Giá trị làm tăng độ lệch chuẩn

Tôi bối rối trước câu nói sau:

"Để tăng độ lệch chuẩn của một tập hợp số, bạn phải thêm một giá trị lớn hơn một độ lệch chuẩn so với giá trị trung bình"

Là gì bằng chứng về điều đó? Tất nhiên tôi biết làm thế nào chúng ta xác định độ lệch chuẩn nhưng phần đó tôi dường như bỏ lỡ bằng cách nào đó. Có ý kiến ​​gì không?

Đối với bất kỳ số y 1 , y 2 , ... , y N với trung bình ˉ y = $N$$y_1,y_2, \ldots, y_N$, phương sai được cho bởi σ $\displaystyle \bar{y} = \frac{1}{N}\sum_{i=1}^N y_i$ Áp dụng(1)vào tập nhất địnhnsốx1,x2,...xn mà chúng tôi mất để thuận tiện trong triển lãm có bìnhˉx=0, ta có σ2=

$$\begin{align} \sigma^2 &= \frac{1}{N-1}\sum_{i=1}^N (y_i-\bar{y})^2\\ &= \frac{1}{N-1}\sum_{i=1}^N \left(y_i^2 - 2y_i\bar{y} + \bar{y}^2\right)\\ &= \frac{1}{N-1}\left[\left(\sum_{i=1}^Ny_i^2\right) - 2N(\bar{y})^2 + N(\bar{y})^2 \right] \\ \sigma^2 &=\frac{1}{N-1}\sum_{i=1}^N \left(y_i^2 - (\bar{y})^2\right) \tag{1} \end{align}$$ $(1)$$n$$x_1, x_2, \ldots x_n$$\bar{x} = 0$ Nếu bây giờ chúng ta thêm vào một quan sát mớixn+1để tập dữ liệu này, sau đó giá trị trung bình mới của tập dữ liệu là $$\sigma^2 = \frac{1}{n-1}\sum_{i=1}^n \left(x_i^2-(\bar{x})^2\right) = \frac{1}{n-1}\sum_{i=1}^n x_i^2$$ $x_{n+1}$ trong khi phương sai mới là σ $$\frac{1}{n+1}\sum_{i=1}^{n+1}x_i = \frac{n\bar{x} + x_{n+1}}{n+1} = \frac{x_{n+1}}{n+1}$$ $$\begin{align} \hat{\sigma}^2 &= \frac{1}{n}\sum_{i=1}^{n+1} \left(x_i^2-\frac{x_{n+1}^2}{(n+1)^2}\right)\\ &= \frac{1}{n}\left[\left((n-1)\sigma^2 + x_{n+1}^2\right) - \frac{x_{n+1}^2}{n+1}\right]\\ &= \left.\left.\frac{1}{n}\right[(n-1)\sigma^2 + \frac{n}{n+1}x_{n+1}^2\right]\\ &> \sigma^2 ~ \text{only if}~ x_{n+1}^2 > \frac{n+1}{n}\sigma^2. \end{align}$$ So $|x_{n+1}|$ needs to be larger than $\displaystyle\sigma\sqrt{1+\frac{1}{n}}$$x_{n+1}$$\bar{x}$$\displaystyle\sigma\sqrt{1+\frac{1}{n}}$, in order for the augmented data set to have larger variance than the original data set. See also Ray Koopman's answer which points out that the new variance is larger than, equal to, or smaller than, the original variance according as $x_{n+1}$ differs from the mean by more than, exactly, or less than $\displaystyle\sigma\sqrt{1+\frac{1}{n}}$.

The puzzling statement gives a necessary but insufficient condition for the standard deviation to increase. If the old sample size is $n$, the old mean is $m$, the old standard deviation is $s$, and a new point $x$ is added to the data, then the new standard deviation will be less than, equal to, or greater than $s$ according as $|x-m|$ is less than, equal to, or greater than $s\sqrt{1+1/n}$.

Leaving aside the algebra (which also works) think about it this way: The standard deviation is square root of the variance. The variance is the average of the squared distances from the mean. If we add a value that is closer to the mean than this, the variance will shrink. If we add a value that is farther from the mean than this, it will grow.

This is true of any average of values that are non-negative. If you add a value that is higher than the mean, the mean increases. If you add a value that is less, it decreases.

I'll get you started on the algebra, but won't take it quite all of the way. First, standardize your data by subtracting the mean and dividing by the standard deviation:

$$Z = \frac{x-\mu}{\sigma} .$$ Note that if $x$ is within one standard deviation of the mean, $Z$ is between -1 and 1. Z would be 1 if $x$ were exactly one sd away from the mean. Then look at your equation for standard deviation: $$\sigma = \sqrt{\frac{\sum_{i=1}^{N}Z_i^2}{N-1}}$$ What happens to $\sigma$ if $Z_N$ is between -1 and 1?