\sum_{i=1}^{n} (x_i - \bar{x}) = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} \bar{x} = \sum_{i=1}^{n} x_i - n\bar{x} = \sum_{i=1}^{n} x_i - n \cdot \frac{1}{n} \sum_{i=1}^{n} x_i = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} x_i = 0. \tag{6} \begin{array}{l} E \left(b ^ {*}\right) = E \left(\frac {S _ {x Y}}{S _ {x x}}\right) = E \left(\sum_ {i = 1} ^ {n} \frac {x _ {i} - \bar {x}}{S _ {x x}} Y _ {i}\right) = \sum_ {i = 1} ^ {n} \frac {x _ {i} - \bar {x}}{S _ {x x}} E \left(Y _ {i}\right) = \sum_ {i = 1} ^ {n} \frac {x _ {i} - \bar {x}}{S _ {x x}} (a + b x _ {i}) \\ = \frac {\sum_ {i = 1} ^ {n} \left(x _ {i} - \bar {x}\right) \left(a + b x _ {i}\right)}{S _ {x x}} = \frac {\sum_ {i = 1} ^ {n} \left(\left(x _ {i} - \bar {x}\right) a + \left(x _ {i} - \bar {x}\right) b x _ {i}\right)}{S _ {x x}} \\ = \frac {\sum_ {i = 1} ^ {n} \left(x _ {i} - \bar {x}\right) a + \sum_ {i = 1} ^ {n} \left(x _ {i} - \bar {x}\right) b x _ {i}}{S _ {x x}} = \frac {a \sum_ {i = 1} ^ {n} \left(x _ {i} - \bar {x}\right) + b \sum_ {i = 1} ^ {n} \left(x _ {i} - \bar {x}\right) x _ {i}}{S _ {x x}} \\ = \frac {a \cdot 0 + b \sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) x _ {i} - b \bar {x} \cdot 0}{S _ {x x}} = \frac {b \sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) x _ {i} - b \bar {x} \sum_ {i = 1} ^ {n} (x _ {i} - \bar {x})}{S _ {x x}} \\ = \frac {b \sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) x _ {i} - b \sum_ {i = 1} ^ {n} \bar {x} (x _ {i} - \bar {x})}{S _ {x x}} = \frac {b (\sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) x _ {i} - \sum_ {i = 1} ^ {n} \bar {x} (x _ {i} - \bar {x}))}{S _ {x x}} \\ = \frac {b \sum_ {i = 1} ^ {n} ((x _ {i} - \bar {x}) x _ {i} - \bar {x} (x _ {i} - \bar {x}))}{S _ {x x}} = \frac {b \sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) ^ {2}}{S _ {x x}} = \frac {b S _ {x x}}{S _ {x x}} = b, \end{array}

där ledningen användes i den andra likheten och (6) i den nionde och tionde likheten. Det följer att $b^{*}$ är väntevärdesriktig vilket skulle visas.