# Why do we use $S^2$ while estimating the variance?

Sorry the title is a bit silly, but I currently confront a problem related to Fisher's information.

Let $$X_1, X_2, \cdots, X_n$$ be of $$N(\mu , \sigma^2 )$$ distribution where $$\mu$$ is known, $$U^2 := n^{-1} \sum_{i=1}^n (X_i - \mu )^2$$, then $$I_{U^2} (\sigma^2 ) > I_{S^2} (\sigma^2 )$$.

In fact, $$U^2$$ is a sufficient statistic.
So, why do we often use $$S^2$$ to estimate the variance instead of using $$U^2$$?

• Because we don't normally know $\mu$? – Glen_b -Reinstate Monica Oct 18 '19 at 14:40
• For normal data: When $\mu$ is unknown and estimated by $\bar X,$ then $S^2 = \frac{1}{n-1}\sum_i(X_i-\bar X)^2$ has $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1),$ which is a bit hard to prove. When $\mu$ is known, then $U^2 = \frac{1}{n}\sum_i(X_i-\mu)^2$ has $\frac{nU^2}{\sigma^2} \sim \mathsf{Chisq}(n),$ which is obvious. – BruceET Oct 18 '19 at 18:43
• So when we know $\mu$, we choose $U$ instead of $S$ ? – j200932 Oct 19 '19 at 10:01