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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

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    $\begingroup$ "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
    Commented Feb 5, 2019 at 13:14
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    $\begingroup$ @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
    Commented 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. $\endgroup$ Commented Feb 5, 2019 at 13:36

1 Answer 1

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1)

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

2)

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

3)

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 $X$.

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

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