Let's assume that we have the following stochastic differential equation:

$dX_t = \mu X_t dt + \sigma X_tdW_t$

and that we have to prove that this is its solution:

$X_t = X_0 \exp\left(\left(\mu -{\frac {\sigma ^{2}}{2}}\right)t+\sigma W_t\right)$

I can solve this thanks to the application of Ito's lemma on this deterministic regular function defined in two variables (time and process),

$f(t, X_t) = \ln(X_t)$

The point is that I can't understand why the partial derivative of that function with respect to t is equal to zero. Why do we use a function defined in t if it is not differentiable in t?


1 Answer 1


$f$ as you've described it is differentiable in $t$ -- the derivative is just equal to zero.

The larger point is that Ito's lemma applies to a broader class of functions $f(t, X_t)$ -- the one we've chosen happens to have $\frac{df}{dt} = 0$ everywhere.

  • $\begingroup$ Yes. So for example if we choose $f(t,X_t)=ln(X_t)+7t$ then there would be a non-zero derivative with respect to t. But the f you have chosen does not have this explicit dependence on t. The subscript on X does not count. If I rewrite your f function as $f(arg1,arg2)= ln(arg2)$ it is clear there is no dependence on arg1, which you call t, it is what is called in programming an unused input to a function. $\endgroup$
    – nbbo2
    Oct 20, 2023 at 17:40

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