3
$\begingroup$

let $B(t)$ is brownian motion.
Find $d(M(t))^2$,where $M(t)=e^{B(t)-\frac{t}{2}}$,

$\endgroup$
3
  • $\begingroup$ @Stat, you cannot take the derivative of a functional of Brownian Motion, not without care at least. $\endgroup$ Dec 20, 2013 at 3:09
  • 1
    $\begingroup$ @peter, this is likely a homework question, can you show us what you have done? You'll need Ito's Lemma for this (hints: $\mu_t = 0$ and $\sigma_t=1$ since the underlying process is Brownian motion). Can you compute $d(M(t))$? What happens when you square that term? $\endgroup$ Dec 20, 2013 at 3:12
  • $\begingroup$ OP: Why not mentioning the accepted full answer here math.stackexchange.com/q/613823? $\endgroup$
    – Did
    Dec 23, 2013 at 1:40

1 Answer 1

2
$\begingroup$

The accepted answer at Math.se is wrong. The question is asking for $dM(t)^2 = dM(t)dM(t)$, the (differential of) quadratic variation of $M(t)$, whereas the answer at Math.se explains how to calculate $d(M(t)^2)$. To find $dM(t)dM(t)$, first you find $dM(t)$ using Ito's formula:

$$dM(t) = \left(\frac{\partial f}{\partial x} + \frac{1}{2} \frac{\partial^2 f}{\partial y^2}\right)\rvert_{t, B(t)} dt + \frac{\partial f}{\partial y}\rvert_{t, B(t)} dB(t)$$ where $f(x,y) = e^{y-x/2}$. This gives $dM(t) = M(t) dB(t)$. Now use the "box calculus" formula for the quadratic variation of an Ito process: if $$dX(t) = \mu(t) dt + \sigma(t) dB(t)$$ then $$dX(t)dX(t) = \sigma(t)^2 dt$$ to get the answer. For more details and derivations of these formulas, see these Columbia notes or these UPenn notes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.