Product rule for integral notation

Question

I am dealing with a form of Ito's Lemma that uses the notation $df$, presumably for a small change in a function. I am trying to calculate a small change in a function that is a product of $f$ with an exponential, ie calculte $d(e^{rt}f)$. Is there are product rule or something similar that would allow me to calculate this, given that I know what $df$ is and $d(e^{rt})$?

Answer 1

The "$df$"-notation is just a shorthand for the corresponding integral equation since for many stochastic processes (e.g. Brownian motion) the process isn't differentiable. So Ito's lemma (in one of its most basic form) says that for a $C^2$ function $f(t,x)$ and a continuous semimartingale $X_t$ we have $$df(t,X_t) = \partial_tf(t,X_t) dt + \partial_x f(t,X_t) dX_t + \frac12 \partial_{xx}f(t,X_t) d \langle X \rangle_t$$ which is defined to mean that $$f(t,X_t) - f(0,X_0) = \int_0^t \partial_tf(s,X_s) ds + \int_0^t \partial_x f(s,X_s) dX_s + \frac12 \int_0^t \partial_{xx}f(s,X_s) d\langle X \rangle_s$$

Assuming that $df = df(X_t)$ where $X_t$ is a semimartingale we can then define $g(t,x) = e^{rt} f(x)$ and by applying Ito's lemma to $g$ we find that (in differential notation) $$dg(t,X_t) = r g(t,x) dt + e^{rt} f'(X_t) dX_t + \frac{1}{2} e^{rt} f''(X_t) d \langle X \rangle_t$$ You might notice that this is the same as $f d(e^{rt}) + e^{rt} df$. However this doesn't always happen. There is a product rule for Ito calculus which can be found by applying the appropriate form of Ito's formula to $f(x,y) = xy$ which is $$d(X_tY_t) = X_t dY_t + Y_t dX_t + d \langle X, Y \rangle_t$$ where $X$ and $Y$ are semimartingales. In your case the last term is $0$ since $Y_t = e^{rt}$ is a finite variation process so $\langle Y, f(X) \rangle_t = 0$ and, in particular, the product rule looks the same as in ordinary calculus in this case.