# Prove that the following process is a Geometric Brownian motion for every constant

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?).

• You mean to say, "prove that the transformation $Y_t = (X_t)^a$ is a geometric Brownian motion" for every $a$? And yes, your application of Itô's lemma is incorrect. If $M_t$ is a continuous square-integrable martingale and $N_t = f(M_t)$ for a nice $f \in C^2$, then $N_t = N_0 + \int_0^t f'(M_s) dM_s + \frac{1}{2}\int_0^t f''(M_s) d[M]_s.$
– snar
Jan 24, 2015 at 19:56
• sorry fixed a typo on the exponent in the derivation. But besides that yes I'm trying to prove that, but I still do not get it, can you please point me towards the error? Thank you Jan 24, 2015 at 20:01
• If $f(x) = x^a$, then $Y_t = f(X_t)$, and $dY_t = f'(X_t) dX_t + \frac{1}{2}f''(X_t) d[X]_t$, not $dY_t = f'(X_t) dB_t + \frac{1}{2}f''(X_t) dt.$ Are you sure the question is correct as stated?
– snar
Jan 24, 2015 at 20:05
• Do you want to prove that $Y_t$ is a geometric Brownian motion? (Obviously, $Y_t$ is not a Brownian motion!)
– saz
Jan 24, 2015 at 20:13
• Yes it is right. But is $d[X_t]=d[B_t]=d[t]$ and also the second derivative seems right ( I fixed it after you pointed the mistake). Might be that the question is wrong. The result is not a g.b.m. right? Jan 24, 2015 at 20:14

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.

• Could you please explain the reasoning behind: $\frac{dX_t}{X_t} = dB_t$ ? Thank you Jan 25, 2015 at 17:07
• Where did I claim this?
– saz
Jan 25, 2015 at 17:09
• the passage where you split the integrals, in the first one, I don't understand where does $X_t^{-1}$ ends up Jan 25, 2015 at 17:13
• Got it...the $dX_t$ got substituted in. Thanks, my bad. I'm checking these answers back and forth several times for reference due to my lack of knowledge on the subject. Jan 25, 2015 at 17:21
• @ClementeCortile You are welcome. And yes, I subsitited $dX_t$ by $X_t \, dt+ X_t \, dB_t$.
– saz
Jan 25, 2015 at 22:16