Having some trouble understanding this problem:
Given the dynamics of the geometric brownian motion $X_t$ where $(B_t)_{t\in\mathbf{R}_{+}}$
$$ dX_t = X_tdt+X_t dB_t,$$
$$X_0=1$$
for which value of $a$ the transformation $Y_t=(X_t)^a$ is a brownian motion
In my try I simply applied the Ito's lemma (possibly in a wrong way) to find:
$$ dY_t=d(X_t)^a= aX_t^adB_t+ \frac{1}{2}a^2X_t^a dt$$
$$ \frac {d(X_t)^a}{X_t^a}= adB_t+ \frac{1}{2}a^2dt$$
Which is not a geometric brownian motion anymore since the parameters are not linear (or am I wrong stating this?).
Best Answer
Let $f(x) := x^a$ for some fixed $a>0$. Then
$$f'(x) = a x^{a-1} \qquad f''(x) = a (a-1) x^{a-2}.$$
Since by Itô's formula
$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s+ \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s$$
we get
$$\begin{align*} Y_t - Y_0 &= f(X_t)-f(X_0) \\ &= a \int_0^t X_s^{a-1} \,d X_s + \frac{1}{2} a (a-1) \int_0^t X_s^{a-2} \, (X_s^2 \, ds) \\ &= a \int_0^t X_s^a \, dB_s + a \int_0^t X_s^a \, ds + \frac{1}{2} a (a-1) \int_0^t X_s^a \, ds \\ &= a \int_0^t Y_s \, dB_s + \left( a + \frac{1}{2} a (a-1) \right) \int_0^t Y_s \, ds. \end{align*}$$
This means that $(Y_t)_{t \geq 0}$ solves the SDE
$$dY_t = \mu Y_t \, dt+ \sigma Y_t \, dB_t$$
with $\mu := \left( a + \frac{1}{2} a (a-1) \right)$ and $\sigma :=a$. Consequently, $(Y_t)_{t \geq 0}$ is a geometric Brownian motion.