$\newcommand{\F}{\mathcal{F}} \newcommand{\powset}[1]{\mathcal{P}(#1)}$
I am reading lecture notes which contradict my understanding of random variables. Suppose we have a probability space $(\Omega, \mathcal{F}, Pr)$, where
-
$\Omega$ is the set of outcomes
-
$\F \subseteq \powset{\Omega}$ is the collection of events, a $\sigma$-algebra
-
$\Pr:\Omega\to[0,1]$ is the mapping outcomes to their probabilities.
If we take the standard definition of a random variable $X$, it is actually a function from the sample space to real values, i.e. $X:\Omega \to \mathbb{R}$.
What now confuses me is the precise definition of the term support.
the support of a function is the set of points where the function is
not zero valued.
Now, applying this definition to our random variable $X$, these lectures notes say:
Random Variables – A random variable is a real valued function defined
on the sample space of an experiment. Associated with each random
variable is a probability density function (pdf) for the random
variable. The sample space is also called the support of a random
variable.
I am not entirely convinced with the line the sample space is also callled the support of a random variable.
Why would $\Omega$ be the support of $X$? What if the random variable $X$ so happened to map some element $\omega \in \Omega$ to the real number $0$, then that element would not be in the support?
What is even more confusing is, when we talk about support, do we mean that of $X$ or that of the distribution function $\Pr$?
This answer says that:
It is more accurate to speak of the support of the distribution than
that of the support of the random variable.
Do we interpret the support to be
- the set of outcomes in $\Omega$ which have a non-zero probability,
- the set of values that $X$ can take with non-zero probability?
I think being precise is important, although my literature does not seem very rigorous.
Best Answer
That looks quite wrong to me.
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.
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.