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

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

• @Stat, you cannot take the derivative of a functional of Brownian Motion, not without care at least. – Cam.Davidson.Pilon Dec 20 '13 at 3:09
• @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? – Cam.Davidson.Pilon Dec 20 '13 at 3:12
• OP: Why not mentioning the accepted full answer here math.stackexchange.com/q/613823? – Did Dec 23 '13 at 1:40

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.