<Dd> E ⁡ (S 2) = E ⁡ (1 n ∑ i = 1 n (X i − μ) 2 − 2 n (X _̄ − μ) ∑ i = 1 n (X i − μ) + (X _̄ − μ) 2) = E ⁡ (1 n ∑ i = 1 n (X i − μ) 2 − 2 n (X _̄ − μ) ⋅ n ⋅ (X _̄ − μ) + (X _̄ − μ) 2) = E ⁡ (1 n ∑ i = 1 n (X i − μ) 2 − 2 (X _̄ − μ) 2 + (X _̄ − μ) 2) = E ⁡ (1 n ∑ i = 1 n (X i − μ) 2 − (X _̄ − μ) 2) = E ⁡ (1 n ∑ i = 1 n (X i − μ) 2) − E ⁡ ((X _̄ − μ) 2) = σ 2 − E ⁡ ((X _̄ − μ) 2) = (1 − 1 n) σ 2 <σ 2 . (\ displaystyle (\ begin (aligned) \ operatorname (E) (S ^ (2)) & = \ operatorname (E) (\ bigg () (\ frac (1) (n)) \ sum _ (i = 1) ^ (n) (X_ (i) - \ mu) ^ (2) - (\ frac (2) (n)) ((\ overline (X)) - \ mu) \ sum _ (i = 1) ^ (n) (X_ (i) - \ mu) + ((\ overline (X)) - \ mu) ^ (2) (\ bigg)) \ \ (8pt) & = \ operatorname (E) (\ bigg () (\ frac (1) (n)) \ sum _ (i = 1) ^ (n) (X_ (i) - \ mu) ^ (2) - (\ frac (2) (n)) ((\ overline (X)) - \ mu) \ cdot n \ cdot ((\ overline (X)) - \ mu) + ((\ overline (X)) - \ mu) ^ (2) (\ bigg)) \ \ (8pt) & = \ operatorname (E) (\ bigg () (\ frac (1) (n)) \ sum _ (i = 1) ^ (n) (X_ (i) - \ mu) ^ (2) - 2 ((\ overline (X)) - \ mu) ^ (2) + ((\ overline (X)) - \ mu) ^ (2) (\ bigg)) \ \ (8pt) & = \ operatorname (E) (\ bigg () (\ frac (1) (n)) \ sum _ (i = 1) ^ (n) (X_ (i) - \ mu) ^ (2) - ((\ overline (X)) - \ mu) ^ (2) (\ bigg)) \ \ (8pt) & = \ operatorname (E) (\ bigg () (\ frac (1) (n)) \ sum _ (i = 1) ^ (n) (X_ (i) - \ mu) ^ (2) (\ bigg)) - \ operatorname (E) (\ bigg () ((\ overline (X)) - \ mu) ^ (2) (\ bigg)) \ \ (8pt) & = \ sigma ^ (2) - \ operatorname (E) \ left (((\ overline (X)) - \ mu) ^ (2) \ right) = \ left (1 - (\ frac (1) (n)) \ right) \ sigma ^ (2) <\ sigma ^ (2). \ end (aligned))) </Dd> <P> In other words, the expected value of the uncorrected sample variance does not equal the population variance σ, unless multiplied by a normalization factor . The sample mean, on the other hand, is an unbiased estimator of the population mean μ . </P> <P> Note that the usual definition of sample variance is S 2 = 1 n − 1 ∑ i = 1 n (X i − X _̄) 2 (\ displaystyle S ^ (2) = (\ frac (1) (n - 1)) \ sum _ (i = 1) ^ (n) (X_ (i) - (\ overline (X)) \,) ^ (2)), and this is an unbiased estimator of the population variance . </P> <P> This can be seen by noting the following formula, which follows from the Bienaymé formula, for the term in the inequality for the expectation of the uncorrected sample variance above: E ⁡ ((X _̄ − μ) 2) = 1 n σ 2 (\ displaystyle \ operatorname (E) (\ big () ((\ overline (X)) - \ mu) ^ (2) (\ big)) = (\ frac (1) (n)) \ sigma ^ (2)) </P>

Why is sample mean an unbiased estimator of population mean