Confirming the definition of the infimum of a random variable

probabilityprobability theory

The infimum of a function is the greatest value such that all $y \in Y, f: X\rightarrow Y$ is greater or equal to.

Random variable is essentially a function that maps events (as in values in a normal function, or X) to a value on the real number line. So in my understanding the infimum of a r.v. is defined as the greatest value on the real number line that is less than all all the values that has being assigned to an event?

This definition then confuses me as I take into account of real distributions such as normal distribution. Specifically, what is the defined events here? If we consider a event as any number on the real line, we then assign a mapping $N: X\rightarrow X$, i.e. the event X is assigned the value of X (which is itself, and considered as the space named $X'$). Hence the infimum of the normal r.v. is a value smaller than all of $X' =X $, which is $-\infty$?

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A random variable is a function, but it's a function from some sample space ($\Omega$) to the real number line. The sample space must also be equipped with a probability measure $(\mathbb P)$.

The typical use of $\inf$ in probability contexts is to have some indexed collection of random variables, not just a single random variable. Additionally, the infimum is usually indexed over the collection and not over the sample space $\Omega$. In this way, $\inf_i\{X_i\}$ is usually a random variable of its own right.

Example: consider a sequence of standard die rolls $X_1, X_2, X_3$, and set $Z = \inf\{X_1, X_2, X_3\}$. If we rolled a 3, 5, and a 4 -- that is, on the event $\{\omega : X_1(\omega) = 3, X_2(\omega) = 5, X_3(\omega) = 4\}$ -- we would have $Z(\omega) = 3$. Similarly, if $X_1 = X_2 = 6$ and $X_3 = 1$, we would have $Z = 1$.

The main reason it's not typically meaningful to index $\inf$ over the domain $\Omega$ (or equivalently, the range of the random variable itself) is because $\Omega$ is rarely explicitly defined at all. Let's use a die roll example; our sample space will be $\Omega = \{A, B, C, D, E, F, G\}$, and our probability measure will be $\mathbb P(A) = \mathbb P(B) = \dots = \mathbb P(F) = 1/6$ and $\mathbb P(G) = 0$. The random variable $X$ will map $A \to 1, B \to 2, C \to 3, D \to 4, E \to 5, F \to 6,$ and $G \to -100$. I claim that this perfectly models a fair 6-sided die roll, because the numbers 1 through 6 all have the appropriate probability.

If you try to use the infimum in the way that you're describing, the only reasonable approach will be something like the infimum of the range of $X$ -- but in this case, that would apparently be $-100$, unless you go out of your way to exclude measure-$0$ sets from $\Omega$. This definition would become immediately problematic when you include things like continuous random variables, where the domain $\Omega$ could be (for instance) the interval $(0, 1)$ with an appropriate mapping $X$, and every single point in the interval has measure 0. We almost never think too hard about a random variable as actually being a function in this way.

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Probability Theory – Precise Definition of the Support of a Random Variable

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.

[Math] Definition of Random Variable on Measure Theory!

The random variable is not limited to values less than $x$. For instance, I can show you that the function $$ X(t) = \frac{1}{t} $$ is a measurable function on $(0, 1)$. Here's how. Let's look at

$$ \{ t \mid X(t) \le 11 \} $$ That's the set of all points in the domain for which $X(t) = 1/t$ is less than 11, which is exactly $$ A = \{t \mid \frac{1}{11} \le t < 1 \} $$

Clear enough? Now look at $A$. Is it a measurable set in the measure space $(0, 1)$? Sure. It's a half-open interval, and those are all measurable sets!

Now there was nothing special about the number 11: I could have chosen $6$ or $113$ or $-12$ (although if I'd chosen $-12$, then the set $A$ woudl have been empty. Fortunately, the empty set is measurable, too!

Now go back and look at the definition above: it says "whenever you build a set like $A$, it's measurable." It does not say that the function $X$ is bounded. Indeed, in my case, we have $$ X(0.1) = 10 \\ X(0.01) = 100 \\ X(0.001) = 1000 $$ and so on, and it's pretty clear that $X$ itself is an unbounded function on $(0, 1)$.

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Probability Theory – Precise Definition of the Support of a Random Variable

[Math] Definition of Random Variable on Measure Theory!

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