Estimating variance given the mean Assume we have a normal distribution with known mean $\mu$. How can we estimate the variance by sampling?
The typical answer to this question is to use the unbiased sample variance estimator i.e. if the data points are indicated by $x_1, \ldots x_n$ the following:
$$\frac{(x_1-\bar{x})^2+\ldots +(x_n-\bar{x})^2}{n-1}$$
Where $\bar{x}$ is the sample mean. Now can we use the actual mean in any meaningful way to get a better estimator of the variance? The first thing that comes to mind would be to replace $\bar{x}$ by $\mu$ and divide it by $n$ instead of $n-1$ (to keep it unbiased). Is this a better estimator? why?
 A: Let's check if $\hat\sigma^2_{you}=\dfrac{\sum_{i = 1}^n (x_i - \mu)^2}{n-1}$ is unbiased for $\sigma^2$.
$$
\mathbb E\Bigg[
\dfrac{\sum_{i = 1}^n (x_i - \mu)^2}{n-1}
\Bigg] \\
=
\dfrac{1}{n-1} \mathbb E\Bigg[
\sum_{i = 1}^n (x_i - \mu)^2
\Bigg] \\=
\dfrac{1}{n-1} \sum_{i = 1}^n\mathbb E\Bigg[
 (x_i - \mu)^2
\Bigg] \\ =
\dfrac{1}{n-1} \sum_{i = 1}^n\mathbb E\Bigg[
 x_i^2  -2\mu x_i + \mu^2
\Bigg] \\=
\dfrac{1}{n-1} \sum_{i = 1}^n\Bigg[\mathbb E\big[x_i^2\big] -2\mu\mathbb E\big[x_i\big]+\mu^2\Bigg]\\=
\dfrac{1}{n-1} \sum_{i = 1}^n\Bigg[\mathbb E\big[x_i^2\big] -2\mu^2+\mu^2\Bigg]\\=
\dfrac{1}{n-1} \sum_{i = 1}^n\Bigg[\mathbb E\big[x_i^2\big] -\mu^2\Bigg]
$$
Now, $\mathbb E\big[x_i^2\big]=\mathbb E\big[x_i\big]^2 + \mathbb Var(x_i)$, so:
$$
\dfrac{1}{n-1} \sum_{i = 1}^n\Bigg[\mathbb E\big[x_i^2\big] -\mu^2\Bigg]\\=
\dfrac{1}{n-1} \sum_{i = 1}^n\Bigg[\mathbb E\big[x_i\big]^2 + \mathbb Var(x_i) -\mu^2\Bigg] \\=
\dfrac{1}{n-1} \sum_{i = 1}^n\Bigg[\mu^2 + \mathbb Var(x_i) -\mu^2\Bigg] \\=
\dfrac{1}{n-1} \sum_{i = 1}^n\mathbb Var(x_i) \\=
 \dfrac{n}{n-1} \mathbb Var(x_i)
$$
Consequently, $\hat\sigma^2_{you}$ is biased for $\sigma^2$!
However, if you redo that calculation with a $n$ denominator, you get an unbiased estimator for $\sigma^2$.
$$\hat\sigma^2_{unbiased}=\dfrac{\sum_{i = 1}^n (x_i - \mu)^2}{n}$$
Whether or not this $n$-denominator is the best estimator is a matter of opinion, but it is unbiased. For better or for worse, most choices about what estimator to use come down to a matter of opinion.
A: Suppose you have a random sample of size $n$ from the
population $\mathsf{Norm}(\mu, \sigma),$ where $\sigma$ is not known and $\mu$ is known.
Let $V = \frac 1n\sum_{i=1}^n (X_i - \mu)^2.$
Then $V$ is a better estimate of the population variance $\sigma^2$ than is $S^2=\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2,$ where $\bar X =\frac 1 n \sum_{i=1}^n X_i.$
Also, a 95% CI for $\sigma^2$ tends to be narrower if we use $V$ than
if we use $S^2.$ [Samples can vary, so this CI is not always narrower.]
In particular, a 95% CI for $\sigma^2$ is based on the
relationship $\frac{nV}{\sigma^2} \sim \mathsf{Chisq}(\nu = n).$
Example: Suppose I have the sample x of size $n = 50$
from  $\mathsf{Norm}(\mu = 20, \sigma = 3),$ where I assume $\mu$ is known and $\sigma$ is not.
set.seed(215)
x = rnorm(50, 20, 3)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  14.21   17.97   19.94   20.30   22.62   29.31 

v = (sum((x-20)^2))/50;  v
[1] 10.69335

CI.1 = 50*v/qchisq(c(.975,.025), 50);  CI.1
[1]  7.486223 16.523827
diff(CI.1)
[1] 9.037604    # width of CI

The formula for this confidence interval is
$\left(\frac{50V}{U}, \frac{50V}{L}\right),$ where $L$
and $U$ cut probabilities $0.025$ from the lower and upper
tails, respectively, of $\mathsf{Chisq}(\nu=50).$
For the data of my example, the CI is $(7.49\, 16.52)$ of width $9.04.$
By contrast, the 95% CI for $\sigma^2$ based on $S^2,$ where $\mu$ is estimated by $\bar X,$ uses the relationship
$\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu=49).$
CI.2 = 49*var(x)/qchisq(c(.975,.025), 49);  CI.2
[1]  7.548087 16.797538
diff(CI.2)
[1] 9.249451   # wider CI

For the data of my example, the CI is $(7.55,\, 16.80)$ of width $9.25 > 9.04.$
