Somewhere I read:

$X$ takes values in the set $S$, and $Y$ and $Y'$ are each uniformly distributed over set $T$

  • Does "take values in the set" mean that $X : S -> ??$?
  • Does "uniformly distributed over set $T$" mean $Y: ?? -> T$?
  • What ?? be in both cases?

I find non formal notation very confusin

  • 1
    $\begingroup$ "takes value in $S$" and "distributed over $T$" both mean that the values realised by the random variable belong to $S$ and $T$, respectively. The originating set is a probabilised set usually denoted $\Omega$, endowed with a $\sigma$ algebra $\mathcal A$ and a probability measure on $\mathcal A$. $\endgroup$
    – Xi'an
    Feb 5, 2019 at 13:14
  • 2
    $\begingroup$ @Xi'an Unfortunately many authors use identical language to mean that $S$ is the domain of $X$! In light of this, absent more context, it seems impossible to determine what the quotation truly is intended to mean. $\endgroup$
    – whuber
    Feb 5, 2019 at 13:22
  • $\begingroup$ @whuber personally speaking, I've never read anywhere "A takes values in B" to mean B is in the domain of A. It feels a rather odd way to define such relationship. $\endgroup$ Feb 5, 2019 at 13:36

1 Answer 1



No, it means $X$ assume values that belong to $S$.


No, it means that the values of $Y$ belong to $T$, and there they are uniformly distributed.


The correct definitions would be:

$X: \Omega \to S \cup Y$ and $Y \in T$, where $\Omega$ is a set of possible outcomes for the random variable $X$.

  • $\begingroup$ in your point (1) does that mean that S is the sample space or the outcome? $\endgroup$
    – ramborambo
    Feb 5, 2019 at 14:19
  • $\begingroup$ @ramborambo sample space, that's what 3) implies. $\endgroup$ Feb 5, 2019 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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