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Briefly I am on a calculus course and right now we are learning Brownian Motion, its properties and proofs. And I have a question such as below in my study set that I cant find a solution. I would apreciate any help.

Show that,

$d(W(t)^n) = \frac{n(n-1)}{2}W(t)^{(n-2)}dt + nW(t)^{(n-1)}dW(t)$

Uskebasi
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skatip
  • 23

1 Answers1

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Let $f(x,t) = x^n$, for $n \geq 2$ $$\frac{\partial f}{\partial t} = 0,\quad \frac{\partial f}{\partial x} = nx^{n-1},\quad \frac{\partial^2 f}{\partial x^2} = n(n-1)x^{n-2}$$

Let also $X_t = W_t$, and apply Ito's lemma to $f(X_t)$ $$df(X_t) = d(W_t^n)$$

Can you take it from here ?

V. Traboulistopopoulos
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  • Then it should be; as nW^{n-1}d(Wt) + dt am I right? of course sigma is 1 and mean is 0 here ? – skatip May 03 '17 at 09:27
  • With my notation, $dX_t = 0\cdot dt + 1 \cdot dW_t$, so yes, but be sure to apply thorougsly Ito's lemma: $df(X_t) = \frac{\partial f}{\partial t}(X_t)dt + \frac{\partial f}{\partial x}(X_t)dX_t + \frac12 \frac{\partial^2 f}{\partial x^2}(X_t)\sigma^2_tdt$ – V. Traboulistopopoulos May 03 '17 at 10:45
  • Thanks & sorry for the confusion derived from my notation. – skatip May 03 '17 at 10:57