## Lecture 24. Burkholder-Davis-Gundy inequalities

In this section, we study some of the most important martingale inequalities: The BurkholderDavisGundy inequalities. Interestingly, the range of application of these inequalities is very large and they play an important role in harmonic analysis and the study of singular integrals (see for instance the nice survey by my colleague Pr. Bañuelos). These inequalities admit several proofs. We present here a proof using Itō’s formula and an interesting domination inequality which is due to Lenglart. For an alternative proof, you may refer to the original approach by Burkholder-Davis-Gundy.

We admit without proof, the following domination inequality which is is due to Lenglart.

Proposition.(Lenglart) Let $(N_t)_{t \ge 0}$ be a positive adapted right-continuous process and $(A_t)_{t \ge 0}$ be an increasing process. Assume that for every bounded stopping time $\tau$, $\mathbb{E} (N_\tau \mid \mathcal{F}_0 ) \le \mathbb{E} (A_\tau \mid \mathcal{F}_0 )$. Then, for every $k \in (0,1)$, $\mathbb{E} \left( \left(\sup_{0 \le t \le T} N_t \right)^k \right) \le \frac{2-k}{1-k} \mathbb{E} \left( A_T^k\right).$

We shall use this lemma to prove the following

Theorem. (Burkholder-Davis-Gundy inequalities) Let $T > 0$ and $(M_t)_{ 0 \le t \le T}$ be a continuous local martingale such that $M_0=0$. For every $0 < p < \infty$, there exist universal constants $c_p$ and $C_p$, independent of $T$ and $(M_t)_{ 0 \le t \le T}$ such that
$c_p \mathbb{E} \left( \langle M\rangle_T^{\frac{p}{2} } \right)\le \mathbb{E}\left(\left(\sup_{0 \le t \le T} |M_t|\right)^p \right) \le C_p \mathbb{E} \left( \langle M\rangle_T^{\frac{p}{2} } \right).$

Proof. By stopping it is enough to prove the result for bounded $M$. Let $q \ge 2$. From Itō’s formula we have
$d |M_t|^q =q|M_t|^{q-1} \mathbf{sgn}(M_t) dM_t +\frac{1}{2} q(q-1) | M_t |^{q-2} d \langle M \rangle _t$
$=q \mathbf{sgn}(M_t)|M_t|^{q-1}dM_t+ \frac{1}{2} q(q-1) | M_t |^{q-2}d\langle M \rangle_t.$
As a consequence of the Doob’s stopping theorem, we get that for every bounded stopping time $\tau$,
$\mathbb{E} \left( |M_\tau|^q \mid \mathcal{F}_0 \right) \le \frac{1}{2} q(q-1) \mathbb{E} \left( \int_0^\tau | M_t |^{q-2} d\langle M \rangle_t \mid \mathcal{F}_0 \right).$
From the Lenglart’s domination inequality, we deduce then that for every $k \in (0,1)$,
$\mathbb{E} \left( \left(\sup_{0 \le t \le T} |M_t|^q \right)^k \right) \le \frac{2-k}{1-k} \left( \frac{1}{2} q(q-1)\right)^k \mathbb{E} \left(\left( \int_0^T | M_t |^{q-2} d\langle M \rangle_t\right)^k \right).$
We now bound
$\mathbb{E} \left(\left( \int_0^T | M_t |^{q-2} d \langle M \rangle _t\right)^k \right)$
$\le \mathbb{E} \left(\left(\sup_{0 \le t \le T} |M_t| \right)^{k(q-2)}\left( \int_0^T d\langle M \rangle _t\right)^k \right)$
$\le \mathbb{E} \left(\left(\sup_{0 \le t \le T} |M_t| \right)^{kq} \right)^{1-\frac{2}{q}} \mathbb{E} \left( \langle M \rangle _T^{\frac{kq}{2}} \right)^{\frac{2}{q}}.$
As a consequence, we obtain:
$\mathbb{E} \left( \left(\sup_{0 \le t \le T} |M_t|^q \right)^k \right) \le \frac{2-k}{1-k} \left( \frac{1}{2} q(q-1)\right)^k \mathbb{E} \left(\left(\sup_{0 \le t \le T} |M_t| \right)^{kq} \right)^{1-\frac{2}{q}} \mathbb{E} \left(\langle M \rangle_T^{\frac{kq}{2}} \right)^{\frac{2}{q}}.$
Letting $p=qk$ yields the claimed result, that is
$\mathbb{E}\left(\left(\sup_{0 \le t \le T} |M_t|\right)^p \right) \le C_p \mathbb{E} \left( \langle M\rangle_T^{\frac{p}{2} } \right).$
We proceed now to the proof of the left hand side inequality. We have,
$M_t^2 =\langle M \rangle_t +2\int_0^t M_s dM_s.$
Therefore, we get
$\mathbb{E} \left( \langle M\rangle_T^{\frac{p}{2} } \right) \le A_p \left( \mathbb{E}\left(\left(\sup_{0 \le t \le T} |M_t|\right)^p \right) + \mathbb{E}\left(\sup_{0 \le t \le T}\ \left| \int_0^t M_s dM_s\right|^{p/2} \right) \right).$
By using the previous argument, we now have
$\mathbb{E}\left(\sup_{0 \le t \le T}\ \left| \int_0^t M_s dM_s\right|^{p/2} \right) \le B_p \mathbb{E}\left( \left( \int_0^T M^2_s d\langle M\rangle_s\right)^{p/4} \right)$
$\le B_p \mathbb{E}\left(\left(\sup_{0 \le t \le T} |M_t|\right)^{p/2} \langle M \rangle_T^{p/4} \right)$
$\le B_p \mathbb{E}\left(\left(\sup_{0 \le t \le T} |M_t|\right)^{p}\right)^{1/2} \mathbb{E} \left( \langle M \rangle_T^{p/2} \right)^{1/2}.$
As a conclusion, we obtained
$\mathbb{E} \left( \langle M\rangle_T^{\frac{p}{2} } \right) \le A_p \left( \mathbb{E}\left(\left(\sup_{0 \le t \le T} |M_t|\right)^p \right) + B_p \mathbb{E}\left(\left(\sup_{0 \le t \le T} |M_t|\right)^{p}\right)^{1/2} \mathbb{E} \left( \langle M \rangle_T^{p/2} \right)^{1/2} \right).$
This is an inequality of the form $x^2 \le A_p \left( y^2 +B_p xy\right)$, which easily implies
$c_p x^2 \le y^2$, thanks to the inequality $2xy \le \frac{1}{\delta} x^2+\delta y^2$, with a conveniently chosen $\delta$ $\square$

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