If we have an adapted function $f(t)$ such that $\int_0^t f(s)ds\,<\infty$, then the Girsanov exponent can be defined:
$$ Z(t):=\exp\left( \int_0^t f(s)dW(s) - \frac{1}{2} \int_0^t f^2(s)ds\right), t \in [0,T] $$
We require conditions for $Z(t)$ to be a true martingale (and Girsanov measure transformation to be applicable). The Novikov condition is sufficient and formulated as
$$ E\left[\exp\left( \frac{1}{2} \int_0^t f^2(s)ds\right)\right]\,<\infty $$
Note that there is an alternative condition specified in Shreve "Stochastic Calculus for Finance II", Theorem 5.4.1, p. 224. I've been taught that the Novikov condition is sufficient though.
Let us consider the following 4 cases:
- $f(t)$ is a constant.
- $f(t)$ is a continuos function of time.
- $f(t)$ is a predicable random process.
- $f(t)$ is a random process.
Am I correct in thinking that the Novikov condition must be met only for the last 2 cases? Sure, we can use it for deterministic functions too, but that's not necessary (?). Instead, we could get away with $\int_0^T f^2(s)ds\,<\infty$ or potentially $\int_0^\infty f^2(s)ds\,<\infty$ (which one?) However at least one source on the web that I found does actually apply Novikov's condition on a deterministic function: http://galton.uchicago.edu/~lalley/Courses/390/Lecture10.pdf . Furthermore, it seems obvious that there are no conditions at all if $f(t)$ is a constant.
I would appreciate if someone could help me make sense out of this.
Thanks!