# Bounding sum of quartic deviations from sample mean

[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.

• It would be sufficient to show $\sum(\bar X-\mu)^4\leq\sum(X_i-\mu)^4$, which certainly seems as though it should be true but doesn't seem all that easy to prove. Jun 22, 2021 at 2:30
• Indeed, that seems true, and amply so, as playing around with code like n <- 10; mu <- 4; x <- rnorm(n, mu); xbar <- mean(x); sum((xbar-mu)^4); sum((x-mu)^4) seems to suggest, so that one might expect the result to be easy to show - apparently, not. Jun 22, 2021 at 6:35
• I think the difficult case will be when $\mu$ is far from the data (which is unlikely but not impossible if it's really the mean) Jun 22, 2021 at 22:10

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.

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

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

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