# Why is a random variable called so despite being a function?

According to my knowledge, its a function $P(X)$ which includes all the possible outcomes a random event.

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$X$ itself is the random variable. – Code-Guru Jul 12 '14 at 4:22
I think a better term would be "random functional," since $X$ is by definition a measurable function. – Clarinetist Jul 12 '14 at 4:25
Because it is neither random nor a variable. Old joke. – André Nicolas Jul 12 '14 at 4:31

If you're in elementary probability instead of measure-theoretic probability, the following will make very little sense. My apologies if this is the case.

I'd say one reason is that we really don't look at the properties of random variables as functions from the underlying set of the probability space (usually denoted $\Omega$). You could change $\Omega$ all around, leaving the structure of $X$ behind, and we wouldn't really care. When I say the structure of $X$, I mean the measure it induces on $\mathbb{R}$, i.e. $\mu(I) = \mathbb{P} \left ( \left \{ \omega \in \Omega : X(\omega) \in I \right \} \right )$. We arguably care a little bit more about $X$ as a function when we start talking about collections of dependent random variables, but really we don't care about it then, either, since then we are just inducing a measure on $\mathbb{R}^n$ instead of $\mathbb{R}$.

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Because we think of it as a variable that take random value intuitively. Formally they are function. Just like why we call a sequence a sequence, or call an arithmetical function an arithmetical function, when they are actually the same thing formally speaking. Just to add to the issue, calling a variable also match the notation. For example, $X=Y+Z$ is NOT the usual function addition, but they are "added" in such a way that make sense when we think as variable.

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I'm really curious now. How would one define "addition" of random variables? I haven't taken a formal measure theory course. – Clarinetist Jul 12 '14 at 4:34
You have seen cumulative distribution functions in elementary probability. These appear again: $X+Y$ is a new random variable whose CDF is $F(z)=\mathbb{P} ( \{ \omega \in \Omega : X(\omega)+Y(\omega) \leq z \} )$. One can assemble another probability space on which this is a random variable (it will be essentially the product of the original probability space with itself), but one rarely works out these details. Instead we prove that every random variable has a CDF and every CDF corresponds to a random variable and then forget about it the underlying details, for the most part. – Ian Jul 12 '14 at 4:38
Oh, I see. Thank you! – Clarinetist Jul 12 '14 at 4:42
I misspoke. What happens is similar, though. We don't add random variables given as functions on the same probability space. We add random variables by starting with a higher dimensional random variable and then adding up the components. There are only one dimensional random variables in this construction at all in the sense that: higher dimensional random variables can be built up from one dimensional ones (if we specify their (in)dependence); one dimensional random variables can be extracted from higher dimensional ones by integrating out all the other dimensions (i.e. marginalizing). – Ian Jul 12 '14 at 13:05