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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?).

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  • $\begingroup$ 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 $N_t = N_0 + \int_0^t f'(M_s) dM_s + \frac{1}{2}\int_0^t f''(M_s) d[M]_s.$ $\endgroup$
    – snar
    Jan 24, 2015 at 19:56
  • $\begingroup$ 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 $\endgroup$ Jan 24, 2015 at 20:01
  • $\begingroup$ 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? $\endgroup$
    – snar
    Jan 24, 2015 at 20:05
  • $\begingroup$ Do you want to prove that $Y_t$ is a geometric Brownian motion? (Obviously, $Y_t$ is not a Brownian motion!) $\endgroup$
    – saz
    Jan 24, 2015 at 20:13
  • $\begingroup$ 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? $\endgroup$ Jan 24, 2015 at 20:14

1 Answer 1

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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.

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  • $\begingroup$ Could you please explain the reasoning behind: $\frac{dX_t}{X_t} = dB_t$ ? Thank you $\endgroup$ Jan 25, 2015 at 17:07
  • $\begingroup$ Where did I claim this? $\endgroup$
    – saz
    Jan 25, 2015 at 17:09
  • $\begingroup$ the passage where you split the integrals, in the first one, I don't understand where does $X_t^{-1}$ ends up $\endgroup$ Jan 25, 2015 at 17:13
  • $\begingroup$ 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. $\endgroup$ Jan 25, 2015 at 17:21
  • $\begingroup$ @ClementeCortile You are welcome. And yes, I subsitited $dX_t$ by $X_t \, dt+ X_t \, dB_t$. $\endgroup$
    – saz
    Jan 25, 2015 at 22:16

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