I frequently work with population studies, where, let's say, age and sex are collected for $N=1,000,000$ individuals in California. I might ask a simple question: what is the average age in this population?

How can I frame this simple question in terms of measure theoretic probability? I would like to define a probability space and then take the expectation of a random variable to get the answer. Then, the simple sample average of age would be an estimate of the expectation of this random variable.

It seems I need to define a "random experiment" which perhaps is drawing a random person from the population I am studying. Since, for my purposes, a random person is simply a tuple of (Age, Sex), then it seems my sample space $\Omega$ should be all tuples of the form $(x,y)$, where $x \in \mathbb R^+$ and $y \in \{\text{M}, \text{F}\}$. As for the $\sigma$-algebra, I can let $E$ be the product sigma algebra of the Borel sigma algebra on $\mathbb R^+$ and $2^{\{M,F\}}$. This approach encounters difficulty though in choosing the probability measure. I would have liked to use the counting measure normalized by $N$. That is, for example, if $A = (65.3243,\text{M}) \in E$,

$$\mathbb P ( A ) = \frac{\text{number of sampled males aged 65.3243 } }{N}.$$

But since I defined age as a real number ( as opposed to a discrete number of ages), this is actually probability 0, which makes no sense.

Another construction might be to model the actual collected data as the sample space. That is, $\Omega$ consists of $N$ elements, one corresponding to each of the data points. Then the $\sigma$-algebra would be $2^\Omega$ and the probability would be that same counting measure defined above.

In my head, the measure theoretic view of probability seems to fall apart whenever I start sampling things. How can I connect that theoretical view with the simple epidemiological questions I ask on a daily basis?

Edit: Having read this highly relevant link, I see another option. I could define a probability space $(\Omega, E, \mathbb P$), where $\Omega$ models the people in California (i.e. for each person in California there is an element in $\Omega$). $E$ is then then $2^\Omega$, and $\mathbb P$ is the counting measure normalized by $N$.

In order to model a random $N$ sized sample of this population, I should define a new probability space $(\Omega^N, E^N, \mathbb P^N)$, where $\Omega^N$ and $E^N$ are the Cartesian products of $\Omega$ and $E$ respectively. But then, I am still not sure how to define the probability measure.

  • $\begingroup$ I believe your question is answered at stats.stackexchange.com/questions/224442/…. Please take a look and if you think not, please explain how your question differs from the ones addressed there. $\endgroup$
    – whuber
    Commented Jun 6, 2018 at 21:09
  • 1
    $\begingroup$ @whuber your link was very close to what I needed--thank you so much for pointing it out. I think I just need a concrete example, like the one I mention in this post. I edited the original post to clarify where I am stuck when applying your framework to this specific example. $\endgroup$ Commented Jun 6, 2018 at 21:58

1 Answer 1


The average age of the population is not a random variable, or at least not an interesting one, even if you make it to be one.

The average age of the fixed size sample, on the other hand, is, and you can construct various probabilistic spaces for it. Here is one particularly simple example, and some thoughts on how can one modify it.

Simple example

Take $\Omega$ to be the set of subsets of the population of size $N$, with sigma algebra $2^\Omega$. Assign equal mass to all the points in $\Omega$.

The sample average is then a random variable $X: \Omega \to \mathbb{R}$, with $$X(\omega) = \frac{1}{N} \sum_{\text{person} \in \omega} \text{person.age}$$

Things to modify

You can modify $\Omega$ and probability measure to better fit reality. Sigma algebra is not important.

Taking $\Omega$ to be set of subsets of the population models the fact that you are sampling $N$ different people. If you change your mind and decide to drop the difference requirement and sample with replacement, you can then take $\Omega$ to be set of multisets (the $N$-fold Cartesian product).

Assigning equal mass to all the points in $\Omega$ models the fact that when you sample, you are equally likely to draw any $N$ individuals. This may be accurate, for example, if you have a full record of Californians and genuinely choose $N$ of them uniformly at random.

This may be not accurate for a different sampling process, and the probability measure is the place to encode that information. For example, if you will only survey adults, then any $\omega$ that contains minors gets $0$ weight, and all other $\omega$ get a slightly higher weight.

  • $\begingroup$ It won't work to "assign equal mass to all the points in $\Omega$" because not all samples possibly can have equal probability. For taking samples of size $n$ (without replication), consider letting $\Omega$ be the set of all $n$-subsets of the population and give it the discrete sigma algebra. You don't have to specify a probability measure: that will be determined by how the samples are selected. $\endgroup$
    – whuber
    Commented Jun 6, 2018 at 20:54
  • $\begingroup$ @whuber Agreed, edited the answer to incorporate your remark. $\endgroup$
    – psarka
    Commented Jun 7, 2018 at 9:06

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