Product Rule for Ito Processes

Question

Is it true that if $X(t)$ is an Ito process and $p(t)$ is non-stochastic, then the ordinary chain rule applies, that is, $$d(X(t)p(t)) = dX(t)p(t) + X(t)p'(t)dt?$$

Answer 1

Note

Let $X_t$ and $Y_t$ be two Ito processes. By application of Ito's lemma, we have $$d(X_t\,Y_t)=Y_t\,dX_t+X_t\,dY_t+d[X_t\,,\,Y_t]$$ Now, if $Y_t$ is of finite variation, then covariation $[X_t\,,\,Y_t]=0$. Let $p(t)$ be differentiable with continuous derivative, thus it has finite variation.In other words $$d[p(t),X(t)]=0$$

Answer 2

This can be obtained directly from Ito's product rule:

$$ d(X(t)Y(t)) = X(t)dY(t) + Y(t)dX(t) + dX(t)dY(t) $$

For illustration, assume your $dX(t)$ and $dp(t)$ has form:

$$ d(X_t) = \mu_1dt + \sigma_1dW_t \\ d(Y_t) = \mu_2dt $$

Since your $p(t)$ is non-stochastic, it only has derivative w.r.t time and thus the final term is :

$$ dX(t)dY(t) = \mu_1dt(\mu_2dt) + \sigma_1dW_t(\mu_2dt) = 0 $$ as a result of cross term and quadratic variation of time.

One other way to understand the product rule is to use the two-dimensional Ito formula and let $f(t,x,y) = xy$.