Random variable

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A Random variable is a measurable map from a probability space to any measurable space

Let [ilmath](\Omega,\mathcal{A},\mathbb{P})[/ilmath] be a probability space and let [ilmath]X:(\Omega,\mathcal{A})\rightarrow(V,\mathcal{U}) [/ilmath] be a random variable


[math]X^{-1}(U\in\mathcal{U})\in\mathcal{A}[/math], but anything [math]\in\mathcal{A}[/math] is [ilmath]\mathbb{P} [/ilmath]-measurable! So we see:

[math]\mathbb{P}(X^{-1}(U\in\mathcal{U}))\in[0,1][/math] which we may often write as: [math]\mathbb{P}(X=U)[/math] for simplicity (see Mathematicians are lazy)


Often a measurable space that is the domain of the RV will be a probability space, given as [math](\Omega,\mathcal{A},\mathbb{P})[/math], and we may write either:

  • [ilmath]X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(V,\mathcal{U}) [/ilmath]
  • [ilmath]X:(\Omega,\mathcal{A})\rightarrow(V,\mathcal{U}) [/ilmath]

With the understanding we write [ilmath]\mathbb{P} [/ilmath] in the top one only because it is convenient to remind ourselves what probability measure we are using.


Note that it is only guaranteed that [math]X^{-1}(U\in\mathcal{U})\in\mathcal{A}[/math] but it is not guaranteed that [math]X(A\in\mathcal{A})\in\mathcal{U}[/math], it may sometimes be the case.

For example consider the trivial [ilmath]\sigma[/ilmath]-algebra [math]\mathcal{U}=\{\emptyset,V\}[/math]

However If you consider [math]X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(V,\{\emptyset,V\})[/math] then this is just the random variable "something happens" underneath it all, or if [math]V=\{2,\cdots,12\}[/math] the event that the sum of the scores is [math]\ge 2[/math].


Discrete random variable

Recall the roll two die example from probability spaces, we will consider the RV [ilmath]X[/ilmath] = the sum of the scores

Recall the die example from probability spaces (which is restated less verbosely here), there:

Component Definition
[ilmath]\Omega[/ilmath] [math]\Omega=\{(a,b)|\ a,b\in\mathbb{N},\ a,b\in[0,6]\}[/math]
[ilmath]\mathcal{A} [/ilmath] [math]\mathcal{A}=\mathcal{P}(\Omega)[/math]
[ilmath]\mathbb{P} [/ilmath] [math]\mathbb{P}(A) = \frac{1}{36}|A|[/math]

Let us define the Random variable that is the sum of the scores on the die, that is [math]X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(\{2,\cdots,12\},\mathcal{P}(\{2,\cdots,12\}))[/math].

It should be clear that [math](\{2,\cdots,12\},\mathcal{P}(\{2,\cdots,12\}))[/math] is a measurable space however we need not consider a measure on it.

Writing [ilmath]X[/ilmath] out explicitly is hard but there are two parts to it:

Warning - the first bullet point is a suspected claim

  • We can look at what generates a space, we need only consider the single events really, that is to say:
    [math]X(A\in\mathcal{A})\cup X(B\in\mathcal{A})=X(A\cup B\in\mathcal{A})[/math], so we need only look at [ilmath]X[/ilmath] of the individual events

TODO: Prove this

  • We can write it more explicitly as:
    [math]X(A\in\mathcal{A})=\{a+b|(a,b)\in A\}[/math]

Example of pitfall

Take [math]X:(\Omega,\mathcal{P}(\Omega),\mathbb{P})\rightarrow(V,\mathcal{U})[/math], if we define [math]\mathcal{U}=\{\emptyset,V\}[/math] then clearly:

[math]X(\{(1,2)\})=\{3\}\notin\mathcal{U}[/math]. Yet it is still measurable.

So an example! [math]\mathbb{P}(X^{-1}(\{5\}))=\mathbb{P}(X=5)=\mathbb{P}(\{(1,4),(4,1),(2,3),(3,2)\})=\frac{4}{36}=\frac{1}{9}[/math]