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In standard mathematical notation, a probability space follows the same structure of a measure space - i.e. a measur-able space, $(S,\mathcal A)$, plus a measure, $\mu$. It is formulated as a triple within parentheses. In the case of a probability space it looks like $(\Omega, \mathcal{F},\mathbb P)$, corresponding to the sample space ($\Omega$), a sigma algebra ($\mathcal F$) of measurable events, and finally the probability measure with a total mass of $1$.

On this probability space a random variable, $X$, is defined as a function or mapping from the subsets of $\Omega$ in the sigma algebra $\mathcal F$ to the real line, $\mathbb R$, expressed as $X: \Omega \rightarrow \mathbb R$. Intuition would have it that the ending of the prior sentence would be $\mathbb P$, which is the last entry in the triple $(\Omega, \mathcal{F},\mathbb P)$. But it, instead an $\mathbb R$. But this is not the actual question...

In the simple case of tossing two die, and counting the notation would make reference to:

  1. Sample space composed of the set of sets $\{\{1,2\},\{1,3\},\{1,4\},\cdots\{6,6\}\}$.
  2. Contrarily to the continuous case where the sample space is $\mathbb R$ and we can't assign probabilities to the immense powerset, we can now seamlessly consider the power set of $\Omega$, i.e. $2^{\Omega}$ as the events included in the sigma algebra in a discrete problem such as the one at hand. So the events would be the same as the sets in the sample space.
  3. The random variable would map each one of these sets to the real line ($\mathbb R$) in the obvious way of $X:\{2,3\}\mapsto 5$.
  4. Finally the cumulative density function (CDF) would be the measure $\mathbb P$ assigned to $X=5$, which I think happens to be $1/9$.

So the questions are:

  1. Why is the random variable $X$ not included in the $(\Omega, \mathcal{F},\mathbb P)$ - for example, $(\Omega, \mathcal{F},\color{blue}{X},\mathbb P) $ - but rather left implicit?
  2. Why is it stated that: A function $X:\Omega \rightarrow \mathbb R$ is measurable with respect to $\mathcal F$ if for any $x \in \mathbb R$ , the set $\{X \leq x\}=\{\omega\in\Omega:X(\omega)\leq x\}$ is measurable, or belongs to $\mathcal F$ - in other words, it is an event? That it needs to be an event is clear from the extrapolation to continuous random variables, but why does the nomenclature recurs back to $\omega$ and $\Omega$ in this expression... Hadn't the sample space been "tamed" into the more malleable events in the sigma algebra? Why not say, $\{X \leq x\}=\{\color{blue}{f}\in \color{blue}{\mathcal F}:X(\color{blue}{f})\leq x\}$?

Quick recap after Matthew's answer: The random variable maps outcomes (in the sample space) to the real line. The probability space maps events to [0,1]. So if we draw from an exponential distribution modeling waiting times with a $\lambda = 30 $ sec., the probability of an outcome of exactly $30$ sec. is $0$, because $30$ is not an event. Yet, evidently we have timed whatever process (random experiment) and mapped it to $\mathbb R$ as a random variable.

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  • $\begingroup$ I don't get the first question. It's not included because there can be different random variables defined on the same probability space. $\endgroup$ – amoeba Apr 21 '16 at 21:52
  • $\begingroup$ Ahhh... It is that easy... It makes total sense... And all of them are constrained to the rules of probability, i.e. $\mathbb P$ can't go beyond 1, disjoint sets are additive, etc... $\endgroup$ – Antoni Parellada Apr 21 '16 at 21:54
  • $\begingroup$ In my answer at stats.stackexchange.com/a/54894/919 I made an effort to motivate the reason for separating random variables from the underlying probability space. $\endgroup$ – whuber Apr 21 '16 at 22:27
  • $\begingroup$ Yes. Your post is superb. I have read it multiple times, including a couple of days ago when you posted an answer on the same topic. The concepts are starting to sink in; the notation, not so much. $\endgroup$ – Antoni Parellada Apr 21 '16 at 22:30
  • $\begingroup$ Out of curiosity, what text(s) or other sources are you working from? $\endgroup$ – Glen_b Apr 22 '16 at 1:59
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  1. Why is the random variable $X$ not included in the $(\Omega, \mathcal{F},\mathbb P)$ - for example, $(\Omega, \mathcal{F},\color{blue}{X},\mathbb P) $ - but rather left implicit?

Because the random variable is additional structure that exists on top of a probability space. The probability space exists independent of the random variable.

For example, if you're rolling two dice. One way to set it up is:

$$ \Omega = \{ (i, j) \mid 1 \leq i, j \leq 6 \} $$ $$ \mathcal{F} = 2^{\Omega} $$ $$ \mathbb{P}(E) = \frac{ |E| }{6 \times 6} $$

That's all we need to define probabilities of various sets of dice rolls, but there is no random variable here. If we want some random varaibles, we can invent them, sums of dice rolls, maximum roll, etc, and the probability space helps us analyse all of these.

  1. Why is it stated that: A function $X:\Omega \rightarrow \mathbb R$ is measurable with respect to $\mathcal F$ if for any $x \in \mathbb R$ , the set $\{X \leq x\}=\{\omega\in\Omega:X(\omega)\leq x\}$ is measurable, or belongs to $\mathcal F$ - in other words, it is an event? That it needs to be an event is clear from the extrapolation to continuous random variables, but why does the nomenclature recurs back to $\omega$ and $\Omega$ in this expression... Hadn't the sample space been "tamed" into the more malleable events in the sigma algebra? Why not say, $\{X \leq x\}=\{\color{blue}{f}\in \color{blue}{\mathcal F}:X(\color{blue}{f})\leq x\}$?

To borrow a phrase from programming, your suggestion does not type check. When you write

$$\{X \leq x\}=\{\color{blue}{f}\in \color{blue}{\mathcal F}:X(\color{blue}{f})\leq x\}$$

The function $X$ maps individual points in the sample space to real numbers, it does not map sets of points to real numbers. It, in some sense, does map sets of points to sets of real numbers, but

$$ \text{set of points} \leq x $$

doesn't really make sense.

In words, the definition you want for a function to be measureable (and hence a random varaible) is

Every set that is formed by collecting together all points $\omega$ such that $X(\omega)$ is less than some value is measurable (i.e. in the sigma algebra).

If you try to translate your proposed definition into words in the same way, I think you will have some difficulties (I tried, and got all tangled up).

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  • $\begingroup$ Matthew, Can you please touch upon two aspect of your answer: 1. What is $\frac{|\omega|}{6\times6}$ in words (of course I get the $36$)? But what is the abolute value of omega (omega being an outcome, I presume), and is the probability defined in this case just because it is so trivial an experiment? $\endgroup$ – Antoni Parellada Apr 21 '16 at 23:24
  • $\begingroup$ And 2. I thought that probabilities can only be assigned to events, not to outcomes, which, despite the fact that the second part of your answer does make sense it sounds at odds with the whole idea of making probabilities work with continuous variables, when it is only the open sets that have assigned probability values (?). $\endgroup$ – Antoni Parellada Apr 21 '16 at 23:27
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    $\begingroup$ @Antoni The $|\,\cdot\,|$ operator doesn't always represent "absolute value" but is a more general magnitude ("size of") operator. For example, with finite sets it usually means "number of elements in". $\endgroup$ – Glen_b Apr 22 '16 at 2:02
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    $\begingroup$ The equation $\mathbb{P}(\omega)=\ldots$ here seems to use $\omega$ to denote an event (subset of $\Omega$, member of $\mathcal{F})$, while typically $\omega$ is used to refer to an outcome (member of $\Omega$). So perhaps the notation should be edited to use some other symbol than $\omega$ for the event. $\endgroup$ – Juho Kokkala Apr 22 '16 at 6:29
  • $\begingroup$ Can it be that the confusion on part (2) is my disregard for the curly brackets? In other words, although inside the brackets there are conditions on what the sample space points need to fulfill, i.e. Need to be less than $x$, it is the $\{\}$ set (or collection of sets(?)) of these outcomes that need to be $\in \mathcal F$ to be measurable. $\endgroup$ – Antoni Parellada Apr 22 '16 at 12:30

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