Intuition for Identically Distributed Random Variables [closed]

Question

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?

Answer 1

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 $X$.