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Given a sample of $n$ data, $x_1, \dots, x_n$. Define the sample mean $$\bar x := \frac{1}{n}(x_1+\cdots+x_n),$$ and sample variance $$s^2 := \frac{1}{n-1} \sum_{i=1}^n (x_i-\bar x)^2.$$ To measure how far away an individual data is from the bulk, define $$t_i := \frac{x_i - \bar x}{s}.$$ The question is how to show that for all $j$ $$\left|t_j\right| < \frac{n-1}{\sqrt{n}}.$$ It is sufficient to show that $$t_j^2 = \frac{(n-1)(x_j - \bar x)^2}{\sum_{i=1}^n (x_i-\bar x)^2} < \frac{(n-1)^2}{n}.$$ This reduced to $$n(x_j - \bar x)^2 < (n-1)\sum_{i=1}^n (x_i-\bar x)^2.$$ Then $$(x_j - \bar x)^2 < (n-1)\sum_{i\neq j}^n (x_i-\bar x)^2.$$ How to proceed from here, please? Thank you!

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  • $\begingroup$ You must at least use $\le$ instead of $<,\;$ otherwise $x_1=0, x_2=2\;$ is a counter-example with $|t_1|=|t_2|=\frac{n-1}{\sqrt{n}}=\frac{1}{\sqrt{2}}$ $\endgroup$ – gammatester Dec 10 '14 at 8:06
  • $\begingroup$ Also see stats.stackexchange.com/questions/172461/…. $\endgroup$ – StubbornAtom May 17 '18 at 11:28
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You have $\displaystyle \frac1n \sum_{j=1}^n t_j = 0$, so $\displaystyle \sum_{j=1}^n t_j = 0$ and $\displaystyle \sum_{j\not =i} t_j = -t_i$, and thus $\displaystyle \sum_{j\not =i} t_j^2 \ge \frac{t_i^2}{n-1}$.

You also have $\displaystyle \frac1{n-1} \sum_{j=1}^n t_j^2 = 1$, so $\displaystyle \sum_{j=1}^n t_j^2 = n-1$ and $\displaystyle \sum_{j\not =i} t_j^2 = n-1-t_i^2$.

Putting these together gives $n-1-t_i^2 \ge \dfrac{t_i^2}{n-1}$ and so $|t_i| \le \dfrac{n-1}{\sqrt{n}}$.

If this had been a population with $\displaystyle s^2 := \frac{1}{n} \sum_{i=1}^n (x_i-\bar x)^2$ then you would have $|t_i| \le {\sqrt{n-1}}$.

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  • $\begingroup$ Thanks. But how to get "thus $\displaystyle \sum_{j\not =i} t_j^2 \ge \frac{t_i^2}{n-1}$", please? $\endgroup$ – LaTeXFan Dec 10 '14 at 22:58
  • $\begingroup$ Also I do not know what you mean in the last line. $\endgroup$ – LaTeXFan Dec 10 '14 at 23:07
  • $\begingroup$ Try the Cauchy–Schwarz inequality where $\displaystyle \left(\sum_{j=1}^m x_j y_j \right)^2 \leq \sum_{i=1}^m x_i^2 \sum_{k=1}^m y_k^2$ where $m=n-1$, $x_j=1$ and $y_j=t_j$ $\endgroup$ – Henry Dec 11 '14 at 0:18
  • $\begingroup$ The last line says that if you use the (slightly lower) population variance instead of the sample variance, then the bound on the standardised values $t_i$ is slightly looser $\endgroup$ – Henry Dec 11 '14 at 0:22
  • $\begingroup$ Do you mean replace $n-1$ by $n$, please? If so, I am afraid there is a typo in the last line. $\endgroup$ – LaTeXFan Dec 11 '14 at 0:25

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