1
$\begingroup$

Question We need to find dZ(t). I know I have to use Ito's formula. But I am confused because in the Ito's formula we have f(y,t) is a twice differentiable function with two variables

But here Z(t) = 1/(2+x(t)), which just has one variable?

So, I am not sure how to proceed. Any tips will be appreciated!

| improve this question | |
$\endgroup$
1
$\begingroup$

It looks like $Z(t)$ is dependent on $t$ through $X_t$ and not directly on $t$, loosely speaking. Assume that $Z_t = f(X_t)$ and use Ito's formula with just one variable.

$$dZ_t = \frac{df}{dX} dX_t + \frac{1}{2} \frac{d^2f}{dX^2} d[X_t,X_t] $$

Can you proceed and finish it?

| improve this answer | |
$\endgroup$
  • $\begingroup$ where did you get 1/2? if we take the beta value from dX(t), we will get (9t^2)/2. Or am I doing it wrong? $\endgroup$ – user13072992 Mar 27 at 20:36
  • $\begingroup$ $[X_t, X_t]$ is the quadratic variation of the Ito process $X_t$. By definition it is $[X_t, X_t] = \int_0^t \beta(s, X_s)^2 ds$, so $\frac{9t^2X_t^2}{2}dt $ is correct. The $\frac{1}{2}$ you get, because you use Taylor expansion to derive the Ito formula. $\endgroup$ – SmurfAcco Mar 28 at 11:06
  • $\begingroup$ The 1/2 comes before second derivative of function with respect to state variable $X_t$ due to Taylor expansion, as @SmurfAcco rightly said. The quadratic variation of $X_t$, $d[X_t,X_t]$ is given by $9t^2X_t^2$. It looks like you have missed $X_t^2$ term. $\endgroup$ – ForumWhiner Mar 28 at 18:21

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.