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$\newcommand{\F}{\mathcal{F}} \newcommand{\powset}[1]{\mathcal{P}(#1)}$ 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.

According to Wikipedia:

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.

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    $\begingroup$ The support of a random variable $X$ with values in $\mathbb{R}^n$ is the set $\{x\in\mathbb{R}^n\mid P_X(B(x,r))>0,\text{for all } r>0\}$ where $B(x,r)$ denotes the ball with center at $x$ and radius $r$. In particular, the support is a subset of $\mathbb{R}^n$. $\endgroup$ – Stefan Hansen Jun 24 '14 at 16:20
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    $\begingroup$ What @StefanHansen said, or the smallest closed set $C$ such that $P_X(C)=1$. $\endgroup$ – Did Jun 24 '14 at 17:16
  • $\begingroup$ @Did your definition is particularly intuitive. $\endgroup$ – jII Jul 3 '14 at 2:51
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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.

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  • $\begingroup$ @StefanHansen For my example (a uniform density in $[0,1]$) $\Omega$ is $\mathbb{R}$ $\endgroup$ – leonbloy Jun 25 '14 at 0:19
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The support of the density $f_x(.)$$ is the range of values of the random variable X for which the density function is positive, i.e.

$\mathcal{R}_x:= \{{x\in \mathcal{R}_X : fx(x) > 0\}}$

Note that $f_x(.)$ is the probability density/mass function.

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  • $\begingroup$ For a density function of a continuous random variable, $f(x)$ is $0$ for every $x$, so this is not a correct definition $\endgroup$ – IceFire Jun 6 '18 at 11:40
  • $\begingroup$ @IceFire That's not accurate. What is actually $0$ is the probability $\mathrm{Pr}(X = x) = 0$ for any $x\in\mathbb{R}$. If the probability density function were $0$ for all $x\in\mathbb{R}$, what use would it be? For continuous random variables, $f_X\colon \mathbb{R} \to \mathbb{R}$ is defined as the unique function such that $\mathrm{Pr}(X \in [a, b]) = \int_a^b f_X(x) \mathrm{d}x$ . Then the fact that $\mathrm{Pr}(X = x) = 0$ is a direct consequence: $\mathrm{Pr}(X = x) = \mathrm{Pr}(X \in [x, x]) = \int_x^x f_X(t) \mathrm{d}t = 0$. So this, albeit not the most formal, is a correct def. $\endgroup$ – Anakhand Mar 3 at 12:56
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I'll start from the beginning to make sure we're using the same definitions.

$\newcommand{\A}{\mathcal{A}} \newcommand{\powset}[1]{\mathcal{P}(#1)} \newcommand{\R}{\mathbb{R}} \newcommand{\deq}{\stackrel{\scriptsize def}{=}} \newcommand{\N}{\mathbb{N}}$ Let $(\Omega, \A, \Pr)$ be a probability space, defined as in the question's body:

  • $\Omega$ is the set of outcomes

  • $\A \subseteq \powset{\Omega} $ is the collection of events, a $\sigma$-algebra

  • $\Pr\colon\ \Omega\to[0,1]$ is the mapping outcomes to their probabilities.

A random variable $X$ is defined as a map $X\colon\; \Omega \to \R$ such that, for any $x\in\R$, the set $\{\omega \in \Omega \mid X(\omega) \le x\}$ is an element of $\A$—that is, the probability map is defined for it. This condition is necessary in order to define the following concepts.

The probability distribution function of a random variable $X$ is defined as the map $$ \begin{align} F_X \colon \quad \R \ &\to\ [0, 1] \\ x\ &\mapsto\ \Pr(X \le x) \deq \Pr(X^{-1}(I_x)) \end{align} $$

We can see that

  • $\Pr(X > x) \deq \Pr(\overline{X^{-1}(I_x)}) = 1 - \Pr(X^{-1}(I_x)) = 1 - F_X(x)$

    where $I_x \deq (-\infty, x]$, and $\overline{A}$ denotes the complement of $A$ in $\Omega$. Notice that this probability is defined since $\A$ is a $\sigma$-algebra (and thus closed under set complement).


  • $\Pr(X < x) \deq \Pr\left(\bigcup\limits_{n\in\N} X^{-1} \left(I_{x-\frac{1}{n}}\right)\right) = \lim_{t \to x^-} \Pr(X \le t) = \lim_{t \to x^-} F_X(t)$

    since $X^{-1} \left(I_{x-\frac{1}{n+1}}\right) \subseteq X^{-1} \left(I_{x-\frac{1}{n}}\right)$ for all $n\in\N$. Note again that that union is valid since it is countable and $\A$ is a $\sigma$-algebra.


  • $\Pr(X = x) \deq \Pr(X^{-1}(I_x) \setminus A_{<x}) = F_X(x) - F(x^-)$

    where $A_{<x} \deq \bigcup_{n\in\N} X^{-1} \left(I_{x-\frac{1}{n}}\right)$ and $F_X(x^-) \deq \lim\limits_{t \to x^-} F_X(t)$.

and so forth.

Now, this function is sufficient to uniquely define a probability measure on $\R$; that is, a map $$ \begin{align} P_X \colon \quad \mathcal{B} \subset \powset{\R} \ &\to \ [0, 1]\\ A \ &\mapsto \ \Pr(X \in A) \deq \Pr(X^{-1}(A)) \end{align} $$ that assigns to any set $A \in \mathcal{B}$ the probability of the corresponding event in $\A$. Here $\mathcal{B}$ is the Borel $\sigma$-algebra in $\R$, which is, loosely speaking, the smallest $\sigma$-algebra containing all of the semi-intervals $(-\infty, x]$. The reason why $P_X$ is defined only on those sets is because we only required $X^{-1}(A) \in \A$ for the semi-intervals A=(-\infty, x]; thus $X^{-1}(A)$ is an element of $\A$ only when, loosely speaking, $A$ is "generated" by those semi-intervals, their complements, and countable unions/intersections thereof (according to the "rules" of a $\sigma$-algebra).

Formally, the support of $X$ can be defined as the smallest closed set $R_X \in \mathcal{B}$ such that $P_X(R_X) = 1$, as Did pointed out in their comment. An alternative but equivalent definition is the one given by Stefan Hansen in his comment. The equivalence can be proven as follows:

Proof
Let $R_X$ be the smallest closed set $R_X \in \mathcal{B}$ such that $P_X(R_X) = 1$. That means that for every $x \in \overline{R_X}$, there exists a radius $r\in\R_+$ such that the open interval (or open ball in the more general case) $(x-r, x+r)$ is contained within $\R \setminus R_X$ (since $R_X$ is closed).

That, in turn, implies that $P_X((x-r, x + r)) = 0$ --- otherwise, $P_X(R_X \cup (x-r,x+r)) = P_X(R_X) + P_X((x-r, x+r)) > P_X(R_X) = 1$, a contradiction.

Conversely, suppose $P_X((x-r, x+r)) = 0$ for some $x\in\R$, $r\in\R_+$. Then $(x-r, x+r) \subseteq \R \setminus R_x$. Otherwise $R_X' \deq R_X \setminus (x-r, x+r)$ would be a closed set smaller than $R_X$ satisfying $P_X(R_X') = 1$.

This proves $\R \setminus R_X = \{x\in\R \mid \exists r \in \R_+\colon P_X((x-r, x+r)) = 0\}$

Negating the predicate, one gets $R_X = \{x\in\R \mid \forall r \in \R_+ P_X((x-r, x+r)) > 0\}$

But more often, different definitions are given.


Alternative definition for discrete r.v.

A discrete random variable can be defined as a random variable $X$ such that $X(\Omega)$ is countable (either finite or countably infinite). Then, for a discrete random variable the support can be defined as

$$R_X \deq \{x\in\R \mid \Pr(X = x) > 0\}\,.$$

Note that $R_X \subseteq X(\Omega)$ and thus $R_X$ is countable. We can prove this by proving its contrapositive:

Suppose $x \in \R$ and $x \notin X(\Omega)$. We can distinguish two cases: either $x < y$ $\forall y \in R_X$, or $x > y$ $\forall y \in R_X$, or neither.

Suppose $x < y$ $\forall y \in R_X$. Then $\Pr(X = x) \le \Pr(X \le x) = \Pr(X^{-1}(I_x)) = \Pr(\emptyset) = 0$, since $\forall \omega\in\Omega\ X(\omega) > x$. Ergo, $x\notin R_X$.

The case in which $x > y$ $\forall y \in X(\Omega)$ is analogous.

Suppose now $\exists y_1, y_2 \in X(\Omega)$ such that $y_1 < x < y_2$. Let $S = \{y\in X(\Omega) \mid y < x\}$, which is. Thus $\sup L$ and exists, and $\lim_{y \to x^-} F_X(y) = F_X(\sup L)$ since $F_X$ is nondecreasing and bounded above. Thus, since $\sup L \le x$, $F_X(x) \ge F_X(\sup L)$ and therefore $\Pr(X=x) = F_X(x) - F_X(x^-) \ge F_X(\sup L) - F_X(x^-) = 0$.


Alternative definition for continuous r.v.

Notice that for absolutely continuous random variables (that is, random variables whose distribution function is continuous on all of $\R$), $\Pr(X = x) = 0$ for all $x\in \R$—since $F_X(x) = F_X(x^-)$. But that doesn't mean that the outcomes of $X^{-1}({x})$ are "impossible", informally speaking. Thus, in this case, the support is defined as

$$ R_X = \{x \in \R \mid f_X(x) > 0\}$$

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A random variable is defined as a function that maps outcomes to numerical quantities, typically real numbers, i.e., X: Ω ↦ R.

The “domain of a random variable” is the set of possible outcomes. In the case of the coin, there are only two possible outcomes, namely heads or tails. Since one of these outcomes must occur, either the event that the coin lands heads or the event that the coin lands tails must have non-zero probability. For an unbiased coin p(H)=p(T)=1/2.

Consider another random variable Y, be the number of heads. So toss a coin twice produce y = 0, 1 and 2 with p(Y) = ¼, ½ and ¼ respectively.

The “support of a real-valued function f ” is the subset of the domain containing those elements which are not mapped to zero. Let f(x) be the p.d.f. of normal distribution with support x∈R , so that f(x)≠0.

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  • $\begingroup$ Welcome to MSE! Please, format your posts using MathJax. $\endgroup$ – mucciolo Nov 4 '17 at 6:50

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