Moment Inequalities and Random Series

Moment estimates let us control both individual sums and entire sequences of partial sums. The first tools below concern deterministic inequalities and approximation. We then use first-crossing events for maximal inequalities and conditional symmetry for sharper moment bounds. In each case, the assumption on dependence determines which cancellations are available.

Lemma 1. Let \(\{x_1,\dots,x_n\}\) and \(\{y_1,\dots,y_n\}\) denote sequences of complex numbers with modulus at most one. Then,

\[|x_1 \cdots x_n - y_1 \cdots y_n| \leq \sum_{1 \leq k \leq n} |x_k - y_k| ~.\]

Proof. Replace the factors one at a time:

\[\prod_{j=1}^nx_j-\prod_{j=1}^ny_j=\sum_{k=1}^n\left(\prod_{j<k}x_j\right)(x_k-y_k)\left(\prod_{j>k}y_j\right).\]

Empty products equal one. Every surrounding product has modulus at most one, so the triangle inequality gives the bound. ◻

Lemma 2 (Clarkson). Let \(x, y \in \mathbb R^n\) or \(\mathbb C^n\). For \(p \in [1,2]\),

\[|x+y|^p + |x-y|^p \leq 2 (|x|^p + |y|^p) ~.\]

Proof. Set \(r=p/2\in[1/2,1]\). Concavity of \(u\mapsto u^r\), the parallelogram identity, and subadditivity for the exponent \(r\leq1\) give

\[\frac{|x+y|^p+|x-y|^p}{2}\leq\left(\frac{|x+y|^2+|x-y|^2}{2}\right)^r=(|x|^2+|y|^2)^r\leq |x|^p+|y|^p.\]

The subadditivity used in the last step is proved in Lemma 3. The argument works for both real and complex Euclidean vectors. ◻

Here absolute values of vectors denote the Euclidean norm.

Lemma 3. For \(r>0\) and real \(a_1,\ldots,a_n\), the following inequality holds with \(c_r=1\) for \(0<r\leq1\) and \(c_r=n^{r-1}\) for \(r\geq1\):

\[\left| \sum_{1 \leq k \leq n} a_k \right|^r \leq c_r \sum_{1 \leq k \leq n} |a_k|^r ~.\]

Proof. The triangle inequality reduces the claim to nonnegative \(a_k\). If \(r\geq1\), convexity gives \((n^{-1}\sum_k a_k)^r\leq n^{-1}\sum_k a_k^r\), and multiplying by \(n^r\) proves the claim. If \(0<r\leq1\) and \(A=\sum_k a_k>0\), put \(b_k=a_k/A\). Then \(0\leq b_k\leq1\), so \(\sum_k b_k^r\geq\sum_k b_k=1\). Multiplication by \(A^r\) gives \(A^r\leq\sum_k a_k^r\). If \(A=0\), all terms vanish and no division is needed. ◻

Lemma 4. For \(0<p<2\) and real \(x\),

\[|x|^p = K(p) \int_{-\infty}^{\infty} \frac{1 - \cos xt}{|t|^{p+1}} \dif t\]

where \(K(p) \triangleq (\int_{-\infty}^{\infty} \frac{1 - \cos u}{|u|^{p+1}} \dif u)^{-1} = \frac{\Gamma(p+1)}{\pi} \sin \frac{p\pi}{2}\), and \(\Gamma(x)\) is the Gamma function.

Evaluation of the constant. For \(t>0\), the defining gamma integral gives \(t^{-p-1}=\Gamma(p+1)^{-1}\int_0^\infty s^p e^{-st}\,ds\). Tonelli applies because \(1-\cos t\geq0\). Also \(\int_0^\infty e^{-st}(1-\cos t)\,dt=1/s-s/(s^2+1)\). Therefore

\[\begin{aligned} \int_{\mathbb R}\frac{1-\cos t}{|t|^{p+1}}\,dt &=\frac{2}{\Gamma(p+1)}\int_0^\infty\frac{s^{p-1}}{1+s^2}\,ds\\ &=\frac{1}{\Gamma(p+1)}\int_0^\infty\frac{u^{p/2-1}}{1+u}\,du =\frac{\Gamma(p/2)\Gamma(1-p/2)}{\Gamma(p+1)} =\frac{\pi}{\Gamma(p+1)\sin(\pi p/2)}. \end{aligned}\]

The substitution is \(u=s^2\). The last two identities are Euler’s beta integral and the gamma reflection formula, applied with \(0<p/2<1\). Taking the reciprocal gives the stated \(K(p)\).

The range of exponents is essential. Near zero, \(1-\cos t=O(t^2)\), so the defining integral for \(K(p)\) is finite exactly when \(p<2\) at that endpoint. At infinity, boundedness of the numerator gives integrability when \(p>0\). Substitution \(u=|x|t\) proves the identity for \(x\ne0\); the case \(x=0\) is immediate.

1. Applications

Theorem 5 (Bernstein). Let \(f\) be a function on \([0,1]\). Define the Bernstein polynomial of degree \(n\) as

\[B_n (x) = \sum_{0 \leq k \leq n} f \left( \frac{k}{n} \right) \binom{n}{k} x^k (1-x)^{n-k} ~.\]

If \(f\) is continuous, then \(B_n (x)\) converges to \(f(x)\) uniformly on \([0,1]\).

Proof. Interpret the Bernstein polynomial as an expectation under a binomial law. Concentration of the sample proportion near its mean then turns uniform continuity into uniform approximation. Let \(M = \sup_{x \in [0,1]} |f(x)|\). Fix some \(\varepsilon > 0\), and let \(\delta = \delta(\varepsilon) = \sup \{ |f(x) - f(y)| : |x-y| \leq \varepsilon \}\). Fix \(n \geq 1\) and \(x \in [0,1]\); let \(X_1,\dots,X_n\) be independent random variables, with respect to some probability space, such that \(\mathop{\mathrm{\mathbf{P}}}(X_i = 1)=x\) and \(\mathop{\mathrm{\mathbf{P}}}(X_i = 0)=1-x\). Set \(S_n = X_1 + \dots + X_n\). Then, since \(\mathop{\mathrm{\mathbf{P}}}(S_n = k) = \binom{n}{k} x^k (1-x)^{n-k}\) by independence, we have that \(\mathbb{E} f(\frac{S_n}{n}) = B_n (x)\). So, the task is to bound the probability that \(|f(\frac{S_n}{n}) - f(x)|\) is small (this is random, as the first term is random).

To do so, note that \(|f(\frac{S_n}{n}) - f(x)| \leq \delta\) on the set \(\{|\frac{S_n}{n} - x| < \varepsilon\}\). Furthermore, \(|f(\frac{S_n}{n}) - f(x)| \leq 2M\) on \(\{|\frac{S_n}{n} - x| < \varepsilon\}^c\), by the boundedness of a continuous function \(f\) on a compact set. Therefore, we have

\[\begin{aligned} |B_n (x) - f(x)| &= \left|\mathbb{E}\left[f\left(\frac{S_n}{n}\right) - f(x)\right] \right| \leq \mathbb{E}\left|f\left(\frac{S_n}{n}\right) - f(x)\right| \\ & \leq \mathop{\mathrm{\mathbf{P}}}\left(\left|\frac{S_n}{n} - x\right| < \varepsilon\right) \delta + 2 \mathop{\mathrm{\mathbf{P}}}\left(\left|\frac{S_n}{n} - x\right| \geq \varepsilon\right) M \leq \delta + \frac{2M \mathop{\mathrm{Var}}(S_n)}{n^2 \varepsilon^2} ~, \end{aligned}\tag{1}\]

For every fixed \(\varepsilon>0\), the preceding bound gives \(\limsup_n\|B_n-f\|_\infty\leq\delta(\varepsilon)\). Now let \(\varepsilon\downarrow0\). Uniform continuity gives \(\delta(\varepsilon)\to0\), proving uniform convergence. ◻

2. Useful Probabilistic (In)equalities

2.1. Useful Preliminary Facts

Theorem 6. If \(\{X_n\}\) is a sequence of uniformly bounded random variables and \(X = \lim\limits_{n\to\infty} X_n\) with probability \(1\), then \(\mathbb{E} X = \lim\limits_{n\to\infty} \mathbb{E} X_n\).

Proof.

If \(|X_n|\leq C\) almost surely for all \(n\), then \(|X|\leq C\) almost surely. For \(A_n=\{|X_n-X|>\varepsilon\}\),

\[\mathbb{E}|X_n-X|\leq2CP(A_n)+\varepsilon.\]

Almost sure convergence implies \(P(A_n)\to0\). Take \(n\to\infty\) and then \(\varepsilon\downarrow0\); the inequality \(|\mathbb{E}X_n-\mathbb{E}X|\leq \mathbb{E}|X_n-X|\) finishes the proof.

◻

Lemma 7. For any \(b \geq 0\) and random variables \(X\) and \(Y\),

\[ \mathop{\mathrm{\mathbf{P}}}(X + Y \leq a-b) - \mathop{\mathrm{\mathbf{P}}}(|Y| > b) \leq \mathop{\mathrm{\mathbf{P}}}(X \leq a) \leq \mathop{\mathrm{\mathbf{P}}}(X + Y \leq a+b) + \mathop{\mathrm{\mathbf{P}}}(|Y| > b) ~.\tag{2}\]

Proof. On \(\{|Y|\leq b\}\), the implication \(X+Y\leq a-b\Rightarrow X\leq a\) follows from \(X=(X+Y)-Y\leq a\). Also \(X\leq a\Rightarrow X+Y\leq a+b\). Thus

\[\{X+Y\leq a-b\}\setminus\{|Y|>b\}\subseteq\{X\leq a\}\subseteq\{X+Y\leq a+b\}\cup\{|Y|>b\}.\]

Taking probabilities proves both inequalities, including \(b=0\). ◻

Lemma 8. Let \((X_i)_{i\geq1}\) be independent random variables. Suppose \(k_n\to\infty\) are integers and \(A_n\in\sigma(X_j:j\geq k_n)\). For every fixed event \(A\) in the underlying probability space,

\[P(A\cap A_n)-P(A)P(A_n)\longrightarrow0.\]

If \(P(A)>0\), this gives \(P(A_n\mid A)-P(A_n)\to0\). Without independence of the observations, the assertion is false: take every \(X_i\) equal to the same nonconstant Bernoulli variable.

Proof.

Let \(\mathscr F_K=\sigma(X_1,\ldots,X_K)\), \(\mathscr F_\infty=\sigma(X_i:i\geq1)\), \(U=\mathbb{E}[1_A\mid\mathscr F_\infty]\), and \(U_K=\mathbb{E}[1_A\mid\mathscr F_K]\). Then \(\mathbb{E}|U-U_K|\to0\) by martingale convergence (equivalently, by approximation of integrable \(\mathscr F_\infty\)-measurable functions by finite-coordinate simple functions). For \(k_n>K\), independence gives \(\mathbb{E}[U_K1_{A_n}]=P(A)P(A_n)\). Since \(\mathbb{E}[1_A1_{A_n}]=\mathbb{E}[U1_{A_n}]\), the absolute difference is bounded by \(\mathbb{E}|U-U_K|\). Let \(n\to\infty\) and then \(K\to\infty\).

◻

2.2. Bounds on Moments of Random Variables

Lemma 9 (Markov). Let \(X\) be a nonnegative random variable. Then, for \(\varepsilon > 0\),

\[\mathop{\mathrm{\mathbf{P}}}(X \geq \varepsilon) \leq \frac{\mathbb{E} X}{\varepsilon} ~.\]

Proof. The pointwise inequality \(\varepsilon1_{\{X\geq\varepsilon\}}\leq X\) gives \(\varepsilon\mathbb{P}(X\geq\varepsilon)\leq\mathbb{E}X\) after integration. This also holds if the expectation is infinite. ◻

Lemma 10 (Chebyshev). Let \(X\) be a real random variable with finite variance. Then, for \(\varepsilon > 0\),

\[\mathop{\mathrm{\mathbf{P}}}(|X-\mathbb{E} X| \geq \varepsilon) \leq \frac{\mathop{\mathrm{Var}}X}{\varepsilon^2} ~.\]

Proof. Apply Markov to the nonnegative variable \((X-\mathbb{E}X)^2\) at threshold \(\varepsilon^2\). Nonnegativity of \(X\) itself is not required. ◻

Lemma 11 (Jensen). For an integrable real random variable \(X\) and a finite convex function \(\varphi:\mathbb R\to\mathbb R\),

\[\varphi(\mathbb{E} X) \leq \mathbb{E} \varphi(X) ~.\]

Proof. At \(m=\mathbb{E}X\), a finite convex function has a supporting slope \(a\) such that \(\varphi(x)\geq\varphi(m)+a(x-m)\) for every \(x\). This follows by choosing \(a\) between the left and right derivatives at \(m\). The affine lower bound is integrable, so the negative part of \(\varphi(X)\) is integrable and its expectation is well defined, possibly \(+\infty\). Taking expectations of the supporting-line inequality proves Jensen. ◻

Lemma 12 (Hölder). For random variables \(X\) and \(Y\),

\[\mathbb{E}|XY| \leq (\mathbb{E}|X|^p)^{\frac1p} (\mathbb{E}|Y|^q)^{\frac1q} ~,\]

provided \(1<p,q<\infty,\quad 1/p+1/q=1\), with finite displayed moments.

Proof. If either norm is zero, the corresponding variable vanishes almost surely. Otherwise divide \(|X|\) and \(|Y|\) by \(A=(\mathbb{E}|X|^p)^{1/p}\) and \(B=(\mathbb{E}|Y|^q)^{1/q}\). Young’s inequality \(uv\leq u^p/p+v^q/q\) follows by minimizing \(u^p/p-uv\) over \(u\geq0\). Integrating it for these normalized variables gives \(\mathbb{E}|XY|/(AB)\leq1/p+1/q=1\). The endpoint version \(\mathbb{E}|XY|\leq\mathbb{E}|X|\,\|Y\|_\infty\) follows directly from the essential-supremum bound. ◻

Corollary 13 (Cauchy-Schwarz and Lyapounov).

(Cauchy-Schwarz)

\[\mathbb{E}|XY| \leq (\mathbb{E}|X|^2)^{\frac12} (\mathbb{E}|Y|^2)^{\frac12} ~.\]
(Lyapounov)

\[(\mathbb{E}|X|^p)^{\frac1p} \leq (\mathbb{E}|X|^q)^{\frac1q}\]

for \(0 < p < q\).

Proof. Cauchy–Schwarz is Hölder with both exponents equal to two. For \(0<p<q\), apply Hölder to \(|X|^p\) and \(1\), with conjugate exponents \(q/p\) and \(q/(q-p)\). This gives \(\mathbb{E}|X|^p\leq(\mathbb{E}|X|^q)^{p/q}\); taking the positive \(p\)th root proves Lyapunov’s inequality. ◻

Lemma 14 (Minkowski). For \(1\leq p<\infty\),

\[(\mathbb{E}|X+Y|^p)^{\frac1p} \leq (\mathbb{E}|X|^p)^{\frac1p} + (\mathbb{E}|Y|^p)^{\frac1p} ~.\]

Proof. It suffices to consider finite right-hand side. At \(p=1\) this is the integrated triangle inequality. For \(p>1\), the elementary bound \(|X+Y|^p\leq2^{p-1}(|X|^p+|Y|^p)\) first establishes integrability. Hölder then gives

\[\mathbb{E}|X+Y|^p\leq\mathbb{E}[(|X|+|Y|)|X+Y|^{p-1}]\leq(\|X\|_p+\|Y\|_p)\|X+Y\|_p^{p-1}.\]

If \(\|X+Y\|_p=0\) the result is immediate; otherwise divide by its \((p-1)\)st power. ◻

2.3. Maximal Inequalities

Theorem 15 (Kolmogorov’s Maximal Inequality). Let \(X_1,\dots,X_n\) be independent random variables with mean zero and finite variances. Then, for any \(\varepsilon > 0\),

\[\begin{aligned} \mathop{\mathrm{\mathbf{P}}}\left( \max_{1 \leq k \leq n} |S_k| \geq \varepsilon \right) \leq \frac{\mathop{\mathrm{Var}}(S_n)}{\varepsilon^2} ~. \end{aligned}\]

Note that this strengthens the result that one can obtain from applying Chebyshev’s inequality to estimating \(\mathop{\mathrm{\mathbf{P}}}(|S_n| \geq \varepsilon)\). The proof separates paths by their first threshold crossing. Independence makes the later increment orthogonal to the value at that crossing; it does not require that \(|S_n|\) remain large on each such path.

Proof. Let \(A_k\) be the set where \(|S_k| \geq \varepsilon\), but where \(|S_j| < \varepsilon\) for \(j < k\). Clearly, \(\{A_k\}\) is a disjoint collection of sets, hence

\[\begin{aligned} \mathbb{E} S_n^2 &\geq \sum_{1 \leq k \leq n} \int_{A_k} S_n^2 \dif \mathop{\mathrm{\mathbf{P}}}\\ &= \sum_{1 \leq k \leq n} \int_{A_k} (S_k^2 + 2 S_k (S_n - S_k) + (S_n - S_k)^2) \dif \mathop{\mathrm{\mathbf{P}}}\\ &\geq \sum_{1 \leq k \leq n} \int_{A_k} S_k^2 + 2 S_k (S_n - S_k) \dif \mathop{\mathrm{\mathbf{P}}}~. \end{aligned}\]

Note that \(A_k\) and \(S_k\) are \(\sigma(X_1,\dots,X_k)\)-measurable, and \(S_n - S_k\) is \(\sigma(X_{k+1},\dots,X_n)\)-measurable. Furthermore, these are mean zero random variables. Therefore, by independence and integrability, we have \(\int_{A_k} S_k(S_n-S_k) \dif \mathop{\mathrm{\mathbf{P}}}= \mathbb{E}[S_n-S_k] (\int_{A_k} S_k) = 0\). Therefore,

\[\begin{aligned} \mathbb{E} S_n^2 &\geq \sum_{1 \leq k \leq n} \int_{A_k} S_k^2 \dif \mathop{\mathrm{\mathbf{P}}}\\ &\geq \sum_{1 \leq k \leq n} \varepsilon^2 \mathop{\mathrm{\mathbf{P}}}(A_k) = \varepsilon^2 \mathop{\mathrm{\mathbf{P}}}\left( \max_{1 \leq k \leq n} |S_k| \geq \varepsilon \right) ~, \end{aligned}\]

which concludes the proof. ◻

Theorem 16. (Etemadi). For independent \(X_1,\ldots,X_n\), partial sums \(S_k\), and \(\varepsilon>0\),

\[P\left(\max_{k\leq n}|S_k|\geq3\varepsilon\right)\leq3\max_{k\leq n}P(|S_k|\geq\varepsilon).\tag{3}\]

Proof.

Let \(A_k\) be the disjoint first-crossing events for \(|S_k|\geq3\varepsilon\). On \(A_k\cap\{|S_n|<\varepsilon\}\), \(|S_n-S_k|>2\varepsilon\). This future increment is independent of \(A_k\). Writing \(q=\max_{j\leq n}P(|S_j|\geq\varepsilon)\), we obtain

\[P(\max_{k\leq n}|S_k|\geq3\varepsilon)\leq P(|S_n|\geq\varepsilon)+\sum_{k<n}P(A_k)P(|S_n-S_k|>2\varepsilon)\leq q+2q\sum_{k<n}P(A_k)\leq3q.\]

The penultimate bound uses \(\{|S_n-S_k|>2\varepsilon\}\subseteq\{|S_n|>\varepsilon\}\cup\{|S_k|>\varepsilon\}\).

◻

2.4. Bounds on Moments of Random Series

We next compare moments of a sum with moments of its summands. A bound with no assumptions on dependence comes first. Let \(X_1,\dots,X_n\) denote random variables (not requiring independence or identically distributed yet). Set \(S_m \triangleq \sum_{1 \leq i \leq m} X_i\) for \(m \leq n\) to be the \(m\)th partial sum of random variables.

We begin with a preliminary bound, called the \(c_r\) inequality (due to Loève).

Lemma 17 (\(c_r\) inequality). For any \(r > 0\),

\[\mathbb{E} \left| \sum_{1 \leq k \leq n} X_k \right|^r \leq c_r \sum_{1 \leq k \leq n} \mathbb{E} |X_k|^r ~,\]

where \(c_r = 1\) if \(r \leq 1\), and \(c_r = n^{r-1}\) for \(r \geq 1\).

Proof. Follows from substituting random variables \(\{X_1,\dots,X_n\}\) into the hypothesis of Lemma 3, and applying expectations to the inequality therein. ◻

To summarize, by writing \(S_n \triangleq \sum_{1\leq k \leq n} X_k\), we have

\[\mathbb{E} |S_n|^p \leq \begin{cases} n^{p-1} \sum_{1 \leq i \leq n} \mathbb{E} |X_i|^p & \text{ if } p > 1 ~, \\ \sum_{1 \leq i \leq n} \mathbb{E} |X_i|^p & \text{ if } p \leq 1 ~. \end{cases}\]

Cancellation improves the bound for \(1\leq p\leq2\). The estimates below do not extend unchanged to \(p>2\): for independent standard normal summands, \(\mathbb{E}|S_n|^p=n^{p/2}\mathbb{E}|X_1|^p\), which cannot be bounded by a fixed constant times \(\sum_i\mathbb{E}|X_i|^p=n\mathbb{E}|X_1|^p\) when \(p>2\).

Definition 18. Let \(X\) and \(Y\) be random variables. \(Y\) conditioned by \(X\) is symmetric (about \(0\)) if, for all \(a \geq 0\),

\[\mathop{\mathrm{\mathbf{P}}}(Y > a \mid X) = \mathop{\mathrm{\mathbf{P}}}(Y < -a \mid X) \quad \mathop{\mathrm{\mathrm{a.s.}}}\]

Lemma 19. Let \(X\) and \(Y\) be random variables such that \(\mathbb{E}|X|^p, \mathbb{E}|Y|^p < \infty\) for \(p \in [1,2]\). If \(Y\) is symmetric about \(0\) conditional on \(X\), then \(\mathbb{E}|X+Y|^p \leq \mathbb{E}|X|^p + \mathbb{E}|Y|^p\).

Proof. Since

\[\begin{aligned} \mathbb{E}|X+Y|^p = \mathbb{E}[\mathbb{E}[|X+Y|^p \mid X]] = \mathbb{E}[\mathbb{E}[|X-Y|^p \mid X]] = \mathbb{E}|X-Y|^p ~, \end{aligned}\]

applying Lemma 2 to the calculation above concludes the proof. ◻

Proposition 20. Let \(\{X_1,\dots,X_n\}\) be random variables such that \(\mathbb{E}|X_i|^p < \infty\) for all \(i \in \{1,\dots,n\}\) and \(p \in [1,2]\). If \(X_{m+1} | S_m\) is symmetric for all \(1 \leq m \leq n-1\), then \(\mathbb{E}|S_n|^p \leq \sum_{1 \leq i \leq n} \mathbb{E}|X_i|^p\).

Proof. Result follows from induction and applications of Lemma 19. ◻

Lemma 21. Let \(X\) have characteristic function \(f(t)=Ee^{itX}\). For \(0<p<2\), the following identity holds (also with value \(+\infty\)):

\[\mathbb{E}|X|^p = K(p) \int_{-\infty}^{\infty} \frac{1 - \Re f(t)}{|t|^{p+1}} \dif t ~.\]

Proof. The integrand is nonnegative, so Tonelli’s theorem permits the exchange of integrals, even if the moment is infinite:

\[\begin{aligned} \mathbb{E}|X|^p &= K(p) \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1 - \cos xt}{|t|^{p+1}} \dif t \dif F(x) = K(p) \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1 - \cos xt}{|t|^{p+1}} \dif F(x) \dif t \\ &= K(p) \int_{-\infty}^{\infty} \frac{1 - \Re f(t)}{|t|^{p+1}} \dif t ~, \end{aligned}\]

as desired. ◻

For \(1\leq p<2\), there is also a characteristic-function proof. Conditional symmetry makes the conditional characteristic function real, so the fractional-moment identity applies. The endpoint \(p=2\) follows directly by expanding the square and using conditional mean zero. Define

\[f_m (t) = \mathbb{E}[\exp(itX_{m+1}) \mid S_m] ~.\]

This is the characteristic function of the random variable \(X_{m+1}\) conditioned by \(S_m\), for all \(1 \leq m \leq n-1\). By assumptions, it is clear that \(f_m (t)\) is real. Therefore, the characteristic function of \(S_{m+1}\) conditioned by \(S_m\) is given by \(\exp(it S_m) f_m(t)\). From the arguments above, we have

\[\begin{aligned} \mathbb{E}[|S_{m+1}|^p \mid S_m] = K(p) \int_{-\infty}^{\infty} \frac{1 - \Re \exp(itS_m) f_m(t)}{|t|^{p+1}} \dif t ~. \end{aligned}\]

By direct calculations,

\[\begin{aligned} 1 - \Re \exp(itS_m) f_m(t) &= 1 - f_m(t) \cos tS_m \\ &= (1 - \cos tS_m) + (1 - f_m (t)) - (1 - \cos tS_m)(1 - f_m (t)) \\ &\leq (1 - \cos tS_m) + (1 - f_m (t)) ~. \end{aligned}\]

It then follows that

\[\begin{aligned} \mathbb{E}[|S_{m+1}|^p \mid S_m] &\leq K(p) \int_{-\infty}^{\infty} \frac{(1 - \cos tS_m) + (1 - f_m (t))}{|t|^{p+1}} \dif t \\ &= |S_m|^p + \mathbb{E}[|X_{m+1}|^p \mid S_m] ~. \end{aligned}\tag{4}\]

Therefore, Proposition 20 follows by taking expectations of the conditional inequality and iterating from \(m=1\).

Remark 22. Let \(X_1,X_2\) be random variables such that \(X_2\) is an independent copy of \(X_1\). Suppose these are identically distributed. Then, if \(f(t)\) is the characteristic function of \(X_1\), then \(X_1 - X_2\) has the characteristic function \(|f(t)|^2\). Applying the identity \(2(1 - \Re f(t)) = (1-|f(t)|^2) + |1-f(t)|^2\) and Lemma 21, we have

\[\begin{aligned} \mathbb{E}|X_1|^p = \frac12 \mathbb{E}|X_1 - X_2|^p + \frac12 K(p) \int_{-\infty}^{\infty} \frac{|1-f(t)|^2}{|t|^{p+1}} \dif t ~,~~\text{for } 0 < p < 2 ~. \end{aligned}\]

For \(1\leq p<2\) and \(\mathbb{E}X_1=0\), the left inequality below follows from the nonnegative integral term; the right follows from Jensen conditional on \(X_1\). At \(p=2\), use \(\mathbb{E}|X_1-X_2|^2=2\mathbb{E}|X_1|^2\). Thus for \(1\leq p\leq2\) and finite \(p\)th moments,

\[\begin{aligned} \frac12 \mathbb{E}|X_1 - X_2|^p \leq \mathbb{E} |X_1|^p \leq \mathbb{E} |X_1 - X_2|^p ~. \end{aligned}\]

Symmetrisation compares a centred variable with the difference of two copies. The next theorem uses this to bound the \(p\)th moment of a partial sum by the sum of the individual moments.

Theorem 23 (von Bahr-Esseen). Let \(X_1,\dots,X_n\) be random variables such that \(\mathbb{E}|X_i|^p < \infty\) for all \(1 \leq i \leq n\) and \(p \in [1,2]\).

Assume

\[\mathbb{E}[X_{m+1} \mid S_m] = 0 ~ \mathop{\mathrm{\mathrm{a.s.}}}\text{ for } 1 \leq m \leq n-1 ~.\]

Then, \(\mathbb{E} |S_n|^p \leq 2 \sum_{1 \leq i \leq n} \mathbb{E}|X_i|^p\).
Assume

\[\mathbb{E}\left[X_i\mid\sum_{j\in J\setminus\{i\}}X_j\right]=0\quad\text{for every nonempty }J\subseteq\{1,\ldots,n\},\ i\in J.\]
This assumption holds for independent centered variables and is inherited by every subcollection. In the proof put \(R_{mi}=\sum_{1\leq j\leq m+1,\ j\ne i}X_j\).

Then, \(\mathbb{E} |S_n|^p \leq (2 - \frac1n) \sum_{1 \leq i \leq n} \mathbb{E}|X_i|^p\).

The conditional-copy construction uses a regular conditional law \(K(s,\cdot)\) of the next real-valued summand given the real sum \(S_m=s\). On an enlarged space, attach an independent uniform variable and use the quantile of \(K(S_m,\cdot)\) to draw \(\tilde X_{m+1}\). Conditional on \(S_m\), the two copies are independent and identically distributed. Thus their difference is conditionally symmetric. Moreover, conditional centering implies \(\mathbb{E}[\tilde X_{m+1}\mid S_m,X_{m+1}]=0\); this is the hypothesis needed for the first Jensen inequality in the proof.

Proof. For the first estimate, introduce a conditional copy to obtain a symmetric difference. For (a), for a fixed \(1 \leq m \leq n-1\), we can introduce \(\tilde X_{m+1}\). Conditional on \(S_m\), take \(\tilde X_{m+1}\) independent of \(X_{m+1}\) with the same conditional distribution. Such a copy can be constructed on an enlarged probability space. Applying Lemmas 19 and 21, we have

\[\begin{aligned} \mathbb{E}|S_{m+1}|^p &= \mathbb{E}|S_m + X_{m+1}|^p \\ &\leq \mathbb{E}|S_m + X_{m+1} - \tilde X_{m+1}|^p \\ &\leq \mathbb{E}|S_m|^p + \mathbb{E} |X_{m+1} - \tilde X_{m+1}|^p \leq \mathbb{E}|S_m|^p + 2 \mathbb{E}|X_{m+1}|^p ~. \end{aligned}\]

Apply induction to conclude.

For the second estimate, repeat the conditional-copy argument with each summand singled out, then average the resulting inequalities. For (b), for a fixed \(i\) such that \(1 \leq i \leq m+1\), \(S_{m+1} = R_{mi} + X_i\). Again, let \(\tilde X_i\) be a random variable such that, conditional on \(R_{mi}\), is an i.i.d. copy of \(X_i\). Applying the same arguments as those for part (a), we have

\[\begin{aligned} \mathbb{E}|S_{m+1}|^p &= \mathbb{E}|R_{mi} + X_i|^p \\ &\leq \mathbb{E}|R_{mi} + X_i - \tilde X_i|^p \\ &\leq \mathbb{E}|R_{mi}|^p + \mathbb{E} |X_i - \tilde X_i|^p ~ \leq \left( 2 - \frac1m \right) \sum_{1 \leq j \leq m+1 : j \neq i} \mathbb{E}|X_j|^p + 2 \mathbb{E}|X_i|^p ~. \end{aligned}\tag{5}\]

Sum (5) over all \(1 \leq i \leq m+1\), we have the inequality

\[\begin{aligned} (m+1) \mathbb{E}|S_{m+1}|^p \leq (2m+1) \sum_{1 \leq i \leq m+1} \mathbb{E}|X_i|^p ~, \end{aligned}\]

and the claim follows from induction. ◻

Remark 24. For \(0<p<q\), Lyapunov’s inequality gives \(\mathbb{E}|S_n|^p\leq(\mathbb{E}|S_n|^q)^{p/q}\). A bound on the higher moment can therefore give a bound on the lower one, though it need not be sharper than a direct estimate. The exponent is \(p/q\).

Remark 25. Conditions required for Theorem 23 are satisfied by requiring \(X_1,\dots,X_n\) to be independent and have mean zero, i.e. \(\mathbb{E} X_i = 0\) for all \(1 \leq i \leq n\).