Difference between revisions of "Random variable"

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A '''Random variable''' is a [[Measurable map|measurable map]] from a [[Probability space|probability space]] to any [[Measurable space|measurable space]]
 
A '''Random variable''' is a [[Measurable map|measurable map]] from a [[Probability space|probability space]] to any [[Measurable space|measurable space]]
  
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Let {{M|(\Omega,\mathcal{A},\mathbb{P})}} be a [[Probability space|probability space]] and let {{M|\Epsilon:\Omega\rightarrow\mathbb{R} }} be a random variable
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Let {{M|(\Omega,\mathcal{A},\mathbb{P})}} be a [[Probability space|probability space]] and let {{M|X:(\Omega,\mathcal{A})\rightarrow(V,\mathcal{U}) }} be a random variable
  
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(that means it is a [[Measurable map|measurable map]] '''FROM''' a probability space to a measurable space recall)
 
  
 
Then:
 
Then:
  
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{{Todo|Finish this because it's iffy}}
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<math>X^{-1}(U\in\mathcal{U})\in\mathcal{A}</math>, but anything <math>\in\mathcal{A}</math> is {{M|\mathbb{P} }}-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]])
 +
 
 +
==Notation==
 +
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:
 +
* {{M|X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(V,\mathcal{U}) }}
 +
* {{M|X:(\Omega,\mathcal{A})\rightarrow(V,\mathcal{U}) }}
 +
 
 +
With the understanding we write {{M|\mathbb{P} }} in the top one only because it is convenient to remind ourselves what probability measure we are using.
 +
 
 +
==Pitfall==
 +
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 [[Sigma-algebra|{{sigma|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>.
 +
 
 +
==Example==
 +
===Discrete random variable===
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{{Begin Example}}
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Recall the roll two die example from [[Probability space|probability spaces]], we will consider the RV {{M|X}} = the sum of the scores
 +
{{Begin Example Body}}
 +
Recall the die example from [[Probability space|probability spaces]] (which is restated less verbosely here), there:
 +
{|class="wikitable" border="1"
 +
|-
 +
! Component
 +
! Definition
 +
|-
 +
| {{M|\Omega}}
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| <math>\Omega=\{(a,b)|\ a,b\in\mathbb{N},\ a,b\in[0,6]\}</math>
 +
|-
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| {{M|\mathcal{A} }}
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| <math>\mathcal{A}=\mathcal{P}(\Omega)</math>
 +
|-
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| {{M|\mathbb{P} }}
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| <math>\mathbb{P}(A) = \frac{1}{36}|A|</math>
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|}
 +
 
 +
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|measurable space]] however we need not consider a measure on it.
 +
 
 +
Writing {{M|X}} 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 {{M|X}} 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>
 +
{{End Example Body}}
 +
{{End Example}}
 +
 
 +
====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>
  
  
 
{{Definition|Measure Theory|Statistics}}
 
{{Definition|Measure Theory|Statistics}}

Latest revision as of 14:41, 20 March 2015

Definition

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


Then:

[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)

Notation

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.

Pitfall

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].

Example

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]

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