How do I successfully communicate an intuitive understanding of what it means for two random variables to be "identically distributed"? The definition is easy to state, but it lacks insight and why it is important. How can it be explained to undergraduates, whom I am teaching, in a way they can understand?

The best I can do is offer a useful example of when identically independent variables come up. When an experiment is repeated, over and over, we can let $X_i$ represent the outcome at stage $i$ of the experiment. Then the $X_i$ will form an identically distributed sequence of random variables. Hence, the intution here is that the $X_i$, are essentially the same random variable, but nonetheless defined different. But I cannot really find a way to explain this in any better way.

Do you have better examples?

closed as unclear what you're asking by Xi'an, mdewey, kjetil b halvorsen, Michael Chernick, DeltaIV Feb 24 at 11:26

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    "Two random variables are identically distributed if their cdf is the same" seems reasonably intuitively clear -- as long as the students know what the (cumulative) distribution function is. What were you seeking other than something like that? – Glen_b Mar 25 '16 at 15:03
  • @Glen_b For example, we have a precise definition for what it means for two random variables to be independent. But we also have the intuitive grasp of it. If I am preforming an experiment and $X,Y$ are two random variables, then I anticipate that they be "independent", if they do not influence one another. This is not a precise definition but it leads me to expect independence without using any kind of formula. Likewise, I wish to find a way to characterize "identically distributed", so that I can say $X,Y$ are ID without the need to check anything, simply having the right intuitive feel. – Nicolas Bourbaki Mar 25 '16 at 20:46

A random variable $X$ is a measurable function from a probability space $\Omega$ into a measurable space $E$. Hence two random variables $X$ and $Y$ are identically distributed if they define the same measure on $E$, that is, if $$\mathbb{P}(\{\omega;\ X(\omega)\in A)=\mathbb{P}(\{\omega;\ Y(\omega)\in A)$$ for all measurable sets $A\in\sigma(E)$.

For instance, if $\Omega=[0,1]$ and $\mathbb{P}$ is the uniform probability on $\Omega$, and if $E=\mathbb{R}$, any bijection $\Psi$ on $E$ with a Jacobian $$\dfrac{\text{d}\Psi}{\text{d}x}$$ equal to plus or minus one will define $$Y=\Psi(X)$$ that is identically distributed with $X$.

An illustration of this identity is through pseudo-random generators: if I want to generate an Exponential $\mathfrak{E}(1)$ random variable, $X$, I can write $X=X(\omega)$ when $\omega\sim\mathcal{U}(0,1)$ if $$X(\omega)=1-\log(\omega)$$ [this is the inverse cdf transform]. Now, if I define $$Y(\omega)=1-\log(1-\omega)$$ this new random variable is also a transform of $\omega$ that is identically distributed with $X$.

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    (-1) It fails to answer the question which is being asked. I already know this. The question is asking for intuition, not a formal definition, and especially intuition which will be helpful for undergraduates. – Nicolas Bourbaki Mar 25 '16 at 9:43
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    Nevertheless, it is an answer in the true Bourbaki tradition :-) – StijnDeVuyst Mar 25 '16 at 11:52
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    @StijnDeVuyst The word "probability" never comes up in any of the thousand of pages in any of their books! – Nicolas Bourbaki Mar 25 '16 at 20:47
  • @Bourbaki Really? I did not know that. Thx. – StijnDeVuyst Mar 27 '16 at 16:37

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