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!


1 Answer 1


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?

  • $\begingroup$ 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? $\endgroup$ Mar 27, 2020 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, 2020 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. $\endgroup$ Mar 28, 2020 at 18:21

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