Suppose $X$ is a uniformly bounded continuous martingale with $X_0=0$. I am trying to show that $$Y_t:=\int_0^t|X_s|^p\text{sgn}(X_s)dX_s$$ is a true martingale, where $p\geq1$. As $Y$ is a local martingale, the necessary and sufficient condition for this is that, for all $t\geq0$, the family $$\mathcal{Y}_t=\{Y_T:T\text{ is a stopping time with }T\leq t\}$$ is uniformly integrable. I imagine that the way to show this is that (I suspect), $(Y_t)_{0\leq t\leq T}$ is uniformly bounded for any non-random $T\geq0$. However, I do not know how to show this – even if $X$ is uniformly bounded, could it not be "moving around" fast enough that the integral is not necessarily bounded? I feel that my understanding of when stochastic integrals are true martingales is lacking, so any advice would be very useful. Thanks!
Does a bounded martingale imply a bounded stochastic integral
probability theorystochastic-analysisstochastic-calculusstochastic-integralsstochastic-processes
Related Solutions
It's not difficult to see that
$$X_t := \exp \left(\sqrt{2} B_t \right)$$
solves the given SDE. (You can either use Itô's formula to check it or use some standard methods for linear SDE's to obtain this solution.) Moreover, by Ito's formula:
$$f(t,X_t)-\underbrace{f(0,x_0)}_{x_0} = \sqrt{2} \int_0^t e^{-s} \cdot X_s \, dB_s + \int_0^t e^{-s} \cdot X_s + (-e^{-s} \cdot X_s) \, ds \\ = \sqrt{2} \int_0^t e^{-s} \cdot e^{\sqrt{2} B_s} \,dB_s \\ \Rightarrow f(t,X_t) = \underbrace{\sqrt{2} \int_0^t e^{\sqrt{2} B_s-s} \,dB_s}_{M_t} + \underbrace{x_0}_{A_t}$$
where $x_0=1$. Let
$$g(s,w) := \sqrt{2} \cdot e^{\sqrt{2} B(s,w)-s}$$
Then $g \in L^2(\lambda_T \otimes \mathbb{P})$, i.e.
$$\int_0^T \int_\Omega g(s,w)^2 \, d\mathbb{P} \, ds <\infty$$
There is a general result which says that this condition implies that $M_t$ is a martingale (and not only a local one). Moreover,
$$\langle M,M \rangle_t = \int_0^t |g(s,w)|^2 \, ds$$
(see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 14.13).
Concerning the integral $\mathbb{E}(e^{-\tau} \cdot X_\tau)$: Remark that
$$\tau = \inf\{t \geq 0; X_t=2-t\} = \inf\{t \geq 0; \sqrt{2} B_t = \ln(2-t)\}$$
Now let
$$\sigma := 2\tau = \inf\{t \geq 0; \underbrace{\sqrt{2} B_{\frac{t}{2}}}_{=:W_t} = \ln (2-t/2)\}$$
where $(W_t)_{t \geq 0}$ is again a Brownian Motion (scaling property). Thus
$$\mathbb{E}(e^{-\tau} \cdot X_\tau) = \mathbb{E}(e^{-\tau+\sqrt{2} B_\tau}) = \mathbb{E}(e^{-\frac{\sigma}{2}+W_\sigma}) \stackrel{\ast}{=} 1$$
In $(\ast)$ we applied the exponential Wald identity (see remark).
Remark Exponential Wald identity: Let $(W_t)_{t \geq 0}$ a Brownian motion and $\sigma$ a $\mathcal{F}_t^W$-stopping time such that $\mathbb{E}e^{\sigma/2}<\infty$, then $\mathbb{E}(e^{W_\sigma-\frac{\sigma}{2}})=1$. (see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 5.14)
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.
Best Answer
I'm not quite sure that $Y_t$ is bounded, but I think you can show it's a true martingale by using the BDG inequalities.
Recall one condition that ensures $Y$ is a martingale is $\mathbb{E}[\langle Y,Y \rangle_t] < \infty$ for all $t$. Let $K$ be such that $|X_s| \le K$ for all $s$. Then we have \begin{align*} \mathbb{E}[\langle Y,Y \rangle_t] &= \mathbb{E}\left[\int_0^t |X_s|^{2p}d\langle X,X \rangle_s\right] \\ &\le K^{2p} \mathbb{E}\left[\int_0^t d\langle X,X \rangle_s\right] \\ &= K^{2p}\mathbb{E}[\langle X,X \rangle_t]. \end{align*}
Now, by the BDG inequality, we know $\mathbb{E}[\langle X,X \rangle_t] \le c\mathbb{E}\left[\sup_{s \le t}|X_s|^2\right] \le c K^2 < \infty$ for some universal $c > 0$.
Hence we conclude \begin{align*} \mathbb{E}[\langle Y,Y \rangle_t] &\le K^{2p}\mathbb{E}[\langle X,X \rangle_t] < \infty. \end{align*} and therefore $Y$ is a true martingale.