# Bessel's correction demonstration

I am currently trying to understand the proof of the Bessel's correction Proof of correctness 2 and there is one step in the demonstration that I do not understand:

$$\operatorname{Var}(\bar x) = \frac{\sigma^2} n$$

$$\operatorname{Var}(x) = \sigma^2 \text{ and } \bar x = \operatorname{E}(x)$$

If anyone can clarify this step I would really appreciate.

UPDATE

I am stuck at the point where:

$$\operatorname{Var}(\bar x) = \operatorname{E} \left(\left(\frac{\sum_{i=0}^n x_i} n - \mu\right)^2\right) = \frac 1 {n^2} \operatorname{E} \left( \left( \sum_{i=0}^n x_i - n\mu\right)^2\right)$$

• $\bar x$ does not stand for $E(x)$: it is the arithmetic mean of the $n$ data. Because the data are assumed independent, the formula for the mean and basic properties of the variance combine to prove this equality. – whuber Dec 17 '16 at 19:49
• So you get $Var(\bar x) = E((\frac{\sum_{i=0}^{n}{x_i}}{n} - \mu)^2) = \frac{\sigma^2}{n}$, but how ? – Yohan Obadia Dec 17 '16 at 19:59
• The appropriate statement would be $E[\bar{x}]=E[x]$. You should consider adding the self study tag, and update the question to show what you know and where you are stuck. – GeoMatt22 Dec 17 '16 at 20:00
• True that was a wrong shortcut for $\bar x = \frac{1}{n}\sum_{i=0}^{n}x_i$. I will update the question. – Yohan Obadia Dec 17 '16 at 20:04
• It is immediate by computing $$\operatorname{Var}(\bar x)=\operatorname{Var}\left(\frac{1}{n}x_1+\frac{1}{n}x_2+\cdots+\frac{1}{n}x_n\right).$$ – whuber Dec 17 '16 at 20:06

The first thing you need to show is this: $$\frac{\sum_{i=1}^n x_i} n - \mu = \frac 1 n \sum_{i=1}^n (x_i - n\mu).$$
In subtracting fractions, use a common denominator: $$\frac{\sum_{i=1}^n x_i} n - \mu = \frac{\sum_{i=1}^n x_i} n - \frac{n\mu} n = \frac{\left(\sum_{i=1}^n x_i\right) - n\mu} n$$ This is $\dfrac 1 n \left(\left(\sum_{i=1}^n x_i \right) - n\mu\right).$
Next, apply an identity concerning variances: $$\operatorname{var}\left( \frac 1 n Y \right) = \frac 1 {n^2} \operatorname{var}(Y).$$
• The step I was stuck in is just after where you stop. How can you go from $\frac{1}nvar(Y)$ to $\frac{\sigma^2}n$. That means that $var(Y) = n\sigma^2$ however I do not see how. – Yohan Obadia Dec 17 '16 at 21:08
• @YohanObadia : $$\operatorname{var}\left( \sum_{i=1}^n (x_i - n\mu) \right) = \operatorname{var} \left( \sum_{i=1}^n x_i \right) = \sum_{i=1}^n \operatorname{var}(x_i) = \sum_{i=1}^n \sigma^2 = n\sigma^2.$$ The second equality follows from independence of $x_i$, $i=1,\ldots, n. \qquad$ – Michael Hardy Dec 17 '16 at 21:39
• The independence of the $x_i$ allows to move from the second member to the third, but how do you get from the first member to the second ? I mean, from $\operatorname{var}\left( \sum_{i=1}^n (x_i - n\mu) \right)$ to $\operatorname{var} \left( \sum_{i=1}^n x_i \right)$ – Yohan Obadia Dec 18 '16 at 12:40