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I am reffering to the formula given at 2:18 in following video

https://www.youtube.com/watch?v=6i7mqDJICzQ

Can someone explain how he arrived from this:

$\bar{X}=\frac{1}{N-1}\sum_{i=1}^Nx_{i}$

to:

$E(\bar{X})=\frac{1}{N-1}\sum_{i=1}^NE(x_{i})$

The first formula is the average of values of $x_i$. The second formula should be the "average of averages", right? If $N$ is the sample size, how can second formula be true?

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    $\begingroup$ Linearity of expectation. $\endgroup$ – Christoph Hanck Dec 25 '16 at 19:53
  • $\begingroup$ @ChristophHanck Could you please explain. $\endgroup$ – Quirik Dec 25 '16 at 23:25
  • $\begingroup$ See your own comment below, I think you got it yourself already. $\endgroup$ – Christoph Hanck Dec 25 '16 at 23:30
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X_bar is a random variable but the denominator customarily would be N rather than N-1. The asymptotics works either way. The second equation would follow by linearity of the expectation as has already been mentioned. Again the formula customarily would have N in the denominator. The x_i actually represent a sequence of random variables. It may be easier to think about the special case when they are independent and identically distributed. Do not consider them as the actual sample values.

E(x_i) is the population mean m. The second equation shows that the sample mean has the same expected value as the individual observations. That does not mean that a realization of the sample mean is always m. Rather if you repeatedly generate samples and compute the sample mean each time the arithmetic average of all these realizations will be close to m and converge to m as the number of repetitions gets large.

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  • $\begingroup$ $N$ would be sample size? What do you mean when you say that $x_i$ should be considered as i.i.d. random values but not as actual sample values? $\endgroup$ – Quirik Dec 25 '16 at 22:09
  • $\begingroup$ This is the key point that may be causing you confusion. I said that they represent a sequence of random variables with the same probability distribution. $\endgroup$ – Michael R. Chernick Dec 25 '16 at 22:40
  • $\begingroup$ Maybe an example will help. Let X be a random variable with the distribution N(0,1) a normal distribution with mean 0 and variance 1. X can take on any value on the real line. X can take on the values when taking a sample. The values tend to be close to 0 and the sample distribution should approximate the normal distribution. Expectation represents the average of the particular random variable. $\endgroup$ – Michael R. Chernick Dec 25 '16 at 22:46
  • $\begingroup$ You make many assumptions that are not necessary for the equation in the question to be true, Michael: the only reason to insist on $1/N$ instead of $1/(N-1)$ is your assumption of an unstated context and that the bar has a conventional meaning; there is no need for the $X_i$ to be independent; and they needn't have a common distribution. The only necessary assumption is that all those expectations actually exist. $\endgroup$ – whuber Dec 25 '16 at 23:41
  • $\begingroup$ @whuber The OP is having trouble understanding the difference between a sample and a random variable. The other two answers that are given fall into the same trap. The 1/N is necessary because otherwise E(X_bar) =N (mu)/(N-1) which is not equal to mu. I realize that it is asymptotically unbiased. $\endgroup$ – Michael R. Chernick Dec 26 '16 at 1:02

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