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I am a bit confused about the way continuous random variables are defined. For example, according to Wikipedia the exponential distribution, for $\lambda > 0,$ has a CDF equal to $F(x) = 1-e^{-\lambda x}$ (well, for non-negative $x$ anyway). So presumably it is possible to define a continuous random variable $X$ with $F$ as its CDF. But to be formal we need to define $X$ as a measurable function from a sample space $(\Omega, \Sigma_{\Omega}),$ with some probability measure $P,$ to the real numbers (with the Borel sigma algebra). So my question is how should $(\Omega, \Sigma_{\Omega}),$ $X$ and $P$ be chosen ? For example, would it be okay to choose $\Omega = [0, \infty),$ and $X$ as the inclusion function, and $P$ as the probability measure on $[0, \infty)$ with the aforementioned CDF ?

Richard Southwell
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  • Ordinarily we think of the underlying probability space as an abstract thing, and only occasionally need to be concerned about formal details of how it and the random variables defined on it are constructed. Specifically we need to be concerned about such matters when we are uncertain whether it is possible to construct a probability space that does what we want. The normal thing that you do to define finitely many independent random variables is to use $[0,1]^n$ with the Lebesgue measure and have $X_i$ be the quantile function of $\omega_i$ where $\omega=(\omega_1,\dots,\omega_n)$. – Ian Jul 28 '21 at 16:44
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    Yes... the only tricky part is showing that the CDF $F$ uniquely defines a probability measure on $[0, \infty)$. For this you need the $\pi$-$\lambda$ theorem: https://en.wikipedia.org/wiki/Dynkin_system –  Jul 28 '21 at 16:45

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Given a univariate random variable $X$ with CDF $F$, the standard probability space on which we can construct $X$ is the unit interval $[0,1]$ equipped with Lebesgue measure. If we define $$ X(\omega):=F^{-1}(\omega),$$ where $F^{-1}$ is an appropriately constructed inverse function of the CDF, you can check that $X$ is measurable and possesses the required distribution. For the exponential distribution the inverse $F^{-1}$ can be determined unambiguously.

For a much more detailed discussion, see this.

grand_chat
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