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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:

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

When we admit that:

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

If anyone can clarify this step I would really appreciate.

UPDATE

I am stuck at the point where:

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

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  • $\begingroup$ $\bar x$ 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. $\endgroup$
    – whuber ♦
    Dec 17, 2016 at 19:49
  • $\begingroup$ So you get $Var(\bar x) = E((\frac{\sum_{i=0}^{n}{x_i}}{n} - \mu)^2) = \frac{\sigma^2}{n}$, but how ? $\endgroup$
    – Yohan Obadia
    Dec 17, 2016 at 19:59
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    $\begingroup$ 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. $\endgroup$
    – GeoMatt22
    Dec 17, 2016 at 20:00
  • $\begingroup$ True that was a wrong shortcut for $\bar x = \frac{1}{n}\sum_{i=0}^{n}x_i$. I will update the question. $\endgroup$
    – Yohan Obadia
    Dec 17, 2016 at 20:04
  • $\begingroup$ It is immediate by computing $$\operatorname{Var}(\bar x)=\operatorname{Var}\left(\frac{1}{n}x_1+\frac{1}{n}x_2+\cdots+\frac{1}{n}x_n\right).$$ $\endgroup$
    – whuber ♦
    Dec 17, 2016 at 20:06

1 Answer 1

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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: $$ \operatorname{var}\left( \frac 1 n Y \right) = \frac 1 {n^2} \operatorname{var}(Y). $$

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  • $\begingroup$ 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. $\endgroup$
    – Yohan Obadia
    Dec 17, 2016 at 21:08
  • $\begingroup$ @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 $x_i$, $i=1,\ldots, n. \qquad$ $\endgroup$
    – Michael Hardy
    Dec 17, 2016 at 21:39
  • $\begingroup$ 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 $\operatorname{var} \left( \sum_{i=1}^n x_i \right)$ $\endgroup$
    – Yohan Obadia
    Dec 18, 2016 at 12:40

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