Previously: (Dumb question: Computing expectation without change of variable formula) I was wondering how to compute $E[X]$ by $\int_{\Omega} X d\mathbb P$ rather than $\int_{\mathbb R} x d \mathcal L_X(x)$, i.e. without the change of variable formula.

Today, I found out (Distribution function of a random variable in Lebesgue measure) an explicit way to write $X(\omega)$ for an exponentially distributed $X$.

Apparently, a random variable $X$ in $(\Omega, \mathcal{A},P) =((0,1), \mathcal{B}, \mu)$ with $\mu$ as the Lebesgue measure given by

$$X(\omega):=\frac{1}{\lambda} \ln \frac{1}{1-\omega}$$

has distribution $F_X(x) = (1-e^{-\lambda x})1_{x \ge 0}$, which we know to be the cdf of an exponentially distributed random variable for $\lambda > 0$. Thus, we can compute

$$E[X] = \int_{\Omega} X d\mathbb P = \int_0^1 \ln \frac{1}{\lambda} \frac{1}{1-\omega} d\mathbb P(\omega) = \int_0^1 \ln \frac{1}{\lambda} \frac{1}{1-\omega} d\mu(\omega) = \int_0^1 \ln \frac{1}{\lambda} \frac{1}{1-\omega} d\omega$$

By change of variable, we can verify that this is the same as $$E[X] = \int_{\mathbb R} x \frac{d}{dx} F_X(x) dx$$

Q1. What's the term for something like $$X(\omega):=\frac{1}{\lambda} \ln \frac{1}{1-\omega}$$ ?

This seems to be an explicit representation of the exponentially distributed random variable $X$. I don't see anything like this on Wiki. By this standard proposition, I want to show$X : ((0,1],\mathcal B,\lambda)\to \mathbb R$ is random variable, every distribution function in a probability space implies the existence of a random variable in $((0,1), \mathcal{B}, \mu)$ with $\mu$ as Lebesgue measure, whose CDF is the distribution function's.

Q2. Also, how does one come up with explicit representations for any distribution function?

(*) By the way, this relates to my deleted questions:

Maths Stackexchange: How do you compute expectation w/o change of variable formula? or Maths Overflow: How do you compute expectation w/o change of variable formula?

Maths Stackexchange: Explicitly representing a random variable in terms of indicator functions or Maths Overflow: Explicitly representing a random variable in terms of indicator functions


1 Answer 1


This is called Skorokhod representation, according to David Williams' Probability with Martingales.(*) For a given cdf $F$, the random variable can be explicitly represented by computing $$X(\omega) = \sup\{y \in \mathbb{R}: F(y) < \omega\}$$

Like for exponential:

$$F(y) < \omega$$

$$\iff 1-e^{-\lambda y} < \omega$$

$$\iff y < \frac{1}{\lambda} \ln(\frac{1}{1-\omega})$$

Thus, $$X(\omega) = \sup(y \in \mathbb{R}: F(y) < \omega) = \sup(-\infty,\frac{1}{\lambda} \ln(\frac{1}{1-\omega})) = \frac{1}{\lambda} \ln(\frac{1}{1-\omega})$$

(*) This can also be called canonical representation (MAT 235A / 235B: Probability Instructor: Prof. Roman Vershynin Prof typeset by Edward D. Kim) or Skorokhod representation of random variables using quantile transforms (Optimal Transport Methods in Economics By Alfred Galichon).

Skorokhod representations relate to quantile functions, similarly defined:

$$Q(p) = \inf\{x \in \mathbb R | F(x) \ge p\}$$

In the wiki page for random variables under distribution functions, it says:

The probability distribution "forgets" about the particular probability space used to define X and only records the probabilities of various values of X. [...] In practice, one often disposes of the space $\Omega$ altogether and just puts a measure on $\mathbb {R}$ that assigns measure 1 to the whole real line, i.e., one works with probability distributions instead of random variables.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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