I am not entirely convinced with the line the sample space is also called the support of a random variable
That looks quite wrong to me.
What is even more confusing is, when we talk about support, do we mean that of $X$ or that of the distribution function $Pr$?
In rather informal terms, the "support" of a random variable $X$ is defined as the support (in the function sense) of the density function $f_X(x)$.
I say, in rather informal terms, because the density function is a quite intuitive and practical concept for dealing with probabilities, but no so much when speaking of probability in general and formal terms. For one thing, it's not a proper function for "discrete distributions" (again, a practical but loose concept).
In more formal/strict terms, the comment of Stefan fits the bill.
Do we interpret the support to be
- the set of outcomes in Ω which have a non-zero probability,
- the set of values that X can take with non-zero probability?
Neither, actually. Consider a random variable that has a uniform density in $[0,1]$, with $\Omega = \mathbb{R}$.
Then the support is the full interval $[0,1]$ - which is a subset of $\Omega$. But, then, of course, say $x=1/2$ belongs to the support. But the probability that $X$ takes this value is zero.
The confussion is exactly that while a random variable maps a sample space onto a measurable space, that measurable space is also a sample space. It has all the requisite properties of one.
So it is permissable to refer to the values of a random variable as outcomes. It may cause some confusion, if you are not very careful, but it is permissable.
So
Yes. It is somewhat common to find the terms "sample space" and "outcome" used to refer to the image of a random variable and its possible values.
Yes. It is possible to define a r.v. where both its domain and its image are the same sample space. One such random variable would be the identity function; though there are others. Consider a sample space consisting of the results shown on a die roll, and the random variable of seven minus the results shown on the die roll. $X:x\mapsto 7-x$. $$X: \{1,2,3,4,5,6\}\mapsto\{1,2,3,4,5,6\}$$
Yes. The image of a random variable can be associated with a sigma algebra and a probability measure. Consider the results of the toss of two fair coins, and the random variable that is the count of heads among them. The random variable's image $\{0,1,2\}$ is the measurable space mapped onto by the random variable, we can build a sigma algebra (its powerset for example) and these events can be assigned a probability measure. $$\mathsf p(A)= \begin{cases}1 & :& A\supseteq \{1,2,3\}\\ 2/3&:& A\cap\{1,2,3\}\in \{\{1,2\}, \{1,3\}, \{2,3\} \}\\ 1/3 &:& A\cap\{1,2,3\}\in \{\{1\}, \{2\}, \{3\} \} \\ 0 &:& A\cap\{1,2,3\}=\emptyset\end{cases}$$ The triple, $(\{1,2,3\}, 2^{\{1,2,3\}}, \mathsf p)$ , fits the criteria required to be a probability space.
Best Answer
$\Omega =\{0,1\}, \mathcal F=\{\emptyset, \Omega\}, X(0)=0,X(1)=1$ is an example since $(X=0) \notin \mathcal F$.