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In my Stats course (can't ask prof, it's an online course), we got to random variables and the notation has changed from using $s^2$, $\bar{x}$, etc...to using what we were initially taught are the symbols used to describe population as opposed to sample; namely $\sigma$, $\mu$, etc...

Why are we using population symbols when talking about random variables? Do we not use random variables for prediction or inference? And if we do, then aren't those samples?

I thought I understood that population is the entire data set we're interested in and that inference is done using a sample of that population, because if we already had the entire data set we're interested in there's no inference to be done since we have all the information we need.

It seems to me that if we want to talk about the probability of something happening, we're interested in the behavior of elements we don't currently have in our data set; that should imply that what we currently have is not the population, but a sample.

In this case, why not use population symbols and for example $\frac{1}{n-1}$ instead of $\frac{1}{n}$?

EDIT: For example, when talking about the variance of a random variable, the formula given to us is: $$\sigma^2 = Var(X) = E[X^2] - (\mu_x)^2 $$

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  • $\begingroup$ Could you edit your post to include a quote where your lecturer uses $\sigma$, $\mu$ etc. in a way you don't understand? $\endgroup$
    – Stephan Kolassa
    Jul 4, 2016 at 18:56
  • $\begingroup$ @StephanKolassa, I edited my post to include an example. Specifically I just don't understand why we're using the greek laters supposedly reserved for describing population tendencies, instead of using sample symbols. I thought any time we are interested in doing prediction/inference we're basically dealing with a sample; so are random variables not used for any of that? $\endgroup$
    – jeremy radcliff
    Jul 4, 2016 at 19:07
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    $\begingroup$ In your example, although $X$ is a random variable, $\operatorname{Var}(X)$ is a number: it is a property of that random variable. Thus, there seems to be no violation of the convention of using lower case Greek letters for properties and other letters for statistics and random variables. And please beware of the "population" metaphor: although it may be linguistically and conceptually convenient, it simply does not apply in many situations where a process or system is being studied or where random variables are used to model other things besides finite samples. $\endgroup$
    – whuber ♦
    Jul 4, 2016 at 19:15
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    $\begingroup$ Just as English often uses the same word to mean different things--and the listener is assumed to develop the correct understanding from context--so it goes in statistics (and even mathematics). In your example, "Var" is also a statistic: a calculation performed on a sample. ("Mean" also behaves similarly.) In your question, though, "$X$" was explicitly stipulated to be a random variable, not a sample. Thus "Var" is understood to mean a property of $X$ qua random variable: the other sense of "Var" as a statistic just doesn't apply. $\endgroup$
    – whuber ♦
    Jul 4, 2016 at 19:23
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    $\begingroup$ @whuber, I think this makes sense. And you're right with the comment you left under Kostia's answer. I think rigor is what I need because I keep finding myself looking for some structural coherence that I can't find given the lack of rigor in the way the concepts are initially explained to me (not prof's fault, it's supposed to be an intro course, so i think the expectation is that you won't go crazy on your own all the time.). I'll keep in mind what you said and keep learning, I'm sure things will fall in place more and more. $\endgroup$
    – jeremy radcliff
    Jul 4, 2016 at 19:31

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You can think of a set of values a random variable can take (called sample space) as a population. For example, if $X\sim\mathcal{N}(\mu,\sigma^2)$ is a normal random variable with mean $\mu$ and variance $\sigma^2$, then the underlying population (sample space) is the set of all real numbers $\mathbb{R}$. Distribution $\mathcal{N}(\mu,\sigma^2)$ is essentially a sampling law: it describes how sample are obtained from the population. If $x_1,\ldots,x_n$ is a sample form $\mathcal{N}(\mu,\sigma^2)$, then the sample mean $\bar{x}$ is an approximation of the "theoretical" mean $\mu$.

Do we not use random variables for prediction or inference?

Yes, we do use random variable for prediction and inference, but this statement is too general. More precisely, in the context of statistical inference, we use random variable to model the variability observed in the data.

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  • $\begingroup$ I guess my confusion is that I've come to associate (perhaps erroneously) the idea of doing inference and predictions with using a sample instead of a population. So in my mind it goes something like "Ok, we're using probability so obviously we want to make predictions about future behavior and likelihood of something happening. Therefore we're using a sample since we're interested in elements no currently in our data set. So...why are we using population symbols? Why is this a population?" $\endgroup$
    – jeremy radcliff
    Jul 4, 2016 at 19:13
  • $\begingroup$ I realize you're trying to give intuitive, informal explanations, but equating sample spaces, populations, and random variables on the one hand, and distributions and "sampling laws" on the other seems to ignore some of the distinctions the OP is attempting to understand. $\endgroup$
    – whuber ♦
    Jul 4, 2016 at 19:13
  • $\begingroup$ @whuber Yes, I agree with your comment. But in this question I simply don't know how to provide intuitive explanation without sacrificing some rigor. I think at this level, it is ok to think of sample space as population, etc. $\endgroup$
    – Kostia
    Jul 4, 2016 at 19:20
  • $\begingroup$ @jeremyradcliff You may find Section 6 from the following notes useful. arxiv.org/pdf/1603.04929v1.pdf Please let me know if this helps. If not I will try my best to provide further explanations. $\endgroup$
    – Kostia
    Jul 4, 2016 at 19:23
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    $\begingroup$ From comments posted by the OP, my impression is that rigor is exactly what he needs. He seems to be suffering from a lack of it. But--to anticipate a possible objection--note that "rigor" does not necessarily mean "couched in opaque mathematical notation." It's possible to be rigorous, clear, and intuitive at the same time. $\endgroup$
    – whuber ♦
    Jul 4, 2016 at 19:25

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