I thought that the sample distribution was an approximation of the distribution of the underlying phenomenon.

But then the book says:

We will denote the sample size by $n$ ($n \le N$) and the values of the sample members by $X_1, X_2, \dots , X_n$. It is important to realize that each $X_i$ is a random variable. In particular, $X_i$ is not the same as $x_i$ : $X_i$ is the value of the $i$-th member of the sample, which is random and $x_i$ is that of the $i$-th member of the population, which is fixed.

I don't understand this distinction. I thought that also $x_i$ should be considered as random; after all, they are all realizazion from an underlying probability distribution. So even the population mean $\mu = \frac 1N\sum x_i$ must be seen as a random variable.

Then I realized we were talking about different experiment (ie the $x_i$ will be considered random when the population is created (so to speak) while will be considered constant and fixed in the contest of the survey we are performing).

Take a look at $Var \ \bar X = \frac{\sigma^2}n\left(\frac{N-n}{N-1}\right)$ If $N=n$, it implies $Var \ \bar X = 0$, that is, if we interview all the population we will find that $\bar X$ is really a constant ($= \mu$).. this brings me to the question:

The sample distribution then is the distribution of what? Apparently isn't the distribution of the underlying phenomenon, but the distribution that arise as having $x_i$ realization and taking a random $n$ between them.. without asking where the $x_i$ come from.

So the sample distribution must be used only as a measure of the accuracy of $\bar X$ to estimate $\mu$, but neither $\bar X$ nor $\mu$ can be seen as estimates of what we really want, that is the $E(Y)$, where the distribution of $Y$ (the underlying distribution for all the population) is what we care about.

I suppose one can do pretty much the same reasoning and conclude that $\mu$ is an unbiased estimator for $E(Y)$ and maybe try to derive it's variance, but then it's not clear to me how to connect all of this with what we actually have ($X_i$).

Also, I think this point should be made more explicit (if it's correct, that is), because it was a source of confusion for me.

• Please have a look on this recent question: stats.stackexchange.com/questions/141416 I don’t get your point of $\overline X$... – Elvis Mar 14 '15 at 15:39
• @Elvis Okay your answer seems to confirm my point about the sample distribution. My problem is that it's not clear to me how to go from $\bar X$ to $E(Y)$.. How to infer the true mean of the underlying probability distribution? To continue your example, when born every Parisinian will have a certain pre-determined height, so his height can be regarded as random variable ($Y$) with a certain probaility distribution.. Shouldn't we care about this probability distribution (and about $E(Y)$) rather that the population mean? – Ant Mar 14 '15 at 15:44
• We might want another example as height is not pre-determined at birth... but let’s admit it. If it were the case, there would be no difference between the pre-determined height at birth and the measured height at adult age. Sampling adults and measuring them is a good way to make inference about the distribution of [pre-determined] height. – Elvis Mar 14 '15 at 15:50
• @Elvis Right. But this inference is a different one than the one we make when we talk about sampling distributions, right? So how one would infer the distribution of pre-determined height by sampling adults? – Ant Mar 14 '15 at 15:52
• I think your (legitimate) problem lies in the fact that in this kind of example the total population is finite. When you sample from a finite population, there is some randomness in the sample but the total population is fixed. It is always a problem to tell something like the distribution of the height of male Parisians is $\mathcal N(175,15^2)$ — which is a continuous distribution, while the true distribution in this experiment is a discrete one. – Elvis Mar 14 '15 at 15:57