Before we start, let’s show that $x_i\not=\mu_x$ for some $i$. We’re given that $x_1, \dots, x_n$ is not totally equal i.e. $x_p\not=x_q$ for some $p$, $q$ in $\{1, \dots, n\}$. Therefore, $\mu_x\not= x_p$ or $\mu_x\not= x_q$ because the relation of equality is transitive. Hence, either $i=p$ or $i=q$, or both, satisfy $x_i \not= \mu_x$

We showed that $x_i\not=\mu_x$ for some $i\in \{1, \dots, n\}$. Let’s begin.

(1)

Let $i\in \{1, \dots, n\}$ be such number s.t. $x_i\not=\mu_x$. We have  

• $mx_i+c = y_i^{\prime}$ , using the definition of $y_i^{\prime}$, since $y=mx+c$ is the least squares line to the data.
• $m\mu_x + c = \mu_y = {1\over n}\sum\limits_{i=1}^ny_i = {1\over n}\sum\limits_{i=1}^ny_i^{\prime} = \mu_{y^{\prime}}$ i.e. \[m\mu_x + c = \mu_{y^{\prime}}\], by Proposition 4.5.5.

Then we have \[m(x_i-\mu_x) = y_i^{\prime} - \mu_{y^{\prime}}\]

and \[m = {{y_i^{\prime}-\mu_{y^{\prime}}}\over{(x_i-\mu_x)}} = {{y_i^{\prime}-\mu_{y^{\prime}}}\over{(x_i-\mu_x)^2}}\]

We showed that if $x_i\not= \mu_x$ for some $i$, then $m = {{y_i^{\prime}-\mu_{y^{\prime}}}\over{(x_i-\mu_x)^2}}$.

Then, let $S\subseteq\{1, \dots, n\}$ be the set of such $i$ s.t. $x_i\not=\mu_x$. We have these statements because the rest containing the factor $x_i-\mu_x$ will be zero:

• $\sum\limits_{i\in S}(x_i - \mu_x)^2 = \sum\limits_{i=1}^n(x_i-\mu_x)^2$
• $\sum\limits_{i\in S}(y_i^{\prime} - \mu_{y^{\prime}})(x_i - \mu_x) = \sum\limits_{i=1}^n(y_i^{\prime} - \mu_{y^{\prime}})(x_i - \mu_x)$

Now we’re going to show that $m = {{\sum\limits_{n=1}^n(y_i^{\prime} - \mu_{y^{\prime}})(x_i - \mu_x)}\over{\sum\limits_{i=1}^n(x_i-\mu_x)^2}}$.

Here, $S$ is non-empty because $x_i\not= \mu_x$ for some $i$.

If $S$ has more than two elements, then for each two $j, k\in S$, $x_j\not=\mu_x$ and $x_k\not= \mu_x$, and we showed that here, \[{{(y_j^{\prime} - \mu_{y^{\prime}})(x_j - \mu_x)}\over{(x_j-\mu_x)^2}}=m={{(y_k^{\prime} - \mu_{y^{\prime}})(x_k - \mu_x)}\over{(x_k-\mu_x)^2}}\]

which shows \[{{(y_j^{\prime} - \mu_{y^{\prime}})(x_j - \mu_x) + (y_k^{\prime} - \mu_{y^{\prime}})(x_k - \mu_x)}\over{(x_j-\mu_x)^2 + (x_k-\mu_x)^2}} = {{(y_j^{\prime} - \mu_{y^{\prime}})(x_j - \mu_x)}\over{(x_j-\mu_x)^2}}=m\] because $(x_j-\mu_x)^2 > 0$ and $(x_k-\mu_x)^2 > 0$ as these $x_i$ and $\mu_x$ are real numbers.

Therefore, it’s sufficient to show by induction that

• $m = {{\sum\limits_{i\in S}(y_i^{\prime} - \mu_{y^{\prime}})(x_i - \mu_x)}\over{\sum\limits_{i\in S}(x_i-\mu_x)^2}} $.
• $\sum\limits_{i\in S}(x_i-\mu_x)^2 > 0$ that is, $\sum\limits_{i=1}^n(x_i-\mu_x)^2>0$.

which means \[m={{\sum\limits_{n=1}^n(y_i^{\prime} - \mu_{y^{\prime}})(x_i - \mu_x)}\over{\sum\limits_{i=1}^n(x_i-\mu_x)^2}}.\blacksquare\]

(2)

Since $y_i^{\prime}, \mu_y, x_i, \mu_x, y_i, x_i$ are scalars, \[\sum\limits_{i=1}^n(y_i^{\prime}-\mu_{y^{\prime}})(x_i-\mu_x) = \sum\limits_{i=1}^nx_iy_i^{\prime} - \mu_{y^{\prime}}\sum\limits_{i=1}^nx_i - \mu_x\sum\limits_{i=1}^ny_i^{\prime} + \sum\limits_{i=1}^n\mu_{y^{\prime}}\mu_x\] , where by Proposition 4.4.5, $\sum\limits_{i=1}^nx_iy_i^{\prime} = \sum\limits_{i=1}^nx_iy_i$ and $\mu_{y^{\prime}} = {1\over n}\sum\limits_{i=1}^ny_i^{\prime} = {1\over n}\sum\limits_{i=1}^ny_i = \mu_y$. Then the formula above is equal to \[\sum\limits_{i=1}^nx_iy_i - \mu_{y}\sum\limits_{i=1}^nx_i - \mu_x\sum\limits_{i=1}^ny_i + \sum\limits_{i=1}^n\mu_{y}\mu_x\], which is \[\sum\limits_{i=1}^n(y_i-\mu_{y})(x_i-\mu_x).\]

Therefore, using the conclusion in (1), \[{{\sum\limits_{n=1}^n(y_i^{\prime} - \mu_{y^{\prime}})(x_i - \mu_x)}\over{\sum\limits_{i=1}^n(x_i-\mu_x)^2}} = {{\sum\limits_{i=1}^n(y_i-\mu_{y})(x_i-\mu_x)}\over{\sum\limits_{i=1}^n(x_i-\mu_x)^2}}\] because $\sum\limits_{i=1}^n(x_i-\mu_x)^2 > 0$. Therefore \[m = {{\sum\limits_{i=1}^n(y_i-\mu_{y})(x_i-\mu_x)}\over{\sum\limits_{i=1}^n(x_i-\mu_x)^2}}.\blacksquare\]

註：裡面的歸納法不是標準的歸納法，我主要是想強調透過這個過程，很容易以歸納法證明該小段。