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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 $\mathbb{R}^n$. $\endgroup$ 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
  • $\begingroup$ The probability map assigns probabilities to the events in $\mathcal{F}$, not each of the atomic outcomes in $\Omega$; i.e., $\Pr \colon\ \mathcal{F} \to [0, 1]$ $\endgroup$
    – Anakhand
    Aug 29 '19 at 15:27
  • $\begingroup$ @Did: How does one define smallest (usually the intersection of all relevant candidates) such that the result is measurable and $P_X C = 1$? $\endgroup$
    – copper.hat
    Oct 2 '19 at 15:14
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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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TL;DR

The support of a r.v. $X$ can be defined as the smallest closed set $R_X \in \mathcal{B}$ such that its probability is 1, as Did pointed out in their comment. An alternative definition is the one given by Stefan Hansen in his comment: the set of points in $\mathbb{R}$ around which any ball (i.e. open interval in 1-D) with nonzero radius has a nonzero probability. (See the section "Support of a random variable" below for a proof of the equivalence of these definitions.)

Intuitively, if any neighbourhood around a point, no matter how small, has a nonzero probability, then that point is in the support, and vice-versa.



I'll start from the beginning to make sure we're using the same definitions.

Preliminary definitions

Probability space

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

  • $\Omega$ is the set of outcomes

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

  • $\Pr\colon\ \mathbf{\A}\to[0,1]$ is the mapping of events to their probabilities. It has to satisfy some properties:

    • $\Pr(\Omega) = 1$   (we know $\Omega \in \A$ since $\A$ is a $\sigma$-algebra of $\Omega$)
    • has to be countably additive

Random variable

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 $\A$, ergo, an element of $\Pr$'s domain to which a probability can be assigned.

We can think of $X$ as a "realisation" of $\Omega$, in that it assigns a real number to each outcome in $\Omega$. Intuitively, this condition means that we are assigning numbers to outcomes in an order such that the set of outcomes whose assigned number is less than a certain threshold (think of cutting the real number line at the threshold and forming the set of outcomes whose number falls on or to the left of that) is always one of the events in $\A$, meaning we can assign it a probability.

This is necessary in order to define the following concepts.

Cumulative Distribution Function of a r.v.

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

where $I_x \deq (-\infty, x]$. (NB: $X^{-1}$ denotes preimage, not inverse; $X$ might well be non-injective.)

For notational clarity, define the following:

  • $\Omega_{\le x} \deq X^{-1}((-\infty, x]) = X^{-1}(I_x)$
  • $\Omega_{> x} \deq X^{-1}((x, +\infty)) = X^{-1}(\overline{I_x}) = \overline{\Omega_{\le x}}$   where $\overline{\phantom{\Omega}}$ denotes set complement (in $\R$ or $\Omega$, depending on the context)
  • $\Omega_{< x} \deq X^{-1}((-\infty, x)) = \displaystyle\bigcup_{n\in\N} X^{-1} \left(I_{x-\frac{1}{n}}\right)$
  • $\Omega_{=x} \deq X^{-1}(x) = \Omega_{\le x} \setminus \Omega_{< x}$

we know all of these are still in $\A$ since $\A$ is a $\sigma$-algebra.

We can see that

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

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

    since $X^{-1} \left(I_{x-\frac{1}{n}}\right) \subseteq X^{-1} \left(I_{x-\frac{1}{n+1}}\right)$ for all $n\in\N$.

  • $\Pr(X = x) \deq \Pr(\Omega_{=x}) = \Pr(\Omega_{\le x} \setminus \Omega_{<x})= \Pr(\Omega_{\le x}) - \Pr(\Omega_{<x}) = F_X(x) - F_X(x^-)$

and so forth.

Note that the limit that defines $F(x^-)$ always exists because $F_X$ is nondecreasing (since if $x< y$, then $\Omega_{\le x} \subseteq \Omega_{\le y}$ and $\Pr$ is $\sigma$-additive) and bounded above (by $1$), so the monotone convergence theorem guarantees that the images by $F_X$ by of any nondecreasing sequence approaching $x$ from the left will also converge, and thus the continuous limit $\lim_{t \to x^-} F_X(t)$ exists.


Probability measure on $\R$ by $X$

The mapping defined by $X$ 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 in our definition we only required $X^{-1}(A) \in \A$ to be true for the semi-intervals of the form $A = (-\infty, x]$; thus $X^{-1}(A)$ is an element of $\A$ only when $A$ is "generated" by those semi-intervals, their complements, and countable unions/intersections thereof (according to the rules of a $\sigma$-algebra).




Support of a random variable

Formal definition

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 support of a random variable $X$ with values in $\R^n$ is the set $\{x\in \R^n \mid P_X(B(x,r))>0, \text{ for all } r>0\}$ where $B(x,r)$ denotes the ball with center at $x$ and radius $r$. In particular, the support is a subset of $\R^n$.

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 $x \in \R \setminus\overline{R_X}$, there exists a radius $r\in\R_{>0}$ 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, if this were strictly positive, $P_X(R_X \cup (x-r,x+r)) = P_X(R_X) + P_X((x-r, x+r)) > P_X(R_X) = 1$, a contradiction.

Conversely, suppose $P_X((x-r, x+r)) = 0$ for some $x\in\R$, $r\in\R_{>0}$. Then $(x-r, x+r) \subseteq \R \setminus R_X$ (and, in particular, $x \in \R \setminus R_X$). 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 $\R \setminus R_X = \{x\in\R \mid \exists r \in \R_{>0}\colon P_X((x-r, x+r)) = 0\}$

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

But more often, different definitions are given.


Alternative definition for discrete random variables

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 random variables

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, by definition of continuity, $F_X(x^-) = F_X(x)$. But that doesn't mean that the outcomes in $X^{-1}({x})$ are "impossible", informally speaking. Thus, in this case, the support is defined as

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

which intuitively can be justified as being the set of points around which we can make an arbitrarily small interval on which the integral of the PDF is strictly positive.

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The support of the density function $f_X(.)$ is the range of values of the random variable X for which the density function is positive. That is,

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

Note that $f_X(.)$ is the probability density/mass function of the random variable X.

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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 '19 at 12:56
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You appear to have misquoted these lecture notes. What they actually say is

Random Variables – A random variable is a process, which when followed, will result in a numeric output. The set of possible outputs is called the support, or sample space, of the random variable. Associated with each random variable is a probability density function (pdf) for the random variable.

It appears to me that the author of the notes is simply using a very old notion of random variable that predates Kolmogorov's definition of it as being a function on the set of outcomes of a probability space, as has now been almost universally adopted by modern mathematicians. The notes appear to have been written for the probability unit of a course in actuarial science. They also seem to me to depart from standard modern mathematical terminology in at least one other respect—namely, in using the term "probability density function" not only in its usual sense, but also to refer to the probability mass function of a discrete distribution. Wikipedia does say, however, that the probability mass function is sometimes also referred to as a "discrete density function" (a term I can't recall ever having previously come across), so perhaps this apparent departure from standard terminology is not quite as egregious as I had originally thought.

What these notes are calling the "support, or sample space" of a random variable is not, as your misquotation has characterised it, the domain of a function constituting the random variable, but the set of possible numerical values which the random variable can have. It is essentially the same thing as what would now be called the support of the random variable's distribution in standard modern mathematical terminology.

If the set of outcomes $\ \Omega\ $ of a probability space is a topological space, with topology $\ \mathscr{T}\ $, say, and the probability $\ \text{Pr}\ $ is a Borel probability measure, then the support $\ supp_{\,\text{Pr}}\ $ of $\ \text{Pr}\ $, as Stefan Hansen and Did have pointed out in the comments, is the closed set equivalently defined as either $$ supp_{\,\text{Pr}}=\left\{x\in\Omega\,|\,\text{Pr}(\mathscr{O})>0\ \text{ for all }\ \mathscr{O}\in\mathscr{T}\ \text{ with }\ x\in\mathscr{O}\right\}\ , $$ or $$ supp_{\,\text{Pr}}=\bigcap\left\{\mathscr{C}\,|\,\text{Pr}(\mathscr{C})=1\ \text{ and }\ \mathscr{C}^{\,c}\in\mathscr{T}\right\}\ . $$ It is easy to show that these definitions are equivalent, and that $\ supp_{\,\text{Pr}}\ $ is closed. While it is not necessarily true that $\ \text{Pr}\big(supp_{\,\text{Pr}}\big)=1\ $ in the most general case, it is true in most cases of practical interest. It is true, for instance, if $\ \Omega\ $ is a $\sigma$-compact metric space.

If $\ X:\Omega\rightarrow\mathbb{R}\ $ is a real-valued random variable, then the distribution $\ \mu_X\ $ of $\ X\ $ is the probability measure $ \mu_X:\left\{S\subseteq\mathbb{R}\,|\,X^{-1}(S)\in\mathcal{F}\,\right\}\rightarrow$$[0,1]\ $ defined by $$ \mu_X(S)=\text{Pr}\left(X^{-1}(S)\right)\ . $$ If $\ X\ $ is Borel measurable, then $\ supp_{\,\mu_X}\ $ is a closed subset of $\ \mathbb{R}\ $ with $\ \mu_X\big(supp_{\,\mu_X}\big)=1\ $, and it is the smallest such closed subset. If $\ X\ $ has a probability density function $\ f\ $, then $\ supp_{\,\mu_X}\ $ will also be the support of $\ f\ $, as that term is normally defined—namely the closure of the set of points $\ x\ $ for which $\ f(x)\ne0\ $. Although the lecture notes you cite don't define random variables as functions on sample spaces, their concept of the "support of a random variable" is essentially the same thing as the set $\ supp_{\mu_X}\ $ I've just described.

I also have my doubts about the extent to which the general definition for the support of a function which you quote from Wikipedia is used in practice. I've previously only ever seen this notion defined as it is in Wikipedia's second sentence:

If the domain of $\ f\ $ is a topological space, the support of $\ f\ $ is ... defined as the smallest closed set containing all points not mapped to zero.

where I have replaced the possibly redundant "instead" with ellipses. Here, for instance, is the online Encyclopedia of Mathematics's definition, here is Wolfram MathWorld's definition, and here is planetmath's definition. As it happens, the support of a function with respect to the discrete topology is exactly the set of points on which the function does not vanish, so Wikipedia's general definition is equivalent to taking the discrete topology as being the default when no other topology has been specified. However, I've never come across an instance in technical mathematical literature where the support of a function has been defined in this way.

As far as I can recall, I have never seen an instance in modern mathematical literature where the support of any probability measure, even when it is the distribution of a random variable, has been called the support of a random variable, and, in my opinion, it would be a poor choice of terminology to use the term in that sense. It's quite conceivable to me, however, that if the sample space on which a random variable $\ X\ $ is defined happens to be a topological space, then studying the support of $\ X\ $, considered as a function on that sample space, in the sense in which the term "support of a function" is defined by the online Encyclopedia of Mathematics, Wolfram MathWorld and planet math, is something one may well wish to do. It seems to me, therefore, that use of the term "support of $\ X\ $" in that sense would be completely unobjectionable.

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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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