Why is the sum of squared deviations divided by n a biased estimator of the population variance?

## The claim Take a random sample `x_1, ..., x_n` from a population with mean `mu` and variance `sigma^2`. Form the sum of squared deviations around the sample mean: ``` SS = sum (x_i - xbar)^2 ``` The result to know is ``` E[SS] = (n - 1) * sigma^2 ``` So `SS / n` has expected value `(n-1)/n * sigma^2`, which is smaller than `sigma^2` — a downward bias — while `SS / (n-1)` has expected value exactly `sigma^2`. ## Why the bias points downward The intuitive argument is a minimisation argument, and it is the one to say out loud first. Consider the function `g(a) = sum (x_i - a)^2`. Differentiating and setting to zero gives `a = xbar`: the sample mean is the unique value that makes the sum of squared deviations as small as possible. The population mean `mu` is some other number (it coincides with `xbar` only by fluke), so ``` sum (x_i - xbar)^2 <= sum (x_i - mu)^2 ``` always, for every sample. The quantity we can compute is guaranteed to be no larger than the quantity we wish we could compute. Averaging over samples, it is strictly smaller. The sample mean has been fitted to this very data, and the data therefore look less spread out around it than they really are around the truth. ## The algebra Add and subtract `mu` inside the square: ``` sum (x_i - mu)^2 = sum ((x_i - xbar) + (xbar - mu))^2 = sum (x_i - xbar)^2 + 2*(xbar - mu)*sum (x_i - xbar) + n*(xbar - mu)^2 ``` The cross term vanishes because `sum (x_i - xbar) = 0` by construction. Rearranged: ``` sum (x_i - xbar)^2 = sum (x_i - mu)^2 - n*(xbar - mu)^2 ``` Now take expectations term by term. Each `E[(x_i - mu)^2] = sigma^2`, giving `n * sigma^2` for the first piece. The second piece is `n` times the variance of the sample mean, which is `sigma^2 / n`, giving `sigma^2`. Hence ``` E[SS] = n*sigma^2 - sigma^2 = (n - 1)*sigma^2 ``` One unit of `sigma^2` was consumed by the act of estimating the centre from the same data. ## Degrees of freedom, said properly The n deviations `x_i - xbar` are not n free numbers: they satisfy one linear constraint, `sum (x_i - xbar) = 0`. Tell me any n-1 of them and the last is determined. So they carry n-1 independent pieces of information about spread, and dividing the sum of their squares by n-1 is dividing by the amount of information actually present. This is the general rule — one degree of freedom is spent per parameter estimated from the data before measuring residual spread — and it is why the correction generalises far beyond this one formula. ## How large is the effect The bias factor `(n-1)/n` is 0.5 at n = 2, 0.9 at n = 10, 0.99 at n = 100 and 0.999 at n = 1000. So the correction is decisive for tiny samples and cosmetic for large ones. Note also that the bias shrinks to zero as n grows: the n-divisor version is biased at every finite sample size yet still converges to `sigma^2` — biased but consistent. ## Three traps **"n-1 makes it bigger, and bigger is safer."** Wrong reasoning even though the arithmetic direction is right. The correction is not a safety margin; it is the exact factor that makes the expected value land on `sigma^2`. **"So the sample standard deviation is unbiased too."** It is not. The n-1 divisor makes the *variance* estimator unbiased; taking a square root is a non-linear operation, and the resulting standard deviation is biased slightly low. Unbiasedness does not pass through square roots. **"You always divide by n-1."** Only when the centre was estimated from the same data. If the population mean `mu` were genuinely known, `sum (x_i - mu)^2 / n` is already unbiased for `sigma^2` — nothing was fitted, so nothing was spent. That contrast is the cleanest way to show an interviewer you understand the mechanism rather than the ritual. ## Saying it in an interview The compact answer: the sample mean is fitted to the data and sits at the point that minimises squared deviations, so deviations around it understate the true spread; the expected sum of squares is `(n-1)*sigma^2`, and dividing by n-1 rather than n corrects that exactly. If pressed, run the add-and-subtract-mu algebra, then connect n-1 to the single constraint the deviations obey.