Marcinkiewicz–Zygmund inequality

Remarks

The proof uses symmetrization and the Khintchine inequality. Let $(\varepsilon_i)$ be independent random signs, independent of the $X_i$. Because the $X_i$ have mean zero, $\mathbb{E}\abs{\sum X_i}^p$ is comparable, with constants depending only on $p$, to $\mathbb{E}\abs{\sum \varepsilon_i X_i}^p$. Conditioning on the $X_i$ and applying Khintchine's inequality to the signs gives the comparison with $\mathbb{E}(\sum \abs{X_i}^2)^{p/2}$.

For $p = 2$ the inequality is an identity with $A_2 = B_2 = 1$: independence and mean zero give $\mathbb{E}\abs{\sum X_i}^2 = \sum \mathbb{E}\abs{X_i}^2$. The mean zero hypothesis is needed: if every $X_i = 1$, the middle term is $N^p$ while the outer terms are of order $N^{p/2}$, so the upper bound fails for large $N$.

Going deeper

For a student

The inequality is the probabilistic counterpart of the Khintchine inequality, which is the special case of random signs times constants. It is a model for the Littlewood–Paley inequality: the dyadic pieces of a function play the role of independent terms, and the square function plays the role of $\left( \sum \abs{X_i}^2 \right)^{1/2}$. Burkholder extended it to martingale differences for $1 < p < \infty$, which links it further with Littlewood–Paley theory. It is used to bound moments of sums of independent variables, for example in proofs of laws of large numbers with rates.