Bounding sum of quartic deviations from sample mean

Question

[Cross-posted here with no answers for a few days]

I came - to the very best of my knowledge from reading the source - across the following statement in The Jackknife and Bootstrap, Shao and Tu, p. 87:

$$\sum_i(X_i-\bar X)^4\leq16\sum_i(X_i-\mu)^4,$$

where $\bar X$ and $\mu$ are the sample mean and expected value of the $X_i$, respectively.

The statement appears in the broader context of their Theorem 3.8, which of course has several conditions. However, since the result seems to be algebraical, I suppose none of these matter for the present question.

Even without the term 16, the statement is well-known (for any $a$, not only $\mu$) when we replace 4 by 2.

My hunch is that the proof uses Pascal's triangle, as there are 16 terms in the r.h.s. when adding and subtracting $\mu$ and multiplying out:

$$\sum_i(X_i-\bar X)^4=\sum_i(X_i-\mu)^4+4(X_i-\mu)^3(\mu-\bar X)+4(X_i-\mu)(\mu-\bar X)^3+6(X_i-\mu)^2(\mu-\bar X)^2+(\mu-\bar X)^4$$

My attempts at bounding the terms on the r.h.s. (other than the first one, which is already in the "right" format), including Hölder's inequality, however have not led anywhere useful.

Answer 1

So, I believe we have a proof - actually provided by my PhD Student Stephan Hetzenecker, who is however not here, so I'll post it in his name!

W.l.o.g., let $\mu = 0$ (otherwise define $Z_i := X_i - \mu$ and show $\sum_i(Z_i-\bar Z)^4 \leq 16 \sum_i Z_i^4 $). Using the binomial theorem, we obtain \begin{align} \tag{1}\label{eq:sum_binom} \sum_i(X_i-\bar X)^4= &\sum_i \left( X_i^4- 4X_i^3\bar X -4X_i\bar X^3 +6X_i^2\bar X^2 +{\bar X}^4\right). \end{align} Let $p \geq 1$. Using Jensen's inequality, we get \begin{align} \tag{2}\label{eq:applyJensen_p} \bar X^p := \left(\frac{1}{n} \sum_i X_i \right)^p \leq \frac{1}{n} \sum_i X_i^p. \end{align}

Hence, applying \eqref{eq:applyJensen_p} $n$ times for $p=4$, \begin{align} \tag{3}\label{eq:bound_barx4} \sum_i \bar X^4 = \sum_i \left(\frac{1}{n} \sum_j X_j \right)^4 \leq \sum_i \frac{1}{n} \sum_j X_j^4 = \sum_j X_j^4 . \end{align} Similarly, using Hölder's inequality with $p=q=2$ and using \eqref{eq:bound_barx4}, \begin{align} \tag{4}\label{eq:bound_barx2} \sum_i X_i^2 \bar X^2 &\leq \left(\sum_i X_i^4 \right)^{1/2} \left(\sum_i \bar X^4 \right)^{1/2}\\& \leq \left(\sum_i X_i^4 \right)^{1/2} \left( \sum_i X_i^4 \right)^{1/2} \\&= \sum_j X_j^4 . \end{align} Furthermore, using \eqref{eq:applyJensen_p} again \begin{align} \tag{5}\label{eq:bound_barx3} - \sum_i X_i\bar X^3 &= - n \bar X^4 \leq n \bar X^4 \\&\leq n \left(\frac{1}{n} \sum_j X_j^4 \right) \\&= \sum_j X_j^4 \end{align} Using Hölder's inequality with $p=4/3$ and $q=4$ and using \eqref{eq:bound_barx4}, \begin{align} \tag{6}\label{eq:bound_barx} -\sum_i X_i^3\bar X &\leq \sum_i \vert X_i^3\bar X \vert \\&\leq \left( \sum_i X_i^4 \right)^{3/4} \left( \sum_i \bar X^4 \right)^{1/4} \\&\leq \sum_i X_i^4 \end{align}

Combining \eqref{eq:sum_binom}, \eqref{eq:bound_barx4}, \eqref{eq:bound_barx2}, \eqref{eq:bound_barx3}, and \eqref{eq:bound_barx}, we obtain \begin{align*} \sum_i(X_i-\bar X)^4 \leq 16 \sum_i X_i^4, \end{align*} which was to be shown.

Answer 2

Let $A = 16\sum\limits_{i=1}^{n}(X_i-\mu)^4-\sum\limits_i(X_i-\bar X)^4$

$=\sum\limits_{i=1}^{n}\left(4(X_i-\mu)^2+(X_i-\bar X)^2\right)\left(2(X_i-\mu)+(X_i-\bar X)\right)\left(2(X_i-\mu)-(X_i-\bar X)\right)$

$=\sum\limits_{i=1}^{n}\left(4a_i^2+b_i^2\right)\left(2a_i+b_i\right)\left(2a_i-b_i\right)$, by letting $X_i-\mu=a_i$ and $X_i-\bar{X}=b_i$

Note that $\sum\limits_{i=1}^{n} b_i = \sum\limits_{i=1}^{n}X_i - n\bar{X}= n\bar{X}- n\bar{X} = 0$

Now, notice if $X_i$s are ordered in an increasing sequence, i.e., w.l.o.g. let $X_1\leq X_2\leq\ldots\leq X_n$, we shall have $a_1\leq a_2\leq\ldots \leq a_n$ and $b_1 \leq b_2 \leq \ldots b_n$, since $\mu, \bar{X}$ are constants.

$\implies 4a_1^2+b_1^2\leq4a_2^2+b_2^2\leq\ldots\leq 4a_n^2+b_n^2$ and $2a_1+b_1\leq 2a_2+b_2\leq\ldots 2a_n+b_n$

$\implies (4a_1^2+b_1^2)(2a_1+b_1)\leq(4a_2^2+b_2^2)(2a_2+b_2)\leq\ldots\leq (4a_n^2+b_n^2)(2a_n+b_n)$, since $4a_i^2+b_i^2 \geq 0$, being sum of squares of real numbers

Similarly, $2a_i-b_i=X_i-2\mu+\bar{X}$

$\implies 2a_1-b_1\leq 2a_2-b_2\leq\ldots\leq 2a_n-b_n$, since $X_1\leq X_2\ldots\leq X_n$

Now, let's use the following inequality from Chebyshev:

Now, applying Chebysev's inequality twice, we have,

$\frac{A}{n}=\frac{1}{n} \sum\limits_{i=1}^{n}\left(4a_i^2+b_i^2\right)\left(2a_i+b_i\right)\left(2a_i-b_i\right)\geq \left(\frac{1}{n} \sum\limits_{i=1}^{n}(4a_i^2+b_i^2)(2a_i+b_i)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i-b_i)\right)$

$\geq \left(\left(\frac{1}{n}\sum\limits_{i=1}^{n}(4a_i^2+b_i^2)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i+b_i)\right)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i-b_i)\right)$

$= \left(\frac{1}{n} \sum\limits_{i=1}^{n}(4a_i^2+b_i^2)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i)\right)$, since $\sum\limits_{i=1}^{n}b_i=0$

$\geq \left(\frac{1}{n} \sum\limits_{i=1}^{n}(4a_i^2+b_i^2)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i)\right)^2 \geq 0$, being product of sum of squares of real numbers

$\implies A = 16\sum\limits_{i=1}^{n}(X_i-\mu)^4-\sum\limits_i(X_i-\bar X)^4\geq 0$

$\implies \sum\limits_i(X_i-\bar X)^4 \leq 16\sum\limits_{i=1}^{n}(X_i-\mu)^4$