Can I see i.i.d. variables as just one? I'm trying to understand the part of this book, page 276 which explains about sample mean:

In statistical inference, a central problem is how to use data to
estimate unknown parameters of a distribution, or functions of unknown
parameters. It is especially common to want to estimate the mean and
variance of a distribution. If the data are i.i.d. random variables $X_1,... ,X_n$
where the mean $E(X_j)$ is unknown, then the most
obvious way to estimate the mean is simply to average the $X_j$ , taking
the arithmetic mean.
For example, if the observed data are $3, 1, 1, 5$, then a simple,
natural way to estimate the mean of the distribution that generated
the data is to use $(3+1+1+5)/4 = 2.5$. This is called the sample mean.

My question is Instead of saying "... i.i.d. random variables $X_1,... ,X_n$ where the mean $E(X_j)$ is unknown" can I write "let X be a random variable and take the $n$ outputs of $X$, where E(X) is unknown". In another words, can I see these $X_1,... ,X_n$ as only an one random variable? since the $X_1,... ,X_n$ are independent and identically distributed?
 A: Yes, rolling one die $n$ times is the same as rolling $n$ dice once, but...

"In an experimental situation, it would be very unusual to observe
only the value of one random variable [...] It would be a modest
experiment indeed if the only datum collected was the body weight of
one person. Rather, the body weights of several people in the
population might be measured. These different weights would be
observations on different random variables, one for each person
measured [...] Thus, we need to know how to describe and use
probability models that deal with more than one random variable at a
time" (Casella & Berger, Statistical Inference, Duxbury, 2002, §4.1,
p. 139)

Rolling dice... ok, but measuring the body weight of one person $n$ times is not the same as measuring the body weights of $n$ people, administering a drug to one person $n$ times is not the same as amministering a drug to $n$ people.
This is why a random sample is defined as $n$ random variables, not as the reiteration of a single random variable.
A: What you are proposing would involve a revision of probability theory, using different terminology and notation.  As it stands, in probability theory a random variable denotes a single "random object" (technically it is a mapping from the sample space to some outcome space, usually a number set).  We sometimes use random vectors which are vectors composed of multiple random scalars, so that is the usual way to accomodate this kind of situation.  In that case we use notation like $\mathbf{X} = (X_1,...,X_n)$ to collectively denote each of the individual random variables under consideration.
What you are proposing is a different method that has some problems.  Aside from being an unecessary re-invention of an existing theory that already works well, there are a few issues with the approach you are suggesting:

*

*You are using the same notational object to describe $n$ different random numbers.  So it then becomes quite difficult for you to express probability results where these outputs are equal to different things.  For example, even simple probability statements where the outputs are equal to different things become hard to express.


*Even assuming you can find an alternative way to express this, it would surely involve some kind of notation differentiating the $n$ outputs, and so it would be tantamount to giving them different notation to begin with.  It is unclear how your method would allow you to make useful probability statements in a way that is no more complicated than in the existing theory.


*Even assuming you can solve this issue, your method still amounts to the implicit idea that seperate "outputs" of a single random variable are IID outcomes of that random variable.  So all this really does is to treat this as implicit instead of as an explicit probability statement on separate objects.
