When is a stochastic integral a martingale?

Question

In what follows, let the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ as well as the chosen filtration $(\mathcal{F}_t)_{t \ge 0}$ be known, and let $f$ denote an arbitrary locally bounded progressively measurable process (i.e. bounded on compact intervals and for all $t$ measurable with respect to $\mathcal{B}([0,t])\otimes \mathcal{F_t}$).

Consider the process $(Y_t)_{t \ge 0}$ defined by the stochastic integral of $f$ with respect to $(X_t)_{t\ge 0}$: $$Y_t = \int_0^t f(s) \mathrm{d}X_s$$ Which of the following are true?

1. $X_t$ is a martingale $\implies$ $Y_t$ is a local martingale but not necessarily a true martingale.

2. $X_t$ is a locally $L^2-$bounded true martingale $\implies Y_t$ is a locally $L^2$-bounded true martingale.

3. $X_t$ is a local martingale $\implies$ $Y_t$ is a local martingale.

4. $X_t$ is a semimartingale $\implies$ $Y_t$ is a semimartingale.

I just want to know which results are true, and then I will supply the proofs on my own later.

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It seems like 2. only holds when we have the additional condition on $f$ that it is locally $L^2$ bounded. In fact, it seems like I may have had it mixed-up: $X_t$ and $Y_t$ need to be globally $L^2$ bounded i.e. square integrable, but $f$ only needs to be locally $L^2$ bounded i.e. square integrable on compact intervals -- do I have this right? I'm still not sure. https://fabricebaudoin.wordpress.com/2012/09/14/lecture-19-stochastic-integrals-with-respect-to-square-integrable-martingales/

EDIT: OK, fixing the assumptions in the manner mentioned above, this is almost certainly true, see Theorem 19 on p.34 of this document: http://math.bu.edu/people/prakashb/Math/stochint.pdf; this also follows from Theorem 6.3 and Lemma 6.4 on p.35 here: https://staff.fnwi.uva.nl/p.j.c.spreij/onderwijs/master/si.pdf; also pp. 137-140 in Revuz and Yor, where being locally bounded also supposedly implies being locally $L^2-$bounded (see the last paragraph of p.140 and consider that $\langle B, B \rangle=t$ and that Brownian motion is a continuous local martingale. So seemingly the condition of being locally bounded is unnecessarily strong in many places where it is used. I am not sure if it can be replaced by locally $L^2-$bounded in all places.

Actually a closer reading of Definition 2.1 on p. 137 of Revuz and Yor implies that 2. holds only for square-integrable integrands $f$, not just locally $L^2-$bounded. If we relax the condition to locally $L^2-$bounded, then only Proposition 2.7 on p.140 is applicable (since martingales are local martingales after all), and we can only conclude that the stochastic integral is a local martingale (although the example given between the end of p.139 and the beginning of p.140 about Brownian motion suggests otherwise -- perhaps if both the integrator and the integrand are locally $L^2$-bounded martingales then we get a martingale, but not square-integrable, as the result?

Also it is worth noting that a corollary of the Martingale representation theorem and the associativity of the stochastic integral (when applicable) allows us to phrase most of the conditions given in terms of integration w.r.t. $\langle M, M \rangle_s$ as conditions in terms of $s$, i.e. $L^2$-boundedness, because the quadratic variation of Brownian motion is $t$.

Answer 1

Well, in the form stated above, none of the statements are true, because you're only assuming $f$ to be progressive and not predictable, and you're not assuming that the integrator $X$ has continuous sample paths.

I'd say that point (4) is neither true nor false but undefined, as the stochastic integral is not necessarily well-defined for integrands which are only progressive and not predictable.

As regards the other three points, as a counterexample, take e.g. $N$ to be a standard Poisson process and let $f(t) = N_t$, $X_t = N_t - t$ and let $(\mathcal{F}_t)$ be the filtration induced by $N$. Then $f$ is locally bounded (by e.g. the sequence of stopping times corresponding to the jump times of $N$), bounded on compacts (because it has cadlag sample paths) and is progressive (because it is cadlag and adapted). Furthermore, the integral is well-defined since $X$ has sample paths of finite variation, so the integral can be defined as a pathwise Lebesgue integral. It holds that $$ Y_t = \int_0^t f(s) dX_s = \int_0^t (N_{s-} + \Delta N_s) dX_s \\ = \int_0^t N_{s-}dX_s + \sum_{0<s\le t}(\Delta N_s)^2 = \int_0^t N_{s-}dX_s + N_t. $$

This functions as a counterexample for points (1-3) because even though $X$ is a locally $L^2$-bounded martingale, $Y$ is not even a local martingale. The problem is that $f$ is not predictable. See also this question for more on this.

If $f$ was assumed predictable, the answers would be:

(1): True. Intuitively, this is because the integral of a predictable process with respect to a local martingale is a martingale, and if $f$ is sufficiently rough, the integral process will not yield integrability, and so a true martingale cannot be expected.

(2): True. Intuitively, this is because the integral process is a local martingale, and by localising so that $f$ is bounded and $X$ is $L^2$-bounded, one obtains $L^2$ boundedness of the integral process.

(3): True. This is almost a defining property of the stochastic integral (depending on the method of construction), but certainly true in any case.

(4): True, also almost by construction, depending on the method of construction.