This question relates to this and this.

When one is modelling the process of (independent) random sampling, it seems to go like this: you start out with a probability space $(\Omega,\mathcal{F},P)$, then you consider $n$ i.i.d. random variables $X_1,\dots,X_n: \Omega \to \mathbb{R}$. Then, a sample is the image of $\omega \in \Omega$ under $(X_1,\dots,X_n)$, that is, $(x_1,\dots,x_n) = (X_1,\dots,X_n)(\omega)$.

This doesn't make intuitive sense to me. As a silly example, if $\Omega$ is a set of people and we wish to estimate their average height, we don't test the height of the same individual using $n$ "different" rulers; instead, we measure the height of $n$ people with the same ruler.

I think there might be an equivalent and more intuitive approach. Instead of considering $n$ random variables, we could consider the product space $(\Omega^n, \mathcal{F}^n,P^n)$ and a random variable $X = X_1$ (as above), which induces $X^n: \Omega^n \to \mathbb{R}^n$. Since $X_1,\dots,X_n$ are i.i.d. it follows that the pushforward measures induced by $(X_1,\dots,X_n)$ and $X^n$ are the same. In fact, for $B_1,\dots,B_n \in \mathcal{B}$ (borel sets) we have $$(X_1,\dots,X_n)^{-1}[B_1 \times \cdots \times B_n] = X_1^{-1}[B_1] \cap \cdots \cap X_n^{-1}[B_n]$$ and $$(X^n)^{-1}[B_1 \times \cdots \times B_n] = X^{-1}[B_1] \times \cdots \times X^{-1}[B_n]$$. From independence and the construction of the product space we have

\begin{align*} P((X_1,\dots,X_n)^{-1}[B_1 \times \cdots \times B_n]) &= P(X_1^{-1}[B_1] \cap \cdots \cap X_n^{-1}[B_n]) \\ &= P(X_1^{-1}[B_1]) \cdots P(X_n^{-1}[B_n]) \\ &= P(X^{-1}[B_1] \times \cdots \times X^{-1}[B_n]) \\ &= P((X^n)^{-1}[B_1 \times \cdots \times B_n]). \end{align*}

Since the pushfoward measures coincide in "basic" borel sets in $\mathbb{R}^n$ they are the same. In that second construction a random sample would be an element of the form $(x_1,\dots,x_n)=(X(\omega_1),\dots,X(\omega_n))$.

My question is, then,

If both constructions are equivalent (I might have made a mistake in my reasoning), why is the first one preferred?


1 Answer 1


I suspect the confusion is just the definition of the sample space. The sample space corresponds to the set out outcomes of an experiment (a experiment which is fixed for a particular $\Omega$). In the example you give about measuring heights it would look something like this:

\begin{align} \Omega = \{(h_1, h_2,\dots, h_n): h_i \in \mathcal{H} \subseteq \mathcal{R}^+, 1\leq i \leq n\} \end{align}

Where $h_i$ is the height of the $i$th person. In the experiment, like you said, we go out and measure the heights of $n$ people. One such realization of this experiment would be a string of heights, one for each n participants. The sample space is all such possible realizations which the set of all such string inside some subset of $\mathcal{R^n}^+$ (coresponding to some min and max possible observable heights).

Now each of the random variables $X_1,\dots, X_n:\Omega \to \mathcal{R}$, which represent the measured height for individual $i$, are the mappings from this sample space to the reals which correspond to picking off the $i$th element of $\omega \in \Omega$, $X_i(\omega) = h_i$. It should then be clear that the image gets us back the sample.

To your point about why not always define in terms of product spaces, the answer is generality in that only some sample spaces can nicely be decomposed into product spaces, especially ones that correspond nicely to an arbitrarily defined set of random variables $X_1,\dots,X_n$. In your height example, of course we could do such a decomposition, but consider the mapping $Y_i:\Omega \to \mathcal{R}$ such that $Y_i = \underset{1 \leq j \leq i}{\max}(X_1,X_2,\dots, X_i)$, where the $X_i$'s are defined as before. These random variables correspond the maximum observed height through participant $i$. If we defined $\Omega^i := \{(h_1,\dots,h_i) : h_k \in \mathcal{H}\subseteq \mathcal{R}^+, 1\leq k \leq i\}$, which you would need to ensure that each variable $Y_i:\Omega^i \to \mathcal{R}$ is in fact a mapping from $\Omega^i$, it would not be true that $\Omega = \Omega^1\times \Omega^2 \times \dots \times \Omega^n$.

It is true that it is often convenient to consider product spaces when possible to do so and there are many examples in statistics which do so, but those who write probability theory or textbooks do not want to rewrite definitions to treat random variables like $X_i$s and $Y_i$s differently when they share most of the same essential properties.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.