如何证明

我一直在尝试建立不平等

$| T_{i} | = \frac{| X_{i} - \bar{X} |}{S} \leq \frac{n - 1}{\sqrt{n}}$ $\left| T_i \right|=\frac{\left|X_i -\bar{X} \right|}{S} \leq\frac{n-1}{\sqrt{n}}$

其中 $\bar{X}$ 是样品平均值和 $S$ 样本标准差，即 $S=\sqrt{\frac{\sum_{i=1}^n \left( X_i -\bar{X} \right)^2}{n-1}}$ 。

很容易看到 $\sum_{i=1}^n T_i^2 = n-1$ ，因此 $\left| T_i \right| < \sqrt{n-1}$ 但这与我一直在寻找的目标不是很接近，也不是一个有用的界限。我已经试验了柯西-舒瓦兹（Cauchy-Schwarz）和三角形不等式，但没有成功。我必须在某个地方缺少一个微妙的步骤。谢谢您的帮助。

self-study descriptive-statistics bounds

— 约翰·克
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Answers:

这是Samuelson的不等式，它需要 $\leq$ 符号。如果您采用Wikipedia版本并对其进行 $n-1$ 的定义重做 $S,$ 则会发现它变成

\frac{| X_{i} - \bar{X} |}{S} \leq \frac{n - 1}{\sqrt{n}}

${{ \left| X_i-\bar X \right| } \over S} \leq {{n-1} \over \sqrt{n}}$

— 索克利
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在本书中，这是一个严格的不平等现象，但是我已经解决了，谢谢。

— JohnK

在通过常规程序简化了问题之后，可以通过将其转换为对偶最小化程序来解决该问题，该程序具有众所周知的答案并带有基本证明。也许这种二元化是问题中提到的“微妙步骤”。不平等也可以在纯机械方式通过最大化建立 $|T_i|$ 通过拉格朗日乘数。

不过，首先，我根据最小二乘的几何形状提供了一种更优雅的解决方案。它不需要初步的简化，几乎是立即的，可以直接直观地了解结果。正如问题中所建议的那样，该问题简化为柯西-舒瓦兹不等式。

几何解

将视为具有通常点积的欧几里得空间中的维向量。让和为的正交投影和到的正交补 $\mathbf{x} = (X_1, X_2, \ldots, X_n)$ $n$ 是基础矢量和。写 $\mathbf{y} = (0,0,\ldots,0,1,0,\ldots,0)$ $i^\text{th}$ $\mathbf{1} = (1,1,\ldots, 1)$ $\mathbf{\hat x}$ $\mathbf{\hat y}$ $\mathbf{x}$ $\mathbf{y}$ $\mathbf{1}$ 。（在统计学术语，它们是相对于该装置的残差。）然后，由于和 $X_i-\bar X = \mathbf{\hat x}\cdot \mathbf{y}$ ， $S = ||\mathbf{\hat x}||/\sqrt{n-1}$

| T_{i} | = \sqrt{n - 1} \frac{| \hat{x} \cdot y |}{| | \hat{x} | |} = \sqrt{n - 1} \frac{| \hat{x} \cdot \hat{y} |}{| | \hat{x} | |}

$|T_i| = \sqrt{n-1}\frac{|\mathbf{\hat x} \cdot \mathbf{y}|}{||\mathbf{\hat x}||} = \sqrt{n-1}\frac{|\mathbf{\hat x} \cdot \mathbf{\hat y}|}{||\mathbf{\hat x}||}$

是的部件在方向。由柯西-施瓦茨，它被精确地最大化时是平行于，为此 $\mathbf{\hat y}$ $\mathbf{\hat x}$ $\mathbf{\hat x}$ $\mathbf{\hat y} = (-1,-1,\ldots,-1,n-1,-1,-1,\ldots,-1)/n$ QED。

T_{i} = \pm \sqrt{n - 1} \frac{\hat{y} \cdot \hat{y}}{| | \hat{y} | |} = \pm \sqrt{n - 1} | | \hat{y} | | = \pm \frac{n - 1}{\sqrt{n}},

$T_i = \pm \sqrt{n-1} \frac{\mathbf{\hat y}\cdot \mathbf{\hat y} }{ ||\mathbf{\hat y}||} = \pm\sqrt{n-1}||\mathbf{\hat y}|| = \pm\frac{n-1}{\sqrt{n}},$

顺便说一句，这个解决方案提供的所有情况下的详尽描述最大化：它们都是形式 $|T_i|$

x = σ \hat{y} + μ 1 = σ (- 1, - 1, \dots, - 1, n - 1, - 1, - 1, \dots, - 1) + μ (1, 1, \dots, 1)

$\mathbf{x} = \sigma\mathbf{\hat y} + \mu\mathbf{1} = \sigma(-1,-1,\ldots,-1,n-1,-1,-1,\ldots,-1) + \mu(1,1,\ldots, 1)$

对于所有实数。 $\mu, \sigma$

该分析很容易地概括为被任何一组回归变量代替的情况。显然，最大的正比于剩余的长度，。 $\{\mathbf{1}\}$ $T_i$ $\mathbf{y}$ $||\mathbf{\hat y}||$

简化版

由于在位置和尺度的变化下是不变的，因此我们可以不失一般性地假设为零，并且它们的平方为。这标识与，因为（均方根）是。最大化等于最大化 $T_i$ $X_i$ $n-1$ $|T_i|$ $|X_i|$ $S$ $1$ 。取不会失去一般性 $|T_i|^2 = T_i^2 = X_i^2$ ，或者说，由于是可更换的。 $i=1$ $X_i$

通过双重配方解决

对偶问题是固定的值，并询问剩余值是多少以使平方和最小，假设。因为给定了，所以这是使最小的问题。 $X_1^2$ $X_j, j\ne 1$ $\sum_{j=1}^n X_j^2$ $\sum_{j=1}^n X_j = 0$ $X_1$ 鉴于 $\sum_{j=2}^n X_j^2$ 。 $\sum_{j=2}^n X_j = -X_1$

该解决方案可以通过多种方式轻松找到。最基本的方法之一就是写

X_{j} = - \frac{X_{1}}{n - 1} + ε_{j}, j = 2, 3, \dots, n

$X_j = -\frac{X_1}{n-1} + \varepsilon_j,\ j=2, 3, \ldots, n$

对于其中。扩展目标函数并使用此零和恒等式简化它会产生 $\sum_{j=2}^n \varepsilon_j = 0$

\sum_{j = 2}^{n} X_{j}^{2} = \sum_{j = 2}^{n} {(- \frac{X_{1}}{n - 1} + ε_{j})}^{2} = \sum {(- \frac{X_{1}}{n - 1})}^{2} - 2 \frac{X_{1}}{n - 1} \sum ε_{j} + \sum ε_{j}^{2} = Constant + \sum ε_{j}^{2},

$\sum_{j=2}^n X_j^2 = \sum_{j=2}^n \left(-\frac{X_1}{n-1} + \varepsilon_j\right)^2 = \\\sum \left(-\frac{X_1}{n-1}\right)^2 - 2\frac{X_1}{n-1}\sum \varepsilon_j + \sum \varepsilon_j^2 \\= \text{Constant} + \sum \varepsilon_j^2,$

立即表现出特有的解决方案是对所有。对于此解决方案， $\varepsilon_j=0$ $j$

(n - 1) S^{2} = X_{1}^{2} + (n - 1) {(- \frac{X_{1}}{n - 1})}^{2} = (1 + \frac{1}{n - 1}) X_{1}^{2} = \frac{n}{n - 1} X_{1}^{2}

$(n-1)S^2 = X_1^2 + (n-1)\left(-\frac{X_1}{n-1}\right)^2 = \left(1 + \frac{1}{n-1}\right)X_1^2 = \frac{n}{n-1}X_1^2$

和

| T_{i} | = \frac{| X_{1} |}{S} = \frac{| X_{1} |}{\sqrt{\frac{n}{(n - 1)^{2}} X_{1}^{2}}} = \frac{n - 1}{\sqrt{n}},

$|T_i| = \frac{|X_1|}{S} = \frac{|X_1|}{\sqrt{\frac{n}{(n-1)^2}X_1^2}} = \frac{n-1}{\sqrt{n}},$

QED.

Solution via machinery

Return to the simplified program we began with:

Maximize X_{1}^{2}

$\text{Maximize } X_1^2$

subject to

\sum_{i = 1}^{n} X_{i} = 0 and \sum_{i = 1}^{n} X_{i}^{2} - (n - 1) = 0.

$\sum_{i=1}^n X_i = 0\text{ and }\sum_{i=1}^n X_i^2 -(n-1)=0.$

The method of Lagrange multipliers (which is almost purely mechanical and straightforward) equates a nontrivial linear combination of the gradients of these three functions to zero:

(0, 0, \dots, 0) = λ_{1} D (X_{1}^{2}) + λ_{2} D (\sum_{i = 1}^{n} X_{i}) + λ_{3} D (\sum_{i = 1}^{n} X_{i}^{2} - (n - 1)) .

$(0,0,\ldots, 0) = \lambda_1 D(X_1^2) + \lambda_2 D\left(\sum_{i=1}^n X_i\right ) + \lambda_3 D\left(\sum_{i=1}^n X_i^2 -(n-1)\right).$

Component by component, these $n$ equations are

\begin{aligned} 0 & = 2 λ_{1} X_{1} + & λ_{2} & + 2 λ_{3} X_{1} \\ 0 & = & λ_{2} & + 2 λ_{3} X_{2} \\ 0 & = \dots \\ 0 & = & λ_{2} & + 2 λ_{3} X_{n} . \end{aligned}

$\eqalign{ 0 &= 2\lambda_1 X_1 +& \lambda_2 &+ 2\lambda_3 X_1 \\ 0 &= & \lambda_2 &+ 2\lambda_3 X_2 \\ 0 &= \cdots \\ 0 &= & \lambda _2 &+ 2\lambda_3 X_n. }$

The last $n-1$ of them imply either $X_2 = X_3 = \cdots = X_n = -\lambda_2/(2\lambda_3)$ or $\lambda_2=\lambda_3=0$ . (We may rule out the latter case because then the first equation implies $\lambda_1=0$ , trivializing the linear combination.) The sum-to-zero constraint produces $X_1 = -(n-1)X_2$ . The sum-of-squares constraint provides the two solutions

X_{1} = \pm \frac{n - 1}{\sqrt{n}}; X_{2} = X_{3} = \dots = X_{n} = \mp \frac{1}{\sqrt{n}} .

$X_1 = \pm\frac{n-1}{\sqrt{n}};\ X_2 = X_3 = \cdots = X_n = \mp\frac{1}{\sqrt{n}}.$

They both yield

| T_{i} | = | X_{1} | \leq | \pm \frac{n - 1}{\sqrt{n}} | = \frac{n - 1}{\sqrt{n}} .

$|T_i| = |X_1| \le |\pm\frac{n-1}{\sqrt{n}}| = \frac{n-1}{\sqrt{n}}.$

— whuber
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Thank you for your addendum, geometry is very powerful and of all three solutions it is the most intuitive to me.

— JohnK

The inequality as stated is true. It is quite clear intuitively that we get the most difficult case for the inequality (that is, maximizing the left hannd side for given $S^2$ ) by choosing one value, say $x_1$ as large as possible while having all the others equal. Let us look at an example with such configuration:

n = 4, x_{1} = x_{2} = x_{3} = 0, x_{4} = 4, \bar{x} = 1, S^{2} = 4,

$n=4, \quad x_1=x_2=x_3=0, x_4=4, \bar{x}=1, S^2=4,$ now

\frac{| x_{i} - \bar{x} |}{S} = {\begin{cases} \frac{1}{2} or \\ \frac{3}{2} \end{cases}

$\frac{|x_i-\bar{x}|}{S}=\begin{cases} \frac12 ~\text{or}~ \\ \frac32 \end{cases}$ depending on

i

$i$ , while the given upper limit is equal to

\frac{4 - 1}{2} = 1.5

$\frac{4-1}{2}=1.5$ which is just enough. That idea can be completed to a proof.

EDIT

We will now prove the claim, as hinted above. First, for any given vector $x=(x_1, x_2, \dots, x_n)$ in this problem, we can replace it with $x-\bar{x}$ without changing either side of the inequality above. So, in the following let us assume that $\bar{x}=0$ . We can also by relabelling assume that $x_1$ is largest. Then, by choosing first $x_1>0$ and then $x_2=x_3=\dots=x_n=-\frac{x_1}{n-1}$ we can check by simple algebra that we have equality in the claimed inequality. So, it is sharp.

Then, define the (convex) region $R$ by

R = {x \in R : \bar{x} = 0, \sum (x_{i} - \bar{x})^{2} / (n - 1) \leq S^{2}}

$R = \{ x\in\mathbb{R} \colon \bar{x}=0, \sum(x_i-\bar{x})^2/(n-1) \le S^2\}$ for a given positive constant

S^{2}

$S^2$ . Note that

R

$R$ is the intersection of a hyperplane with a sphere centered at the origin, so is a sphere in

(n - 1)

$(n-1)$ -space. Our problem can now be formulated as

max_{x \in R} max_{i} | x_{i} |

$\max_{x\in R} \max_i |x_i|$ since an

x

$x$ maximizing that will be the most difficult case for the inequality. This is a problem of finding the maximum of a convex function over a convex set, which in general are difficult problems (minimums are easy!). But, in this case the convex region is a sphere centered on the origin, and the function we want to maximize is the absolute value of the coordinates. It is obvious that that maximum is found at the boundary sphere of

R

$R$ , and by taking

| x_{1} |

$|x_1|$ maximal, our first test case is forced.

— kjetil b halvorsen
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@JohnK you can delete your comments now, the post is corrected

— kjetil b halvorsen

Although this answer shows that the inequality (assuming it is true, which it is) is tight, it isn't evident how that single calculation could be "completed to a proof." Could you provide some indication of how that would be done?

— whuber

Will, but tomorrow, now I have to prepare tomorrows class.

— kjetil b halvorsen

Thank you--I appreciate your careful formulation of the problem. But your "proof" seems to come to the statement that "it is obvious that." You could always apply Lagrange multipliers to finish the job, but it would be nice to see an approach that (a) actually is a proof and (b) provides insight.

— whuber

@whuber If you have the time, I would appreciate it if you can post your Lagrange multipliers solution. I think the inequality overall is not as famous as it should be.

— JohnK