Burkholder-Davis-Gundy inequalities

Work on a filtered probability space satisfying the usual conditions. \(B\) denotes Brownian motion, and stochastic integrands are predictable with the stated integrability. A local martingale need not be a true martingale: the moment conditions in each claim are essential. See Brownian motion and martingale inequalities.

The Burkholder–Davis–Gundy inequalities compare the size of a continuous local martingale path with its quadratic variation.

Theorem 1. Let \(M_0=0\), \(M_t^*=\sup_{s\leq t}|M_s|\), and \(A_t=\langle M\rangle_t\). For \(2\leq p<\infty\) and every finite stopping time \(T\), we will prove

\[\mathbb{E}(M_T^*)^p\leq c_p\mathbb{E} A_T^{p/2}, \qquad \mathbb{E} A_T^{p/2}\leq C_p\mathbb{E}(M_T^*)^p.\]

First stop at \(\tau_K=\inf\{t\geq0:M_t^*+A_t\geq K\}\wedge K\). Continuity makes both the stopped maximum and bracket bounded, so every stochastic integral below is a true square-integrable martingale. Write \(S=T\wedge\tau_K\). Doob’s inequality and Itō’s formula give

\[\begin{aligned} \mathbb{E}(M_S^*)^p&\leq\left(\frac p{p-1}\right)^p\mathbb{E}|M_S|^p,\\ |M_S|^p&=p\int_0^S |M_s|^{p-1}\operatorname{sgn}(M_s)\,\mathrm dM_s +\frac{p(p-1)}2\int_0^S|M_s|^{p-2}\,\mathrm dA_s. \end{aligned}\]

For \(p=2\), the last integrand is one. Taking expectations and applying Hölder when \(p>2\),

\[\mathbb{E}(M_S^*)^p\leq c_p\mathbb{E}[(M_S^*)^{p-2}A_S] \leq c_p\bigl(\mathbb{E}(M_S^*)^p\bigr)^{(p-2)/p} \bigl(\mathbb{E} A_S^{p/2}\bigr)^{2/p}.\]

If the maximum has zero moment, the claim is immediate; otherwise divide and raise to the power \(p/2\). The same result follows directly from the preceding identity at \(p=2\). This proves the first estimate uniformly in \(K\).

For the reverse estimate put \(r=p/2\). When \(r=1\), \(\mathbb{E} A_S=\mathbb{E} M_S^2\leq\mathbb{E}(M_S^*)^2\). For \(r>1\), integrate the identity \(\mathrm d(M^2)=2M\,\mathrm dM+\mathrm dA\) against the adapted finite-variation process \(A^{r-1}\). Its stochastic-integral term has mean zero, and integration by parts gives

\[\begin{aligned} \mathbb{E} A_S^r &=r\mathbb{E}\int_0^S A_s^{r-1}\,\mathrm d(M_s^2)\\ &=r\mathbb{E}\left[A_S^{r-1}M_S^2-\int_0^S M_s^2\,\mathrm d(A_s^{r-1})\right]\\ &\leq r\mathbb{E}[A_S^{r-1}(M_S^*)^2]\\ &\leq r\bigl(\mathbb{E} A_S^r\bigr)^{(r-1)/r} \bigl(\mathbb{E}(M_S^*)^{2r}\bigr)^{1/r}. \end{aligned}\]

Division yields \(\mathbb{E} A_S^r\leq r^r\mathbb{E}(M_S^*)^{2r}\), also covering the zero case. Finally \(\tau_K\uparrow\infty\) almost surely, and both stopped maxima and stopped brackets increase to their values at \(T\). Monotone convergence gives both estimates without any boundedness assumption. No limit in the fixed stopping time \(T\) is needed.

The following discrete estimates explain a complementary method based on tail comparisons. They are proved in full below, using the stopped square-sum estimate in Lemma 1 of the square-function notes. In this discrete discussion set \(X_0=0\), \(Y_i=X_i-X_{i-1}\), \(S_m=(\sum_{i=1}^mY_i^2)^{1/2}\), and \(X_m^*=\max_{i\leq m}|X_i|\).

Lemma 2 (Maximal inequality; truncated submartingales). Suppose \(X\) is a nonnegative submartingale. For \(\theta>0\), put \(M=\max(\theta S_n,X_n^*)\) and \(\beta=\sqrt{1+2\theta^2}\). Then, for every \(\lambda>0\),

\[\lambda\mathbb P(M>\beta\lambda)\leq3\mathbb{E}[X_n\mathbb1_{\{M>\lambda\}}].\]

Consequently, if \(1<p<\infty\) and \(q=p/(p-1)\), then

\[\|S_n\|_p\leq9\sqrt p\,q\|X_n\|_p.\tag{1}\]

Proof. Since \(\beta>1\),

\[\lambda\mathbb P(M>\beta\lambda) \leq\lambda\mathbb P(X_n^*>\lambda) +\lambda\mathbb P(\theta S_n>\beta\lambda,\ X_n^*\leq\lambda).\tag{2}\]

The first-crossing proof of Doob’s inequality bounds the first term by \(\mathbb{E}[X_n\mathbb1_{\{X_n^*>\lambda\}}]\). For the second term, activate the process only after its square function crosses the smaller threshold. Define

\[\nu=\inf\{1\leq j\leq n:\theta S_j>\lambda\},\qquad T_m=X_m\mathbb1_{\{\nu\leq m\}},\qquad T_0=0,\quad Z_i=T_i-T_{i-1}.\]

Again an empty infimum is infinity. This activation uses information available at time \(m\); it is not multiplication by a terminal event. Nonnegativity gives

\[\mathbb{E}[T_m\mid\mathscr F_{m-1}] \geq\mathbb1_{\{\nu\leq m-1\}}\mathbb{E}[X_m\mid\mathscr F_{m-1}] \geq T_{m-1}.\]

On \(\{\theta S_n>\beta\lambda,\ X_n^*\leq\lambda\}\), the crossing time is finite, \(\theta^2S_{\nu-1}^2\leq\lambda^2\), and \(|Y_\nu|\leq\lambda\) because the process is nonnegative. Also \(Z_i=Y_i\) after \(\nu\). Therefore

\[\theta^2\sum_{i=\nu+1}^nZ_i^2 >\bigl(\beta^2-1-\theta^2\bigr)\lambda^2 =\theta^2\lambda^2.\]

In particular \(\sum_i Z_i^2>\lambda^2\), while \(\max_iT_i\leq\lambda\). Let \(\rho\) be the first crossing of \(\lambda\) by \(T\), or \(n+1\) if there is none. Markov’s inequality and Lemma 1 yield

\[\lambda\mathbb P\left(\sum_i Z_i^2>\lambda^2,\ \max_iT_i\leq\lambda\right) \leq\lambda^{-1}\mathbb{E}\sum_{i<\rho}Z_i^2 \leq2\mathbb{E} T_n \leq2\mathbb{E}[X_n\mathbb1_{\{M>\lambda\}}].\]

Adding the two estimates proves the tail bound. Integrate it against \(p\lambda^{p-2}\). When \(X_n\in L^p\), every \(X_i\in L^p\) by conditional Jensen, so all moments below are finite:

\[\mathbb{E} M^p\leq3q\beta^p\mathbb{E}[X_nM^{p-1}] \leq3q\beta^p\|X_n\|_p\|M\|_p^{p-1}.\]

Thus \(\theta\|S_n\|_p\leq\|M\|_p\leq3q\beta^p\|X_n\|_p\). Choose \(\theta=p^{-1/2}\); then \(\beta^p=(1+2/p)^{p/2}<e<3\). If \(\|X_n\|_p=\infty\), the asserted upper bound is automatic. ◻

A good-lambda inequality compares a large excursion of one variable with a smaller excursion of another. Integrating the tail estimate gives a comparison of moments.

Lemma 3 (Good lambda inequality). Let \(X,Y\) be nonnegative random variables. Suppose constants \(\beta>1\), \(\delta>0\), and \(\varepsilon>0\) satisfy, for every \(\lambda>0\),

\[\mathbb P(X>\beta\lambda,Y\leq\delta\lambda)\leq\varepsilon\mathbb P(X>\lambda).\tag{3}\]

If \(p>0\) and \(\varepsilon\beta^p<1\), then

\[\mathbb{E} X^p\leq\frac{\beta^p\delta^{-p}}{1-\varepsilon\beta^p}\mathbb{E} Y^p.\]

Proof. Split according to whether \(Y\leq\delta\lambda\) to obtain \(\mathbb P(X>\beta\lambda)\leq\varepsilon\mathbb P(X>\lambda)+\mathbb P(Y>\delta\lambda)\). Integrating only up to \(R\) avoids assuming the conclusion’s finiteness:

\[\mathbb{E}(X\wedge\beta R)^p \leq\varepsilon\beta^p\mathbb{E}(X\wedge R)^p +\beta^p\delta^{-p}\mathbb{E}(Y\wedge\delta R)^p.\]

The first truncated moment on the right is at most the one on the left. Absorb it and let \(R\uparrow\infty\) by monotone convergence. This proves the estimate in Bass (1995), Lemma 6.9 directly. ◻

Theorem 4 (Burkholder). For a martingale \(X\) and \(1<p<\infty\), there are positive constants depending only on \(p\) such that

\[C_1(p)\mathbb{E} S_n^p\leq\mathbb{E}|X_n|^p\leq C_2(p)\mathbb{E} S_n^p.\]

Proof. Write \(X_n^+=\max(X_n,0)\) and \(X_n^-=\max(-X_n,0)\). For \(1\leq i\leq n\), set \(T_i=\mathbb{E}[X_n^+\mid\mathscr F_i]\) and \(U_i=\mathbb{E}[X_n^-\mid\mathscr F_i]\), with \(T_0=U_0=0\). These are nonnegative submartingales, and their differences \(W_i=T_i-T_{i-1}\), \(Z_i=U_i-U_{i-1}\) satisfy \(Y_i=W_i-Z_i\). Minkowski’s inequality and Lemma 2 give

\[\|S_n\|_p\leq\left\|\Bigl(\sum_iW_i^2\Bigr)^{1/2}\right\|_p+ \left\|\Bigl(\sum_iZ_i^2\Bigr)^{1/2}\right\|_p \leq18\sqrt p\,q\|X_n\|_p.\]

For the reverse bound, assume \(0<\|X_n\|_p<\infty\), and put \(R_n=\operatorname{sgn}(X_n)|X_n|^{p-1}/\|X_n\|_p^{p-1}\). Then \(\|R_n\|_q=1\). Let \(R_i=\mathbb{E}[R_n\mid\mathscr F_i]\) for \(i\geq1\), \(R_0=0\), and \(D_i=R_i-R_{i-1}\). Products involving differences from distinct time steps have zero expectation by conditioning just before the later difference. Thus

\[\|X_n\|_p=\mathbb{E}[R_nX_n]=\mathbb{E}\sum_iD_iY_i \leq\left\|\Bigl(\sum_iD_i^2\Bigr)^{1/2}\right\|_q\|S_n\|_p \leq18\sqrt q\,p\|S_n\|_p.\]

The lower estimate just proved also applies to a sequence started at index one with a possibly nonzero mean; its proof used only the nonnegative conditional-expectation submartingales. This justifies its use for \(R\). Zero norms are immediate. If \(S_n\in L^p\), then \(|X_n|\leq\sqrt n S_n\), so the finiteness assumption needed for duality holds; otherwise the upper bound is automatic. We may take \(C_1=(18\sqrt p\,q)^{-p}\) and \(C_2=(18\sqrt q\,p)^p\). ◻

Bass, Richard F. 1995. Probabilistic Techniques in Analysis. Springer Series in Statistics. New York: Springer-Verlag.

Reading: Le Gall (2016), Brownian Motion, Martingales, and Stochastic Calculus.