Bessel's Correction I'm trying to understand the intuition behind Bessel's correction where 
$\sum (x_i - \overline{x})^2 / (n-1)$.
My difficulty is stemming from the fact that the sample mean leads to a standard deviation that, when comparing it to hypothetical population means, is ALWAYS smaller. 
Why is it that the mean of a sample results in producing a standard deviation that will never be lower by using another mean? 
 A: I think, the most clear argument in favour of making such a correction is that: $\frac{\sum\limits_{i=1}^n (x_{i} - \bar x)^2}{\sigma^2} \sim \chi^{2}_{n-1}$ (Fisher's Lemma). So, $\mathbb{E} \sum\limits_{i=1}^n (x_{i} - \bar x)^2 = \sigma^2 \cdot (n - 1) \Rightarrow \ $ the unbiased estimate of $\sigma^2$ would be $\frac{\sum\limits_{i=1}^n (x_{i} - \bar x)^2}{n - 1}$ 
A: If I understand you correctly, you're asking

Why does using $\bar x$ in $\sum(x_i-\bar x)^2/(n-1)$ give a smaller result than using any other value?

You can find the answer by looking at the quadratic function
$$
f(z)=\sum_{i=1}^n\frac{(x_i-z)^2}{n-1}
$$
If we dfferentiate it (with respect to $z$), we see that the derivative, which is a linear function in $z$ and thus only has one root, has a root at $z=\bar x$. Which means that $f(z)$ has its minimum there.
A: The use of $n-1$ instead of $n$ fixes the bias in estimation of the population variance.
Indeed:
$$\sigma^2=\frac{\sum(X-\mu)^2}{N};s^2=\frac{\sum(X-\bar{X})^2}{n-1}\\
\mathbb E(s^2)=\mathbb E\left(\frac{\sum (X-\bar{X})^2}{n-1}\right)=\frac{1}{n-1}\mathbb E(\sum X^2-n\bar{X}^2)=\\
\frac1{n-1}\left[\sum \mathbb E(X^2)-n\mathbb E(\bar{X}^2)\right]=\\
\frac1{n-1}\left[n(\sigma^2+\mu^2)-n\left(\frac{\sigma^2}{n}+\mu^2\right)\right]=\frac{1}{n-1}\cdot [(n-1)\sigma^2]=\sigma^2,$$
where:
$$Var(X)=\sigma^2=\mathbb E(X^2)-(\mathbb E(X))^2=\mathbb E(X^2)-\mu^2\Rightarrow \mathbb E(X^2)=\sigma^2+\mu^2\\
Var(\bar{X})=\mathbb E(\bar{X}^2)-(\mathbb E(\bar{X}))^2\Rightarrow \mathbb E(\bar{X}^2)=Var\left(\frac{\sum X}{n}\right)+\left(\mathbb E\left(\frac{\sum X}{n}\right)\right)^2=
\frac{\sigma^2}{n}+\mu^2.$$
