Martingale Square-Function Inequalities

Let \((M_n,\mathcal F_n)\) be a discrete-time martingale, \(d_n=M_n-M_{n-1}\) its differences, and \(M_n^*=\max_{k\leq n}|M_k|\) its running maximum. The maximal inequalities give the first comparison between the terminal value and the path maximum.

To measure the accumulated size of a martingale’s increments, we add their squares and take a square root. The inequalities below compare this square function with the terminal value and the running maximum. Set \(X_0=0\), \(Y_i=X_i-X_{i-1}\), \(S_m=(\sum_{i=1}^mY_i^2)^{1/2}\), and \(X_m^*=\max_{0\leq i\leq m}|X_i|\). Martingales are indexed from zero with respect to \((\mathscr F_i)\). Nonnegative submartingales can be extended to time zero by setting their initial value to zero.

Lemma 1 (Stopping time and squared differences). Let \(X_0,\ldots,X_n\) be an integrable martingale or a nonnegative submartingale, with \(X_0=0\). For \(\varepsilon>0\), set \(\tau=\inf\{1\leq i\leq n:|X_i|>\varepsilon\}\), using \(\tau=n+1\) if there is no crossing. Then

\[\mathbb{E} S_{\tau-1}^2+\mathbb{E} X_{\tau-1}^2\leq2\varepsilon\mathbb{E}|X_n|.\]

Proof. Extend the process by \(X_{n+1}=X_n\), so \(Y_{n+1}=0\). Summation of \(X_i^2-X_{i-1}^2=2X_{i-1}Y_i+Y_i^2\) gives, for every \(1\leq m\leq n+1\),

\[S_{m-1}^2+X_{m-1}^2=2X_{m-1}X_m-2\sum_{i=1}^{m}X_{i-1}Y_i.\]

Substitute \(m=\tau\). The multiplier \(X_{i-1}\mathbb1_{\{\tau\geq i\}}\) is predictable and bounded in absolute value by \(\varepsilon\). Hence each summand has expectation zero for a martingale, and nonnegative expectation for a nonnegative submartingale. All products are integrable. Bounded optional sampling for \(|X|\) now gives

\[\mathbb{E}(S_{\tau-1}^2+X_{\tau-1}^2) \leq2\mathbb{E}[X_{\tau-1}X_\tau] \leq2\varepsilon\mathbb{E}|X_\tau| \leq2\varepsilon\mathbb{E}|X_n|.\]

The constant extension includes the no-crossing case in the same argument. ◻

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 \(\|S_n\|_p\leq9\sqrt p\,q\|X_n\|_p\).

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

\[\lambda\mathbb P(M>\beta\lambda) \leq\lambda\mathbb P(X_n^*>\lambda) +\lambda\mathbb P(\theta S_n>\beta\lambda,\ X_n^*\leq\lambda).\tag{1}\]

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. Suppose first that \(X_n\in L^p\). The nonnegative submartingale inequality and conditional Jensen give \(X_i^p\leq(\mathbb E(X_n\mid\mathscr F_i))^p\leq\mathbb E(X_n^p\mid\mathscr F_i)\). Therefore each \(X_i\), and hence the finite sums defining \(S_n\) and \(M\), has a finite \(p\)-th moment. Tonelli and the substitution \(t=\beta\lambda\) now give the full integration step:

\[\begin{aligned} \mathbb EM^p &=p\beta^p\int_0^\infty\lambda^{p-1}\mathbb P(M>\beta\lambda)\,d\lambda\\ &\leq3p\beta^p\int_0^\infty\lambda^{p-2}\mathbb E(X_n\mathbf1_{\{M>\lambda\}})\,d\lambda\\ &=3p\beta^p\mathbb E\left(X_n\int_0^M\lambda^{p-2}\,d\lambda\right) =3q\beta^p\mathbb E(X_nM^{p-1}). \end{aligned}\]

The integral at zero is finite because \(p>1\). Hölder therefore yields

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

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 \(\lambda>0\),

\[\mathbb P(X>\beta\lambda,Y\leq\delta\lambda)\leq\varepsilon\mathbb P(X>\lambda).\tag{2}\]

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) 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

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

For the orthogonality identity, expand \(R_nX_n=\sum_{i,j}D_iY_j\). If \(i<j\), then \(D_i\) is \(\mathscr F_{j-1}\)-measurable and \(\mathbb E(Y_j\mid\mathscr F_{j-1})=0\), so \(\mathbb E(D_iY_j)=0\). If \(i>j\), condition instead on \(\mathscr F_{i-1}\), using \(\mathbb E(D_i\mid\mathscr F_{i-1})=0\). These products are integrable by Hölder. The first increment \(D_1\) need not have conditional mean zero at time zero; no off-diagonal term conditions on that increment, so the argument remains valid.

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 \(C_2=(18\sqrt q\,p)^p\). ◻

Theorem 5 (Maximal inequality; martingales). Let \(X_0=0,X_1,\ldots,X_n\) be an integrable martingale. For each \(p>0\) there is \(C_p<\infty\) such that

\[\mathbb{E}(X_n^*)^p\leq C_p\left[\mathbb{E}\max_{1\leq i\leq n}|Y_i|^p+ \mathbb{E}\left(\sum_{i=1}^n\mathbb{E}[Y_i^2\mid\mathscr F_{i-1}]\right)^{p/2}\right].\]

Proof. Put \(v_i=\mathbb{E}[Y_i^2\mid\mathscr F_{i-1}]\) and \(V_k=\sum_{i=1}^kv_i\), \(J_k=\max_{i\leq k}|Y_i|\), and \(Q=\max(\sqrt{V_n},J_n)\). Fix \(\beta>1\) and \(0<\delta<\beta-1\). Define predictable indicators and a martingale transform by

\[I_k=\mathbb1_{\{\lambda<X_{k-1}^*\leq\beta\lambda,\ J_{k-1}\leq\delta\lambda,\ V_k\leq\delta^2\lambda^2\}}, \qquad T_i=\sum_{k=1}^i I_kY_k.\]

On \(\{X_n^*>\beta\lambda,Q\leq\delta\lambda\}\), take the first crossing of \(\lambda\), then the first crossing of \(\beta\lambda\). The first value has absolute value at most \((1+\delta)\lambda\). All increments between these crossings are retained in \(T\), so \(T^*_n>(\beta-1-\delta)\lambda\). Orthogonality of martingale differences gives

\[\mathbb{E} T_n^2=\mathbb{E}\sum_{k=1}^n I_kv_k \leq\delta^2\lambda^2\mathbb P(X_n^*>\lambda).\]

The conditional second moments \(v_k\) may take the value infinity. On the predictable set where \(I_k=1\), however, \(V_k\leq\delta^2\lambda^2\), so every retained \(v_k\) is finite and their sum is at most \(\delta^2\lambda^2\). The product \(I_kv_k\) means \(v_k\) on that set and zero elsewhere. Hence the transform is square-integrable even when \(X\) is not. No increment is retained unless \(X_n^*>\lambda\), which proves the displayed bound. Applying the weak maximal inequality to \(T_i^2\) yields

\[\mathbb P(X_n^*>\beta\lambda,Q\leq\delta\lambda) \leq\frac{\delta^2}{(\beta-1-\delta)^2}\mathbb P(X_n^*>\lambda).\]

Choose \(\beta=2\) and then \(\delta\) small enough for Lemma 3. Since \(Q^p\leq V_n^{p/2}+J_n^p\), the result follows. ◻

Theorem 6 (Rosenthal). For \(2\leq p<\infty\), a martingale with \(X_0=0\) and \(X_n\in L^p\) satisfies

\[C_1(p)\left[\sum_{i=1}^n\mathbb{E}|Y_i|^p+\mathbb{E} V_n^{p/2}\right] \leq\mathbb{E}|X_n|^p \leq C_2(p)\left[\sum_{i=1}^n\mathbb{E}|Y_i|^p+\mathbb{E} V_n^{p/2}\right].\]

Proof. The upper bound follows from Theorem 5 and \(\max_i|Y_i|^p\leq\sum_i|Y_i|^p\). For the lower bound, put \(r=p/2\geq1\), \(A_k=V_k\), and \(B=\sum_iY_i^2\). The deterministic inequality for successive increments of an increasing sequence gives

\[A_n^r\leq r\sum_{i=1}^n A_i^{r-1}(A_i-A_{i-1}).\]

The deterministic bound follows by applying the fundamental theorem of calculus to each nonnegative increment:

\[A_i^r-A_{i-1}^r=\int_{A_{i-1}}^{A_i}r t^{r-1}\,dt \leq r A_i^{r-1}(A_i-A_{i-1}),\]

then summing. For \(r=1\), the derivative is identically one, including at zero.

Because \(A_i\) is \(\mathscr F_{i-1}\)-measurable, conditioning replaces \(A_i-A_{i-1}\) by \(Y_i^2\) in expectation. For \(r>1\), Hölder’s inequality then gives

\[\mathbb{E} A_n^r\leq r\mathbb{E}[A_n^{r-1}B] \leq r\|A_n\|_r^{r-1}\|B\|_r, \qquad \|A_n\|_r\leq r\|B\|_r.\]

All these moments are finite by conditional Jensen and the finite number of summands. For \(r=1\), \(\mathbb{E} A_n=\mathbb{E} B\) directly. Finally \(\sum_i|Y_i|^p\leq S_n^p\), and Theorem 4 controls \(\mathbb{E} S_n^p\) by \(\mathbb{E}|X_n|^p\). Combining these bounds proves the lower estimate. ◻

Bass, Richard F. 1995. Probabilistic Techniques in Analysis. Probability and Its Applications. New York: Springer-Verlag.