Example:Joint probability distribution

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Example

Define a probability space, (S,Ω,P) as follows:

  • Let S:={0,1,2,3}N,
  • let Ω:=σ(S), and
  • let P:ΩR be uniform over S, that is to say: P:R|R||S|
    • specifically in this case: P(R):=14|R|.



Experiment mappings

We define two random variables:

  • X:S{A,B}[Note 1]
    • By X(0):=A, X(1):=A, X(2):=B and X(3):=B
      • X:0,1A and X:2,3B
        TODO: Is this better?
    • Think of X as categorising the elementary events as "low" (0 and 1) and "high" (2 and 3)
  • Y:S{C,D}Again:[Note 1]
    • By Y:0C and Y:1,2,3D
    • Think of Y as an "is zero" measurement


Note: Our experiment here has 2 measurements, X is "high or low" and Y is "zero or non zero"

PAGE NOTES

We get the following:

  • Events: A={0,1} (giving P[A]=24=12), B={2,3} (giving P[B]=24=12), C={0} (giving P[C]=14) and D={1,2,3} (giving P[D]=34) as events in Ω, thus:
    • AC[Note 2]={0} so P[AB]=14
    • AD={1} so P[AD]=14
    • BC={}= so P[BC]=0
    • BD={2,3} so P[BD]=12
  • Notice as well that P[AB]=P[A]+P[B][Note 3]=1 and P[CD]=P[C]+P[D]=1 for the same reasoning

Giving the following table:

A↓ B↓ Σ
C → 14 0 14
D → 14 24=12 34
Σ 12 12 1

But there are others, eg the following is if (or for) independent X and Y where the individual distributions of X and Y are the same, ie P[X=A]=12 still for example

A↓ B↓ Σ
C → 14×12=18 14×12=18 28=14
D → 34×12=38 34×12=38 68=34
Σ 48=12 48=12 1

In general:

A↓ B↓ Σ
C → α:=P[AC] β:=P[BC] α+β=P[C]
D → γ:=P[AD] δ:=P[BD] γ+δ=P[D]
Σ α+γ=P[A] β+δ=P[B] 1

Solving the equations that arise and parameterising α as t we get:

A↓ B↓ Σ
C → t4 1t4 14
D → 2t4 1+t4 34
Σ 12 12 1
By using the constraints we see t[0,1] produces a "valid something" (see the
TODO: XXX note just below
)

This is our case for t=1, for t=12 it is the independent example

We have essentially parameterised all
TODO: of something, note that there are only 24 =(4C2)x(4C1) such RVs, would t=22 make sense?

So what have we parameterised I wonder?

Final thoughts

Thinking about it.... really only t=0 and t=1 are valid values as we must deal in a whole-number of quarters Thus there are only 2 "joint distributions" for any such set up - if you think about it, Y always maps 1 element of S to either C or D and the remaining elements of S to the other one, as X maps 2 of S to one and the other two to its other value, one of these will always never be from the single Y one - so out of the 24 distributions of (X,Y) - there are actually only 2 "situations" (experiments?) described by them!

Notes

  1. Jump up to: 1.0 1.1 Let Z:ST where T is any finite set, note that for UP(T)=σ(T) that Z1(U)=VUZ1(V) - which is a finite union. As S is a finite set in this example, we see that σ(S)=P(S), which gives us sS[{s}σ(S)].
    • Thus as Z is a function, we see it must be measurable, both the codomains of X and Y are finite, thus they're both measurable
  2. Jump up Meaning "A and C"
  3. Jump up As X is a function A and B as events in Ω must be disjoint!