# What does the term sampling distribution mean conceptually and rigorously?

Sorry if this is basic, but I was reviewing hypothesis testing after 6 years and came across khan academy's video explaining it. At some point at minute 4.03 the speaker says "sampling distribution", what does he mean rigorously with that? From the spoken video and what he write it seems that he mean the distribution of the sample mean. i.e. if we have i.i.d. random variable $X_i$ and we define a new r.v.:

$$M_m = \frac{1}{m} \sum^m_{i=1} X_i$$

then it seems that he refers to the sampling distribution as the distribution of $M_m$. It seems confusing because the sampling distribution for me in my head should be the distribution from where we obtain samples, i.e. the distribution of $X_i$ which is of course the population distribution.

I know this is probably mostly a conceptual question but why does he refer to the sampling distribution as $M_m$? It seems weird because we obtain samples from the population not from the distribution according to $M_m$. I think I am probably wrong so I wanted to understand why.

• maybe section 2 of his answer may be of help: stats.stackexchange.com/questions/167972/…, it is in the context of confidence intervals, but the concept of sampling distribution is the same for confidence intervals as for hypothesis testing – user83346 Aug 20 '16 at 9:41
• This definition is unfortunately correct. But I'm hesitant to post an answer that just says "yes, this is a bad naming convention" -- hopefully someone with knowledge of history will come along and explain what our statistical forefathers were thinking with this one – shadowtalker Aug 20 '16 at 11:14
• But it's not that bad. Think of it as "the distribution of this statistic in a sample" – shadowtalker Aug 20 '16 at 11:16
• The $X_i$ are random variables. A function of random variables is itself a random variable. So if the $X_i$ have some distribution, a statistic like $\bar{X}$ also has a distribution -- which is a function of the distribution(s) from which the data were drawn. When the distribution of some statistic like $M_m$ is based on a random sample, that distribution is called a sampling distribution – Glen_b Aug 20 '16 at 23:06
• @Glen_b I think from my question its clear that $M_m$ is a random variable and has a distribution so I never doubted that. Just putting that out there. However, your suggestion seems that any "statistic" that is computed from random samples is a sampling statistic. So any function $s_j$ that has the honor of being called a "statistic" that processes random sample forms a sampling statistic by definition i.e. If we have $$S_m = s_j(X_1,...,M_m)$$ the we call the distirbution of $S_m$ the sampling statistic. Right? Or is there something important I missed? – Pinocchio Aug 21 '16 at 0:41

## 1 Answer

You are correct in your understanding. However, I can try to give some justification of why this term is used the way it is.

The "population" $X$ is an abstract concept, represented by a hypothetical distribution $p[X=x]$ which in principle assigns some probability to each numeric value $x$ that the random variable $X$ might take on. This population distribution is essentially never known. But we can try to learn about this distribution from measurements $X_1,\ldots,X_n$ which constitute a "sample" of the population.

Of course any statistic we compute from a sample, $Y=f[X_{1:n}]$, (e.g. $Y=\bar{X}$) is itself a random variable. The important point is that the "population" distribution of $Y$ is now a function of the population distribution $p[X=x]$ but also depends on the sampling scheme. For example $p[Y=y]$ will change if we have a different sample size $n$, or if we have a biased sampling scheme (e.g. due to detection tolerance).

If we just knew that $Y$ is some random variable, then we might refer to its distribution as a "population" distribution (or just "distribution", more casually). But if we know that $Y$ is actually some statistic computed from a sample of $X$, then we would refer to its distribution as a "sampling" distribution.

Hopefully this was not more confusing!