# Welcome to Randal Douc's wiki

A collaborative site on maths but not only!

• Theatre
• Research
• Teaching

### Miscellanous

world:marcinkiewicz


2017/10/07 23:39 ·

# $g$-lemma

Let $X$ be a random variable non-negative a.s. and $g \colon \rset_+ \rightarrow \rset_+$ an increasing differentiable function such that $g(0)=0$. Then, \begin{equation*} \PE\lrb{g(X)} = \int_{\rset_+} g'(x) \PP\lr{X \geq x} \rmd x \in \rset_+ \cup \lrc{+\infty}. \end{equation*}

Click to display ⇲

Click to hide ⇱

Write, using that for $x\in \rset_+$, $g'(x) 1_{X\geq x}\geq 0$, \begin{equation*} \int_{\rset_+} g'(x) \underbrace{\PP\lr{X \geq x}}_{\PE\lrb{1_{X\geq x}}} \rmd x = \int_{\rset_+} \PE \lrb{g'(x) 1_{X\geq x}} \rmd x = \PE \lrb{\int_{\rset_+} g'(x) 1_{X\geq x} \rmd x} = \PE \lrb{\int_{0}^X g'(x)\rmd x} = \PE \lrb{g(X) - \underbrace{g(0)}_{=0}}. \end{equation*}

# Convexity inequality

Let $p \geq 1$, $n \in \nset$ and $\lr{X_i}_{1\leq i \leq n}$ real-valued random variables. Then, by convexity \begin{equation*} \PE \lrb{\lrav{\sum_{i=1}^n X_i}^p} \leq n^{p-1} \sum_{i=1}^n \PE \lrb{\lrav{X_i}^p}. \end{equation*}

# Marcinkiewicz–Zygmund inequality

Let $p \geq 2$, $n \in \nset$ and $\lr{X_i}_{1\leq i \leq n}$ centered independent real-valued random variables in $L^p$. Then, there exists a universal constant $C_p$ depending only on $p$ such that \begin{equation*} \PE \lrb{\lrav{\sum_{i=1}^n X_i}^p} \leq C_p \eqsp n^{p/2-1} \sum_{i=1}^n \PE \lrb{\lrav{X_i}^p}. \end{equation*}

## Proof

Set $S_n \eqdef \sum_{i=1}^n X_i$. Let $x > 0$. We first establish an upper-bound for $\PP \lr{\lrav{S_n} \geq x}$.

Let $y > 0$ and define for all $i \in [1;n]$, $Z_i \eqdef X_i 1_{X_i < y}$ and $T_n \eqdef \sum_{i=1}^n Z_i$. Then, $$\label{eq:s_n_t_n} \PP \lr{S_n \geq x} \leq \PP \lr{T_n \geq x} + \PP \lr{S_n \neq T_n} \leq \PP \lr{T_n \geq x} + \sum_{i=1}^n \PP \lr{X_i \geq y}.$$ Let $h > 0$. The Chernoff bound and the independence of the $\lr{Z_i}_{1\leq i \leq n}$ by independence of the $\lr{X_i}_{1\leq i \leq n}$ both provide $$\label{eq:t_n_only} \PP \lr{T_n \geq x} \leq e^{-hx} \PE\lrb{e^{h T_n}} = e^{-hx} \prod_{i=1}^n \PE\lrb{e^{h Z_i}}.$$ Using the Taylor formula with the exponential function yields that the function defined on $\rset$ by $s \mapsto \frac {e^s-1-s} {s^2} = \frac 1 {s^2} \int_0^s (u-s)e^u \rmd u= \int_0^1 (u-1)e^{su} \rmd u$ is increasing, and together with $Z_i \leq y$ for all $i \in [1;n]$, we deduce \begin{equation*} e^{h Z_i} \leq 1 + h Z_i + Z_i^2 \frac {e^{hy}-1-y} {y^2}. \end{equation*} The fact that $y>0$ implies $Z_i \leq X_i$ and thus $\PE \lrb{Z_i} \leq \PE \lrb{X_i} = 0$. Combining with $\PE \lrb{Z_i^2} = \PE \lrb{X_i^2 1_{X_i < y}} \leq \PE \lrb{X_i^2}$ yields for all $i \in [1;n]$, \begin{equation*} \PE \lrb{e^{h Z_i}} \leq 1 + \PE \lrb{X_i^2} \frac {e^{hy}-1-y} {y^2}. \end{equation*} Together with \eqref{eq:s_n_t_n} and \eqref{eq:t_n_only} this provides $$\label{eq:s_n_step_1} \PP \lr{S_n \geq x} \leq \sum_{i=1}^n \PP \lr{X_i \geq y} + \exp \lrb{-hx + B_n \frac {e^{hy}-1-y} {y^2}},$$ where $B_n \eqdef \sum_{i=1}^n \PE \lrb{X_i^2} < \infty$. Note that $B_n = 0$ implies that the $\lr{X_i}_{1 \leq i \leq n}$ are all equal to zero a.s., a situation where the inequality is trivially true, and we can thus assume $B_n > 0$. The argument of the exponential in \eqref{eq:s_n_step_1} is then minimized in $h$ at $h_{\min} \eqdef \frac 1 y \log \lr{1 + \frac {xy} {B_n}}$, with \begin{equation*} -h_{\min}x + B_n \frac {e^{h_{\min}y}-1-y} {y^2} = - \frac x y \log \lr{1 + \frac {xy} {B_n}} + \frac {B_n} {y^2} \lrb{\frac {xy} {B_n} \underbrace{- \log \lr{1 + \frac {xy} {B_n}}}_{\leq 0}} \leq \frac x y - \frac x y \log \lr{1 + \frac {xy} {B_n}}. \end{equation*}

Click to display ⇲

Click to hide ⇱

The function defined on $\rset_+^*$ by $h \mapsto -hx + B_n \frac {e^{hy}-1-y} {y^2}$ is continuous, diverges to infinity when $h \rightarrow +\infty$, and its derivative $h \mapsto -x + \frac {B_n} y \lr{e^{hy}-1}$ has a unique zero $h_{\min}$ on $\rset_+^*$ defined by $e^{h_{\min}y}-1 = \frac {xy} {B_n}$.

With $y = \frac x r$ where $r > 0$, combining with \eqref{eq:s_n_step_1} yields \begin{equation*} \PP \lr{S_n \geq x} \leq \sum_{i=1}^n \PP \lr{X_i \geq \frac x r} + e^r \lr{1 + \frac {x^2} {r B_n}}^{-r}. \end{equation*} Considering $\lr{-X_i}_{1 \leq i \leq n}$ provides a similar inequality for $-S_n$, and using the fact that $x > 0$ we deduce \begin{equation*} \PP \lr{\lrav{S_n} \geq x} = \PP \lr{S_n \geq x} + \PP \lr{-S_n \geq x} \leq \sum_{i=1}^n \PP \lr{\lrav{X_i} \geq \frac x r}+ 2 e^r \lr{1 + \frac {x^2} {r B_n}}^{-r}. \end{equation*}

Using the $g$-lemma with $g \colon x \mapsto x^p$ we deduce \begin{align*} \PE \lrb{\lrav{S_n}^p} = p \int_{\rset_+} x^{p-1} \PP\lr{\lrav{S_n} \geq x} \rmd x &\leq \sum_{i=1}^n p \int_{\rset_+} x^{p-1} \PP\lr{\lrav{X_i} \geq x} \rmd x + 2p e^r \int_{\rset_+} \frac {x^{p-1}} {\lr{1 + \frac {x^2} {r B_n}}^r} \rmd x \\ &= r^p \sum_{i=1}^n \PE \lrb{\lrav{X_i}^p} + 2p e^r B_n^{p/2} \int_0^{+\infty} \frac {u^{p/2-1}} {\lr{1+\frac u r}^r} \rmd u \quad \in \rset_+ \cup \lrc{+\infty}, \end{align*} with the change of variables $u = \frac {x^2} {B_n}$. The integral is finite iff $r>p/2$, and we can choose $r = p$ to deduce the Rosenthal inequality: $$\label{eq:rosenthal} \PE \lrb{\lrav{S_n}^p} \leq c_p \lr{\sum_{i=1}^n \PE \lrb{\lrav{X_i}^p} + \lr{\sum_{i=1}^n \PE \lrb{X_i^2}}^{p/2}},$$ where $c_p \eqdef \max(p^p, 2p e^p \int_{\rset_+} \frac {u^{p/2-1}} {\lr{1+\frac u p}^p} \rmd u)$ only depends on $p$. Finally, by Jensen inequality as $p \geq 2$, and by convexity, \begin{equation*} \lr{\frac 1 n \sum_{i=1}^n \PE \lrb{X_i^2}}^{p/2} = \PE \lrb{\frac 1 n \sum_{i=1}^n X_i^2}^{p/2} \leq \PE \lrb{\lr{\frac 1 n \sum_{i=1}^n X_i^2}^{p/2}} \leq \PE \lrb{\frac 1 n \sum_{i=1}^n \lrav{X_i}^p}. \end{equation*}

which together with the Rosenthal inequality \eqref{eq:rosenthal} yields the Marcinkiewicz–Zygmund inequality: \begin{equation*} \PE \lrb{\lrav{\sum_{i=1}^n X_i}^p} \leq C_p \eqsp n^{p/2-1} \sum_{i=1}^n \PE \lrb{\lrav{X_i}^p}, \end{equation*} where $C_p \eqdef 2 c_p$.

# Generalized Marcinkiewicz–Zygmund inequality

Let $d \in \nset^*$ and $\norm{\cdot}$ a norm on $\rset^d$. Let $n \in \nset^*$ and $\lr{X_i}_{1 \leq i \leq n}$ independent random variables of $L^p(\rset^d)$ with $2 \leq p < \infty$. Then, \begin{equation*} \mathbb{E}\lrb{\norm{\sum_{i=1}^n \lr{X_i-\mathbb{E}\lrb{X_i}} }^p} \leq C_{p, \norm{}} \times n^{p/2-1} \times \sum_{i=1}^n \mathbb{E}\lrb{\norm{X_i}^p} , \end{equation*} where $C_{p, \norm{}}$ is a constant depending only on $p$ and on the choice of the norm $\norm{\cdot}$.

Click to display ⇲

Click to hide ⇱

First, notice that the result only needs to be proved for centered random variables. Indeed, by convexity, for any random variable $X$, \begin{equation*} \mathbb{E}\lrb{\norm{X-\mathbb{E}\lrb{X}}^p} \leq 2^{p-1} \mathbb{E}\lrb{\norm{X}^p + \norm{\mathbb{E}\lrb{X}}^p} \leq 2^p \mathbb{E}\lrb{\norm{X}^p} . \end{equation*} Moreover, by equivalence of norms in finite dimension, the result only needs to be proved for the norm $\norm{\cdot}_p$ on $\rset^d$. Using the Marcinkiewicz–Zygmund inequality in dimension 1 provides \begin{align*} \mathbb{E}\lrb{\norm{\sum_{i=1}^n X_i }_p^p} &= \mathbb{E}\lrb{\sum_{j=1}^d \lrav{ \sum_{i=1}^n X_i(j) }^p} \\ &= \sum_{j=1}^d \mathbb{E}\lrb{\lrav{ \sum_{i=1}^n X_i(j) }^p} \\ &\leq \sum_{j=1}^d C_p \times n^{p/2-1} \times \sum_{i=1}^n \mathbb{E}\lrb{\lrav{X_i(j)}^p} \\ &= C_p \times n^{p/2-1} \times \sum_{i=1}^n \mathbb{E}\lrb{\sum_{j=1}^d \lrav{X_i(j)}^p} \\ &= C_p \times n^{p/2-1} \times \sum_{i=1}^n \mathbb{E}\lrb{\norm{X_i}_p^p} \eqsp. \end{align*}