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]]
−	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	+	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
−	(that means it is a [[Measurable map|measurable map]] '''FROM''' a probability space to a measurable space recall)
	Then:		Then:
−	{{Todo|Finish this because it's iffy}}	+	<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>
		+
		+	==Example==
		+	===Discrete random variable===
		+	Recall the die example from [[Probability space|probability spaces]] (which is restated less verbosely here), there:
		+	{|class="wikitable" border="1"
		+	|-
		+	! Component
		+	! Definition
		+	|-
		+	| {{M|\Omega}}
		+	| <math>\Omega=\{(a,b)|\ a,b\in\mathbb{N},\ a,b\in[0,6]\}</math>
		+	|-
		+	| {{M|\mathcal{A} }}
		+	| <math>\mathcal{A}=\mathcal{P}(\Omega)</math>
		+	|-
		+	| {{M|\mathbb{P} }}
		+	| <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|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>
		+
		+	====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}}

---

Revision as of 09:02, 19 March 2015

Definition

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

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

Then:

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

$$\mathbb{P}(X^{-1}(U\in\mathcal{U}))\in[0,1]$$ which we may often write as: $$\mathbb{P}(X=U)$$ 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

$$(\Omega,\mathcal{A},\mathbb{P})$$ , and we may write either:

• $X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(V,\mathcal{U})$
• $X:(\Omega,\mathcal{A})\rightarrow(V,\mathcal{U})$

With the understanding we write $\mathbb{P}$ 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

$$X^{-1}(U\in\mathcal{U})\in\mathcal{A}$$ but it is not guaranteed that $$X(A\in\mathcal{A})\in\mathcal{U}$$ , it may sometimes be the case.

For example consider the trivial $\sigma$-algebra

$$\mathcal{U}=\{\emptyset,V\}$$

Example

Discrete random variable

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

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

Let us define the Random variable that is the sum of the scores on the die, that is

$$X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(\{2,\cdots,12\},\mathcal{P}(\{2,\cdots,12\}))$$ .

It should be clear that

$$(\{2,\cdots,12\},\mathcal{P}(\{2,\cdots,12\}))$$ is a measurable space however we need not consider a measure on it.

Writing $X$ 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:

  $$X(A\in\mathcal{A})\cup X(B\in\mathcal{A})=X(A\cup B\in\mathcal{A})$$ , so we need only look at $X$ of the individual events

---

TODO: Prove this

---

• We can write it more explicitly as:

  $$X(A\in\mathcal{A})=\{a+b|(a,b)\in A\}$$

Example of pitfall

Take

$$X:(\Omega,\mathcal{P}(\Omega),\mathbb{P})\rightarrow(V,\mathcal{U})$$ , if we define $$\mathcal{U}=\{\emptyset,V\}$$ then clearly:

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

So an example!

$$\mathbb{P}(X^{-1}(\{5\}))=\mathbb{P}(X=5)=\mathbb{P}(\{(1,4),(4,1),(2,3),(3,2)\})=\frac{4}{36}=\frac{1}{9}$$