Decomposition of a martingale into a semimartingale The definition of semimartingale follows from Wikipedia.
I want to find a semimartingale representation for the process$$ F_t=\int_0^t f(t-s)dB_s,$$where $f$ is a continuous bounded variation function and $B_s$ the standard Brownian motion.
I tried using integration by parts and found$$ F_t=f(0)B_t-\int_0^tB_s\dfrac{\partial f(t-s)}{\partial s}ds,$$which gives $$dF_t=f(0)B_t+B_tf'(0)dt, $$but $B_tf'(0)$ is obviously not of bounded variation. Can someone enlighten me?
 A: Assuming $f$ is an L2 converging deterministic integrand function. To transform $F$ into a semimartingale, one could apply the Girsanov theorem and change the probability measure of your original process $F$.
To apply the Girsanov theorem, you need to define an adapted process $\Theta(t)$ such that, defining the Radon-Nikodym derivative
$$
Z_t = \exp{\{-\int_0^t{\Theta(s)dB_s}-\frac{1}{2}\int_0^t{\Theta(s)^2ds}\}},
$$
$$
\Rightarrow\mathbb{E}\int_0^t{\Theta(s)^2Z_t^2ds}<\infty.
$$
Then, defining
$$
\tilde{B}_t=B_t+\int_0^t{\Theta(s)ds},
$$
$\tilde{B_t}$ is a Brownian motion.
Going back to your process,
$$
F_t=\int_0^t f(t-s)dB_s
$$
assuming that $f(t-s)$ is a deterministic function of time, integrable, and $\int_0^t f(t-s)^2dt<\infty$, I suggest to write
$$
F_t=\int_0^s f(t-u)dB_u+\int_s^t f(t-u)dB_u.
$$
Then, fixing a terminal $t$ for $f$, one could write the following new function:
$$
g(s|t)=g(s)=\int_0^s f(t-u)dB_u.
$$
Because of the assumed properties of $f$, in particular being of bounded variation, square integrable deterministic function of $u$, given $t$, $g(s)$ is an Ito integral and a martingale. For this reason, it is possible to write it in differential form as:
$$
dg(u)=f(t-u)d\tilde{B}_u.
$$
But from the Girsanov theorem expressed in differential form we have
$$
dB_u=d\tilde{B}_u-\Theta(s)du,
$$
so that
$$
dg(u)=f(t-u)dB_u=f(t-u)[d\tilde{B}_u-\Theta(u)du],
$$
which returns the semimartingale under the new probability measure as
$$
dg(u)=-\Theta(u)f(t-u)du + f(t-u)d\tilde{B}_u,
$$
which has an integral representation
$$
g(s)=g(0)-\int_0^s \Theta(u)f(t-u)du+\int_0^s f(t-u)dB_u.
$$
Thus, since, as $s\rightarrow t$, $g(s)=g(s|t)\rightarrow F_t$, one could write a semimartingale representation for $F$ as
$$
F_t=g(0)-\int_0^t \Theta(s)f(t-s)ds+\int_0^t f(t-s)dB_s,
$$
or
$$
F_t=F_0-\int_0^t \Theta(s)f(t-s)ds+\int_0^t f(t-s)dB_s.
$$
