Why do you divide by n-1 for sample variance

When you calculate variance from a sample, you first compute the sample mean $\bar{x}$ and then measure how each observation deviates from that mean. Those deviations are not independent of the data because the mean itself is a function of the same observations. As a result, the raw sum of squared deviations tends to underestimate the true variability of the underlying population.

Why the divisor is n‑1

The expectation of the sum of squared deviations equals $(n-1)\sigma^2$, where $\sigma^2$ is the population variance. Dividing by $n-1$ therefore yields an estimator $s^2$ whose expected value is exactly $\sigma^2$. If you divided by $n$ you would obtain $\frac{n-1}{n}\sigma^2$, which is systematically too low. This bias disappears only as the sample size grows, but for small samples the correction is essential.

Key reasons for using n‑1:

• Ensures unbiasedness of the variance estimator
• Accounts for the loss of one degree of freedom when estimating the mean
• Provides a more accurate estimate of population spread, especially for small n

Deriving the unbiased divisor:

1. 1Write the sample variance with a generic divisor d: $s_d^2 = \frac{1}{d}\sum_{i=1}^n (x_i-\bar{x})^2$
2. 2Take expectations: $E[\sum (x_i-\bar{x})^2] = (n-1)\sigma^2$
3. 3Set $E[s_d^2] = \sigma^2$ and solve for d, giving d = n‑1

Worked example with real numbers:

If you mistakenly divide the same sum of squares by $n$ you obtain $5$ for the example above, which underestimates the true population variance. The bias becomes larger as the sample size shrinks; with $n=2$ the difference between dividing by 2 and by 1 is a factor of two. Remember that Bessel's correction restores unbiasedness by acknowledging the one lost degree of freedom.