clarification on Variance formula [duplicate]

It is possible to calculate the variance using two formulas. one is dividing by n and the other is dividing by n-1. Thus, my question is that is the result we get by dividing (n) and (n-1) the same? Example

• Obviously not the same. Commented Aug 1 at 9:54
• @Lucenaposition, if the numerator is zero, then it doesn't matter what the denominator is. Otherwise sure. Commented Aug 1 at 14:04
• You definitively should look at this video on Youtube (youtube.com/watch?v=xslIhnquFoE), and visit the website (which explains the "mistery of n-1" in 3 parts) here (civil.uwaterloo.ca/brodland/EasyStats/EasyStats/…). This hould answer al your questions, and more. Commented Aug 1 at 18:07

If the observations are all equal, the two calculations give the same result (zero). Otherwise, the calculations will differ.

PROOF

(This proof assumes at least two (possibly equal) $$X_i$$. Without that assumption, the $$n-1$$ formula does not even make sense.)

Define $$t = \overset{n}{\underset{i = 1}{\sum}}\left(X_i - \bar X\right)^2$$. Then define $$\hat\sigma^2_p=\dfrac{t}{n-p}$$.

First, let all $$X_i$$ be equal. Then $$\bar X = X_i$$, so $$t=0$$, and $$\hat\sigma^2_0 = \hat\sigma^2_1 = 0$$. This proves the first part.

Next, let the $$X_i$$ not all be equal. Then $$t \ne 0$$. If $$\hat\sigma^2_p = \hat\sigma^2_{p^{\prime}}$$, then $$\dfrac{t}{n-p} = \dfrac{t}{n-p^{\prime}}$$. Cross-multiply.

$$tn - tp^{\prime} = tn - tp\\ n - p^{\prime} = n - p\\ p = p^{\prime}$$

This shows that, for $$t\ne0$$, $$\hat\sigma^2_p = \hat\sigma^2_{p^{\prime}}$$ implies $$p = p^{\prime}$$.

Taking the contrapositive, for $$t\ne0$$, $$p\ne p^{\prime}$$ implies $$\hat\sigma^2_p \ne \hat\sigma^2_{p^{\prime}}$$. As $$0\ne 1$$, $$\hat\sigma^2_0 \ne \hat\sigma^2_1$$.

• The proof can be made much simpler division (where defined) is bijective (because its inverted by re-multiplication) and $n\ne n-1,$ QED.
– whuber
Commented Aug 1 at 14:38