0
$\begingroup$

I am trying to calculate $d(X_t^2)$ for Geometric Brownian Motion. I know that for GBM we have $dX_t=\mu X_tdt + \sigma X_tdW_t$, where $W_t$ is a Wiener process.

I am trying to work towards the solution which is: $2X_tdX_t+(dX_t)^2 = 2\mu X_t^2dt + 2\sigma X_t^2dW_t + \sigma^2 X_t^2dt$, but I am unsure how to get there.

I know that when we apply stochastic calculus to GBM we have: $dS_t=S_t[(\mu+\sigma^2/2)dt +\sigma dW_t]$, but when I use this in combination with what we have above and using Ito's Lemma, I'm not getting the required answer.

Any help would be great!

$\endgroup$

1 Answer 1

1
$\begingroup$

Recall that Ito's lemma tells you that if $f$ is a $C^2$ function then $$df(X_t) = f'(X_t) dX_t + \frac12 f''(X_t) d \langle X \rangle_t$$ where $d\langle X \rangle_t$ is the object you denote with the formal notation $(dX_t)^2$ ($\langle X \rangle_t$ is the quadratic variation of $X$). Applying this with $f(x) = x^2$, you get that $$dX_t^2 = 2X_t dX_t + d \langle X \rangle_t$$ Now $dX_t = \mu X_t dt + \sigma X_t dW_t$ and from this we also get that $d \langle X \rangle_t = \sigma^2 X_t^2 dt$. Substituting these in to the equation above gives the solution.

$\endgroup$
1
  • $\begingroup$ very much appreciated, thank you! $\endgroup$
    – ho_howdy
    May 9, 2020 at 23:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .