View source for Example:Joint probability distribution

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__TOC__
==Example==
Define a [[probability space]], {{M|(S,\Omega,\mathbb{P})}} as follows:
* Let {{M|S:\eq\{0,1,2,3\}\subseteq\mathbb{N} }},
* let {{M|\Omega:\eq}}[[Sigma-algebra generated by|{{M|\sigma(S)}}]], and 
* let {{M|\mathbb{P}:\Omega\rightarrow\mathbb{R} }} be uniform over {{M|S}}, that is to say: {{M|\mathbb{P}:R\mapsto\frac{\vert R\vert}{\vert S\vert} }}
** specifically in this case: {{M|\mathbb{P}(R):\eq\frac{1}{4}\vert R\vert}}.


<div style="float:right;overflow:hidden;">
{| class="wikitable" border="1"
|-
| <center><mm>\xymatrix{\mathbf{X} & \mathbf{S} & \mathbf{Y} }</mm><hr/><mm>\xymatrix{ A & 0 \ar[l] \ar[r] & C \\ & 1 \ar[ul] \ar[dr] & \\ B & 2 \ar[l] \ar[r] & D\\ & 3 \ar[ur] \ar[ul] }</mm></center>
|-
! Experiment mappings
|}
</div>
We define two {{link|random variable||s}}:
* {{M|X:S\rightarrow\{A,B\} }}<ref group="Note" name="rvmeasurability"/>
** By {{M|X(0):\eq A}}, {{M|X(1):\eq A}}, {{M|X(2):\eq B}} and {{M|X(3):\eq B}}
*** {{M|X:0,1\mapsto A}} and {{M|X:2,3\mapsto B}} {{XXX|Is this better?}}
** Think of {{M|X}} as categorising the {{link|elementary event||s}} as "low" ({{M|0}} and {{M|1}}) and "high" ({{M|2}} and {{M|3}}) 
* {{M|Y:S\rightarrow\{C,D\} }}<sup>Again:</sup><ref group="Note" name="rvmeasurability"/>
** By {{M|Y:0\mapsto C}} and {{M|Y:1,2,3\mapsto D}}
** Think of {{M|Y}} as an "is zero" measurement


'''Note:'''
Our experiment here has 2 measurements, {{M|X}} is "''high or low''" and {{M|Y}} is "''zero or non zero''"
==PAGE NOTES==
We get the following:
* Events: {{M|A\eq\{0,1\} }} (giving {{M|\mathbb{P}[A]\eq\frac{2}{4}\eq\frac{1}{2} }}), {{M|B\eq\{2,3\} }} (giving {{M|\mathbb{P}[B]\eq\frac{2}{4}\eq\frac{1}{2} }}), {{M|C\eq\{0\} }} (giving {{M|\mathbb{P}[C]\eq\frac{1}{4} }}) and {{M|D\eq\{1,2,3\} }} (giving {{M|\mathbb{P}[D]\eq\frac{3}{4} }}) as events in {{M|\Omega}}, thus:
** {{M|A\cap C}}<ref group="Note">Meaning "{{M|A}} and {{M|C}}"</ref>{{M|\eq\{0\} }} so {{M|\mathbb{P}[A\wedge B]\eq\frac{1}{4} }}
** {{M|A\cap D\eq\{1\} }} so {{M|\mathbb{P}[A\wedge D]\eq\frac{1}{4} }}
** {{M|B\cap C\eq\{\}\eq\emptyset }} so {{M|\mathbb{P}[B\wedge C]\eq 0 }}
** {{M|B\cap D\eq\{2,3\} }} so {{M|\mathbb{P}[B\wedge D]\eq\frac{1}{2} }}
* Notice as well that {{M|\mathbb{P}[A\cup B]\eq\mathbb{P}[A]+\mathbb{P}[B]}}<ref group="Note">As {{M|X}} is a function {{M|A}} and {{M|B}} as events in {{M|\Omega}} must be [[disjoint]]!</ref>{{M|\eq 1}} and {{M|\mathbb{P}[C\cup D]\eq\mathbb{P}[C]+\mathbb{P}[D]\eq 1}} for the same reasoning

Giving the following table:
{| class="wikitable" border="1" style="min-width:15em;"
|-
! 
! A↓
! B↓
! {{M|\mathbf{\Sigma} }}
|-
! C →
| style="text-align:center;" | {{M|\frac{1}{4} }}
| style="text-align:center;" | {{M|0}}
! {{M|\frac{1}{4} }}
|-
! D →
| style="text-align:center;" | {{M|\frac{1}{4} }}
| style="text-align:center;" | {{M|\frac{2}{4}\eq\frac{1}{2} }}
! {{M|\frac{3}{4} }}
|-
! {{M|\mathbf{\Sigma} }}
! {{M|\frac{1}{2} }}
! {{M|\frac{1}{2} }}
| style="text-align:center;" | {{M|1}}
|} 
But there are others, eg the following is if (or for) independent {{M|X}} and {{M|Y}} where the individual distributions of {{M|X}} and {{M|Y}} are the same, ie {{M|P[X\eq A]\eq\frac{1}{2} }} still for example
{| class="wikitable" border="1" style="min-width:15em;"
|-
! 
! A↓
! B↓
! {{M|\mathbf{\Sigma} }}
|-
! C →
| style="text-align:center;" | {{M|\frac{1}{4}\times\frac{1}{2}\eq\frac{1}{8} }}
| style="text-align:center;" | {{M|\frac{1}{4}\times\frac{1}{2}\eq\frac{1}{8} }}
! {{M|\frac{2}{8}\eq\frac{1}{4} }}
|-
! D →
| style="text-align:center;" | {{M|\frac{3}{4}\times\frac{1}{2}\eq\frac{3}{8} }}
| style="text-align:center;" | {{M|\frac{3}{4}\times\frac{1}{2}\eq\frac{3}{8} }}
! {{M|\frac{6}{8}\eq\frac{3}{4} }}
|-
! {{M|\mathbf{\Sigma} }}
! {{M|\frac{4}{8}\eq\frac{1}{2} }}
! {{M|\frac{4}{8}\eq\frac{1}{2} }}
| style="text-align:center;" | {{M|1}}
|} 
In general:
{| class="wikitable" border="1" style="min-width:15em;"
|-
! 
! A↓
! B↓
! {{M|\mathbf{\Sigma} }}
|-
! C →
| style="text-align:center;" | {{M|\alpha:\eq\mathbb{P}[A\cap C] }}
| style="text-align:center;" | {{M|\beta:\eq\mathbb{P}[B\cap C]}}
! {{M|\alpha+\beta\eq\mathbb{P}[C] }}
|-
! D →
| style="text-align:center;" | {{M|\gamma:\eq\mathbb{P}[A\cap D]}}
| style="text-align:center;" | {{M|\delta:\eq\mathbb{P}[B\cap D]}}
! {{M|\gamma+\delta\eq\mathbb{P}[D] }}
|-
! {{M|\mathbf{\Sigma} }}
! {{M|\alpha+\gamma\eq\mathbb{P}[A] }}
! {{M|\beta+\delta\eq\mathbb{P}[B] }}
| style="text-align:center;" | {{M|1}}
|} 

Solving the equations that arise and parameterising {{M|\alpha}} as {{M|t}} we get:
{| class="wikitable" border="1" style="min-width:15em;"
|-
! 
! A↓
! B↓
! {{M|\mathbf{\Sigma} }}
|-
! C →
| style="text-align:center;" | {{M|\frac{t}{4} }}
| style="text-align:center;" | {{M|\frac{1-t}{4} }}
! {{M|\frac{1}{4} }}
|-
! D →
| style="text-align:center;" | {{M|\frac{2-t}{4} }}
| style="text-align:center;" | {{M|\frac{1+t}{4} }}
! {{M|\frac{3}{4} }}
|-
! {{M|\mathbf{\Sigma} }}
! {{M|\frac{1}{2} }}
! {{M|\frac{1}{2} }}
| style="text-align:center;" | {{M|1}}
|} 

By using the constraints we see {{M|t\in[0,1]}} produces a "valid something" (see the {{XXX|XXX note just below}})

This is our case for {{M|t\eq 1}}, for {{M|t\eq\frac{1}{2} }} it is the independent example 

We have essentially parameterised all {{XXX|of something, note that there are only 24 {{C|1==(4'''C'''2)x(4'''C'''1)}} such RVs, would {{M|t\eq\frac{\sqrt{2} }{2} }} make sense?}}

So what have we parameterised I wonder? 
===Final thoughts===
Thinking about it.... really only {{M|t\eq 0}} and {{M|t\eq 1}} are valid values as we must deal in a whole-number of quarters
{{Green highlight|Thus there are only 2 "joint distributions" for any such set up}} - if you think about it, {{M|Y}} always maps 1 element of {{M|S}} to either {{M|C}} or {{M|D}} and the remaining elements of {{M|S}} to the other one, as {{M|X}} maps 2 of {{M|S}} to one and the other two to its other value, one of these will always never be from the single {{M|Y}} one - so out of the 24 distributions of {{M|(X,Y)}} - there are actually only 2 "situations" (experiments?) described by them! 
==Notes==
<references group="Note">
<ref name="rvmeasurability">Let {{M|Z:S\rightarrow T}} where {{M|T}} is any [[finite set]], note that for {{M|U\in}}[[power set|{{M|\mathcal{P}(T)}}]]{{M|\eq\sigma(T)}} that {{M|Z^{-1}(U)\eq\bigcup_{V\in U}Z^{-1}(V) }} - which is a finite union. As {{M|S}} is a finite set in this example, we see that {{M|\sigma(S)\eq\mathcal{P}(S)}}, which gives us {{M|\forall s\in S[\{s\}\in \sigma(S)]}}.
* Thus as {{M|Z}} is a function, we see it must be measurable, both the {{link|codomain|function|s}} of {{M|X}} and {{M|Y}} are finite, thus they're both measurable</ref>
</references>
[[Category:Probability]][[Category:Statistics]]