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Let $x_1, \ ... \, x_n \in \Bbb R$ and $b \in \Bbb R$. Assume that $z_k = x_k + b, \ k = 1, \ ... \, n$ and $\bar z = \bar x + b$.

Show that

$s_z^2 = s_x^2.$

Here,

$$\bar x = {1 \over n} \sum_{k=1}^n x_k$$

and

$$s_x^2 = {1 \over n - 1} \sum_{k=1}^n (x_k - \bar x)^2.$$

The same definition applies for $\bar z$ and $s_z^2$.

In the end, I must reach

$${1 \over {n - 1}} (\sum_{k=1}^n x_k^2 - 2\sum_{k=1}^n x_k \bar x + \sum_{k=1}^n \bar x^2) $$

by definition of $s_x^2$. But some factors seem to disappear. In the following, I will use the fact that $\bar x = {1 \over n} \sum_{k=1}^n x_k \Rightarrow \sum_{k=1}^n x_k = n\bar x$. For the sake of simplicity, I will cut out the factor ${1 \over n - 1}$ and only concentrate on the sums.

\begin{align} & \sum_{k=1}^n z_k^2 - 2\sum_{k=1}^n z_k \bar z + \sum_{k=1}^n \bar z^2\\ & = \sum_{k=1}^n (x_k + b)^2 - 2(\bar x + b)\sum_{k=1}^n (x_k + b) + \sum_{k=1}^n (\bar x + b)^2 \\ & = \sum_{k=1}^n (x_k + b)^2 - 2(\bar x + b)n(\bar x + b) + n (\bar x + b)^2 \\ & = \sum_{k=1}^n (x_k + b)^2 - 2n(\bar x + b)^2 + n (\bar x + b)^2 \\ & = \sum_{k=1}^n (x_k + b)^2 - n(\bar x + b)^2 \\ & = \sum_{k=1}^n (x_k^2 + 2x_kb + b^2) - n\bar x^2 -2n {\bar x}b -nb^2 \\ & = (\sum_{k=1}^n x_k^2) + 2n \bar x b + nb^2 - n\bar x^2 -2n {\bar x}b -nb^2 \\ & = \sum_{k=1}^n x_k^2 - n\bar x^2 \\ & = \sum_{k=1}^n x_k^2 - \sum_{k=1}^n \bar x^2 \end{align}

I don't see how to go from here to the desired result.

Borol
  • 491

1 Answers1

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Note that

$${1 \over {n - 1}} \left(\sum_{k=1}^n x_k^2 - 2\sum_{k=1}^n x_k \bar x + \sum_{k=1}^n \bar x^2\right) = {1 \over {n - 1}} \left(\sum_{k=1}^n x_k^2 - 2n \bar x^2 + \sum_{k=1}^n \bar x^2\right) = {1 \over {n - 1}} \left(\sum_{k=1}^n x_k^2 - n\bar x^2\right) $$

Intuitively, you can argue that when $x_i$ changes by a constant $b$, then sample mean changes by $b$. But since the spread of the distribution does not change, the variance of the sample remains the same.

Roby5
  • 4,287