Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am readin Ross's "A first course in probability" and I got to the chapter that talks about random variables.

I am tryng to understand the exact meaning of something of the form $g(X)$ where g is a real valued function and X is a r.v.

I can't figure the "form" of g, that is, I don't understand what to write in $g: ?\to ??$.

I saw an example that made me think $g:\mathbb{R}\to\mathbb{R}$, is this correct ?

share|cite|improve this question
You are right, $g$ would be a function from the reals to the reals, although it could be also on other sets and to other sets. But if $X$ is itself from $\Omega \to \mathbb{R}$, $g$ would need to be from $\mathbb{R}$. – Raskolnikov Apr 23 '12 at 9:27
up vote 3 down vote accepted

Some textbooks have the following framework. If $(\Omega,\mathscr{M},\mu)$ is a probability space and $X:\Omega\to\mathbb{R}$ is a r.v. (i.e., $\mu$-measurable function) and $g:\mathbb{R}\to\mathbb{R}$ (e.g. a continuous function), then $g(X):\Omega\to\mathbb{R}$ is a r.v. which is attained by the composition of $X$ and $g$, i.e. $g\circ X$.

share|cite|improve this answer
why continuous ? by definition ? I don't see why if $g$ is not continuous $g(x)$ is not a r.v ... – Belgi Apr 23 '12 at 9:30
You usually need to consider something from $g$ in order for $g(X)$ to be $\mu$-measurable. If $g$ is continuous, then $g(X)$ will be $\mu$-measurable function aswell, i.e. a random variable. – T. Eskin Apr 23 '12 at 9:31
Here's small explanation. In this case, $g(X)$ is $\mu$-measurable if the preimage $g(X)^{-1}(U)=X^{-1}(g^{-1}(U))$ is $\mu$-measurable for all open $U\subset \mathbb{R}$. If $g^{-1}(U)$ is open, (which would require continuity), then the measurability of $X$ implies that $g(X)$ is measurable. Unless we know anything about $g$, then we don't know anything about $g^{-1}(U)$. – T. Eskin Apr 23 '12 at 9:39
You could also merely ask $g$ to be measurable wrt. the Borel algebra. – Najib Idrissi Apr 23 '12 at 10:04
Yeah, that would also work for example. – T. Eskin Apr 23 '12 at 10:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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