$M = (M_t)_{t\geq 0}$ obtained by Itô's formula is a martingale

Question

Here I defined a non-negative stochastic process. Now, taking $F(t,x) = tx^2$, I would like to find that the continuous local martingale ``part'' obtained by Itô's formula is indeed a Martingale. To do so, I obtained, by applying Itô's formula, and $$ d X_t = 3dt + 2\sqrt{X_t}d B_t, \quad\quad d\langle X,X\rangle_t = 4X_t\langle B,B\rangle_t, $$ the following formula: $$ F(t,X_t) = \int_0^2 [X_s^2 + 10s X_s ] ds + 4\int_0^t sX_s \sqrt{X_s}d B_s. $$

Hence, I defined $M = (M_t)_{t\geq 0}$ by $$ \int_0^t sX_s \sqrt{X_s}d B_s. $$ To see that it is a martingale, it is sufficient to see that $E\sqrt{\langle M,M\rangle_s}< \infty$, but, by (concave function version) Jensen's inequality, $E\sqrt{\langle M,M\rangle_s}\leq \sqrt{E\langle M,M\rangle_s}$, and then it is enough to see that $E\langle M,M\rangle_s<\infty$. Hence, $$ E\langle M,M\rangle_s = E\int_0^t s^2 X_s^3 ds. $$ Since $X_s\geq 0$, we can apply Fubini's theorem, and hence $$ E\int_0^t s^2 X_s^3 d s = \int_0^t s^2 EX_s^3 ds. $$ My question is: How can I asssure that $EX_s^3 <\infty$? Is there any other way of doing it? Thank you in advance and I hope you find it interesting.

Answer 1

There are general statements on the existence of moments of solutions to SDEs. If you are lucky, you have one of those results at your disposal to deduce that $\sup_{s \leq t} \mathbb{E}(|X_s|^3)<\infty$ for all $t>0$. If this is not the case, then you need to do some further calculations.

In the following, I will consider $\mathbb{E}(X_t^4)$ (rather than $\mathbb{E}(|X_t|^3)$) since this saves us the trouble to deal with the modulus. By Itô's formula, we have

$$X_t^4 = 8 \int_0^t X_s^3 \sqrt{X_s} \, dB_s +30 \int_0^t X_s^3 \, ds.$$

Now we want to take the expectation, but since we do not yet know whether the expectation is finite, we first need to some stopping. To this end, define

$$\tau_r := \inf\{t \geq 0; |X_t| \geq r\},$$

note that $\tau_r \uparrow \infty$ as $r \to \infty$ and $|X_{t \wedge \tau_r}| \leq r$. Since

$$t \mapsto \int_0^{t \wedge \tau_r} X_s \sqrt{X_s} \, dB_s$$

is a martingale (because of the stopping!), and, hence, has expectatio zero, we get

$$\mathbb{E}(X_{t \wedge \tau_r}^4) = 30 \mathbb{E} \left( \int_0^{t \wedge \tau_r} X_s^3 \, ds \right).$$

Using the elementary estimate

$$x^3 \leq |x|^3 \leq 1+x^4, \qquad x \in \mathbb{R},$$

for $x=X_s$, we find that

$$\mathbb{E}(X_{t \wedge \tau_r}^4) \leq 30 \mathbb{E} \left( \int_0^{t \wedge \tau_r} (1+X_s^4) \, ds \right) \leq 30 \int_0^t \mathbb{E}(1+X_{s \wedge \tau_r}^4) \, ds.$$

If we define $u(t) := \mathbb{E}(1+X_{t \wedge \tau_r}^4)$ for fixed $r>0$, then

$$u(t) \leq 1+ 30 \int_0^t u(s) \, ds$$

and so Gronwall's lemma yields

$$u(t) \leq c e^{Mt}$$

for suitable constants $c,M>0$, which do not depend on $r>0$. Consequently, we have shown that

$$\mathbb{E}(1+X_{t \wedge \tau_r}^4) \leq c e^{Mt}, \qquad t \geq 0.$$

By Fatou's lemma, this implies

$$\mathbb{E}(1+X_{t}^4) \leq c e^{Mt}, \qquad t \geq 0.$$

In particular,

$$\sup_{t \leq T} \mathbb{E}(X_t^4) < \infty, \qquad T>0,$$

which also yields

$$\sup_{t \leq T} \mathbb{E}(X_t^3) < \infty, \qquad T>0.$$