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Proof that the Sample Variance (using n-1) is an Unbiased Estimator of the Population Variance

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The Formal Theorem

Let Y1,Y2,…,Yn Y_1, Y_2, \dots, Y_n be independent and identically distributed random variables from a population with mean μ \mu and variance σ2 \sigma^2 . The sample variance, defined as S2=1n−1∑i=1n(Yi−Yˉ)2 S^2 = \frac{1}{n-1} \sum_{i=1}^n (Y_i - \bar{Y})^2 , where Yˉ=1n∑i=1nYi \bar{Y} = \frac{1}{n} \sum_{i=1}^n Y_i , is an unbiased estimator of the population variance σ2 \sigma^2 . That is,
E[S2]=σ2 E[S^2] = \sigma^2

Analytical Intuition.

Imagine trying to estimate the 'spread' of an entire city's population (the population variance, σ2 \sigma^2 ) by measuring the heights of just a small group of people (a sample). If we used the average height of our sample to calculate the 'spread' within that sample, we'd consistently underestimate the true spread of the entire city. This is because our small sample is less likely to contain extreme values. The n−1 n-1 in the denominator acts like a 'correction factor', slightly inflating our estimate of the sample spread. This inflation precisely counteracts the underestimation tendency, ensuring that, on average, our sample variance calculation accurately reflects the true population variance. It's like a subtle nudge that keeps our estimate honest.
CAUTION

Institutional Warning.

The key confusion lies in the denominator. Using n n leads to a downward bias because the sample mean Yˉ \bar{Y} is used instead of the true population mean μ \mu .

Academic Inquiries.

01

Why do we use n−1 n-1 and not n n for sample variance?

Using n−1 n-1 (Bessel's correction) is crucial because the sample mean Yˉ \bar{Y} is used in the calculation of the sample variance. Since Yˉ \bar{Y} is itself an estimate derived from the sample, it tends to be closer to the sample observations than the true population mean μ \mu . This proximity leads to smaller squared deviations from Yˉ \bar{Y} compared to deviations from μ \mu , resulting in a downward bias if n n were used. The n−1 n-1 corrects for this bias.

02

What does it mean for an estimator to be 'unbiased'?

An estimator is unbiased if its expected value is equal to the true value of the parameter it is estimating. In simpler terms, if you were to take many, many samples and calculate the sample variance for each, the average of all those sample variances would converge to the true population variance.

03

Can the sample variance ever be biased?

The sample variance calculated with n−1 n-1 is unbiased for the population variance. However, if you were to calculate the 'population variance' from a sample using n n in the denominator, that estimator *would* be biased (specifically, it would be a biased estimator of the population variance).

04

Is the proof of unbiasedness applicable to any distribution?

Yes, the proof that S2 S^2 is an unbiased estimator of σ2 \sigma^2 relies on the independence of the random variables and the definition of variance, not on the specific distribution of those variables, as long as they have finite mean and variance.

Standardized References.

  • Definitive Institutional SourceCasella, George, and Roger L. Berger. Statistical Inference. Cengage Learning, 2001.

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Institutional Citation

Reference this proof in your academic research or publications.

NICEFA Visual Mathematics. (2026). Proof that the Sample Variance (using n-1) is an Unbiased Estimator of the Population Variance: Visual Proof & Intuition. Retrieved from https://nicefa.org/library/applied-statistics/proof-that-the-sample-variance--using-n-1--is-an-unbiased-estimator-of-the-population-variance

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