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I often see the sentence "let $X_1, X_2, \ldots$ be a sequence of i.i.d. random variables with a certain distribution". But given a random variable $X$ on a probability space $\Omega$, how do I know that there is a sequence of INDEPENDENT random variables of the same distribution on $\Omega$?

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This is Kolmogorov's consistency theorem. See for example Varadhan's Probability Theory page 41 for a nice explanation. It's available on Google books. – Stefan Hansen Nov 28 '12 at 12:29
@Stefan Hansen : To apply Kolmogorov's consistency theorem, you need the law to be a probability measure on a Polish space. Nevertheless, this result holds in full generality, and can be seen as a particular case of Ionescu-Tulcea's theorem. – Ahriman Nov 28 '12 at 12:56
I have also to mention than the sequence is defined on a new probability space, and not on the original space $\Omega$ where $X$ is defined. This is not a real issue, because of the philosophy of probability theory : this is the laws of random variables which are of interest, rather than the probability space on which they are defined. – Ahriman Nov 28 '12 at 12:59
up vote 13 down vote accepted

The easiest way to show the existence in our case is to construct such a probability space. The intuition is that it is easy to define the joint probability measure on some "simple" sets using the iid property and then one leverages the celebrated Caratheodory extension theorem.

More precisely, let $(E,\mathcal E)$ be the measurable space where our random variables are supposed to take values and let $\mu$ be a probability measure on $(E,\mathcal E)$ representing the distribution of such random variables. Define $\Omega = E^{\mathbb N_0}$ to be the space of countable trajctories over $E$ and let $\mathcal F$ be its product $\sigma$-algebra. Define the probability measure $\mathsf P$ on $(\Omega,\mathcal F)$ just based on the independence, i.e. for any $A_0,\dots,A_N\in \mathcal E$ we put $$ \mathsf P(X_0\in A_0,\dots,X_n\in A_n):=\mu(A_0)\times \dots\times \mu(A_n). $$ So far the measure $\mathsf P$ is only defined on the collection $\mathcal A$ of measurable rectangles, i.e. subset of $\Omega$ of the form $A_0\times\dots\times A_n\times \Omega\times\Omega\times\dots$ where $A_i\in \mathcal E$. Finite unions of elements of $\mathcal A$ form the algebra, say $\mathcal B$. Clearly, $\mathsf P$ is a finite pre-measure on $\mathcal B$ and hence by the Caratheodory extension theorem we obtain the unique measure $\mathsf P$ on $\mathcal F = \sigma(\mathcal B)$.

As Ahriman has pointed out, if you are given a random variable $X:\Omega\to E$ it may not be possible to construct the whole sequence on $\Omega$ as the latter may be quite a poor space, so you would have to go for a richer space. For example, $E$ always can be considered as a sample space for the distribution over it, by taking $\mathrm id_E$ being a random variable. But in case $E = \{a,b\}$ and you have $\mu(a) = 0.4$ and $\mu(b) = 0.6$ it is only possible to construct one and only one random variable defined on $E$ which has $\mu$ as a distribution.

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+1. Nice answer. – Stefan Hansen Nov 28 '12 at 13:13

Since they're i.i.d., you can just use a product measure on a product space $\Omega\times\Omega\times\Omega\times\cdots$.

And for many purposes, you can take $\Omega=\mathbb R$ and let the measurable subsets of $\Omega$ be the Borel sets.

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