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If there is a sample with sequence $ x_1 , x_2 , ... , x_n$ for example, that is taken randomly from a population, what exactly are these elements $ x_1 , x_2 , ... , x_n$? What do they represent? Every time I search for an explanation, all they do is take some numbers $ 2,5,3,1,6,3$ for example and estimate the mean $\mu$ of the population, but what do these elements $ 2,5,3,1,6,3$ represent?

I'm asking this question because I'm trying to prove the expected value of the sample mean $\bar x$

$ \bar x = \frac {\sum x_i}{n} = \frac {x_1 + x_2 + ... + x_n}{n}$

$E( \bar x) = \frac {1}{n} [E(x_1) + E(x_2) + ... + E(x_n)]$

I'm stuck here because I don't know what $x_1,x_2,...,x_n$ represent and what does it mean to take the expected value of them, I know each expected value of them should be equal to $\mu$ but I don't know why, hopefully someone enlightens me, thank you.

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  • $\begingroup$ Your first sentence answers your question. They are $n$ values chosen randomly from the population. $\endgroup$ – John Douma Mar 8 '19 at 15:11
  • $\begingroup$ Then what does it mean to take the expected value of a distinct number? $\endgroup$ – cazanova Mar 8 '19 at 15:15
  • $\begingroup$ You can find that in any Probability textbook. Another way to find out would be here. $\endgroup$ – John Douma Mar 8 '19 at 15:31
  • $\begingroup$ These $x_i$ are the observations you make in your sample, and these all came from some underlying distribution with mean $\mu$. Therefore we can expect any observation to be equal to $\mu$. $\endgroup$ – WaveX Mar 8 '19 at 16:30
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    $\begingroup$ The sample mean $\overline X$ has a probability distribution of its own (the sampling distribution). In any case, $E(\overline X)$ should make sense as the expected value of the random variable $\overline X$. $\endgroup$ – StubbornAtom Mar 8 '19 at 18:29
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A lecturer wants to know the average age of the people in a large auditorium. It's obviously impractical to ask everyone his age age, so he selects $10$ people at random and asks them their ages. The $10$ people chosen are the elements of a random sample. The average of their ages is the sample mean.

EDIT

Suppose we have a set of $1000$ distinct integers. We select $10$ integers at random and compute their average. The $10$ integers chosen are the elements of the sample, and their average is the sample mean. But we could perform this experiment again, choosing another random sample of $10$ numbers (after replacing the first sample) and computing their average. Say we do this repeatedly. Each time we do this, we get a value of a random variable $X.$ You are being asked to show that $E(X)$ is the average of all the numbers in the population, so that taking the sample mean is actually a reasonable way of estimating the population average.

In answer to your second comment, $x_1$ is a random variable whose value is the first number selected at random, so it makes perfect sense to ask for its expectation. When we say $x_1=3$ we mean that in a particular experiment, the first number selected was $3,$ not that $x_1$ is a symbol representing the integer $3.$

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  • $\begingroup$ That's nice, suppose the set $x_i$ $ i = 0,1,2..,10$ are the ages we collected, and I can calculate the sample mean $ \bar x$ , now how can I get the expected value of the sample mean? Normally when we take the expected value of a discrete random variable, it's $E(x) = \sum x_i f(x_i)$ , but in my question, $x_i$ are just single numbers, how can I take the expectation? $\endgroup$ – cazanova Mar 8 '19 at 16:44
  • $\begingroup$ @cazanova I'm not entirely sure I understand you, but I'll edit my answer to try to explain this. $\endgroup$ – saulspatz Mar 8 '19 at 16:51
  • $\begingroup$ In this equation: $E( \bar x) = \frac {1}{n} [E(x_1) + E(x_2) + ... + E(x_{10})]$ You said these $x_i$ will represent a random sample of the ages. In this context, what is $E(x_1) , E(x_2) , ...$ ? To evaluate them, they are supposed to be random variables with probability distributions, but $x_i$ are just single elements taken from the population, if I'm missing something please inform me, thank you. $\endgroup$ – cazanova Mar 8 '19 at 16:57
  • $\begingroup$ I understand now the logic, but only one problem left, how are we sure that $E(x_1) = E(x_2) .... = E(x_{10}) = \mu$ ? I'm sorry $\endgroup$ – cazanova Mar 8 '19 at 17:14
  • $\begingroup$ That is what you are asked to prove. But it's very simple. Just use the formula you wrote down for the expectation in your first comment. $\endgroup$ – saulspatz Mar 8 '19 at 17:17

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