1
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ $\bar x$ 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 '16 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 '16 at 19:59
  • 1
    $\begingroup$ 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 '16 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 '16 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 '16 at 20:06
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '16 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 '16 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 '16 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.