term of $\frac{dS^p}{S^p}$ has two parts: $p \left(\alpha - \frac{1}{2} \sigma^2 \right)$ from regular differentiation $\frac{1}{2} p^2 \sigma^2$ is the Ito term. When you sum them up you get $p \left(\alpha + \frac{1}{2} (p-1) \sigma^2 \right)$

2

Well, if you assume Fx is a Brownian Motion $W_t$ then $\frac{1}{X_t} = -\frac{1}{X^2_t} \bullet X_t + \frac{2}{X^3_t} \bullet \langle X\rangle_t = -\frac{1}{X^2_t} \bullet X_t + \frac{2}{X^3_t} \bullet \sigma^2 X^2_t t$. So $d\bigl(\frac{1}{X_t}\bigr) = -\frac{1}{X_t} [(\Delta r + 2\sigma^2 ) dt + \sigma dW_t ]$ Setting $\frac{1}{X_t} = M_t$, we see it ...

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