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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 $U^2$?

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    $\begingroup$ Because we don't normally know $\mu$? $\endgroup$ – Glen_b -Reinstate Monica Oct 18 '19 at 14:40
  • $\begingroup$ 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. $\endgroup$ – BruceET Oct 18 '19 at 18:43
  • $\begingroup$ So when we know $\mu$, we choose $U$ instead of $S$ ? $\endgroup$ – j200932 Oct 19 '19 at 10:01

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