Difference between revisions of "Random variable"
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==Example== | ==Example== | ||
===Discrete random variable=== | ===Discrete random variable=== | ||
| + | {{Begin Example}} | ||
| + | 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: | Recall the die example from [[Probability space|probability spaces]] (which is restated less verbosely here), there: | ||
{|class="wikitable" border="1" | {|class="wikitable" border="1" | ||
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* We can write it more explicitly as: | * We can write it more explicitly as: | ||
*: <math>X(A\in\mathcal{A})=\{a+b|(a,b)\in A\}</math> | *: <math>X(A\in\mathcal{A})=\{a+b|(a,b)\in A\}</math> | ||
| + | {{End Example Body}} | ||
| + | {{End Example}} | ||
====Example of pitfall==== | ====Example of pitfall==== | ||
Revision as of 14:18, 20 March 2015
Contents
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]
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]
