# Example:Joint probability distribution

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## Example

Define a probability space, [ilmath](S,\Omega,\mathbb{P})[/ilmath] as follows:

• Let [ilmath]S:\eq\{0,1,2,3\}\subseteq\mathbb{N} [/ilmath],
• let [ilmath]\Omega:\eq[/ilmath][ilmath]\sigma(S)[/ilmath], and
• let [ilmath]\mathbb{P}:\Omega\rightarrow\mathbb{R} [/ilmath] be uniform over [ilmath]S[/ilmath], that is to say: [ilmath]\mathbb{P}:R\mapsto\frac{\vert R\vert}{\vert S\vert} [/ilmath]
• specifically in this case: [ilmath]\mathbb{P}(R):\eq\frac{1}{4}\vert R\vert[/ilmath].

 Experiment mappings $\xymatrix{\mathbf{X} & \mathbf{S} & \mathbf{Y} }$$\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] }$

We define two random variables:

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

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

## PAGE NOTES

We get the following:

• Events: [ilmath]A\eq\{0,1\} [/ilmath] (giving [ilmath]\mathbb{P}[A]\eq\frac{2}{4}\eq\frac{1}{2} [/ilmath]), [ilmath]B\eq\{2,3\} [/ilmath] (giving [ilmath]\mathbb{P}[B]\eq\frac{2}{4}\eq\frac{1}{2} [/ilmath]), [ilmath]C\eq\{0\} [/ilmath] (giving [ilmath]\mathbb{P}[C]\eq\frac{1}{4} [/ilmath]) and [ilmath]D\eq\{1,2,3\} [/ilmath] (giving [ilmath]\mathbb{P}[D]\eq\frac{3}{4} [/ilmath]) as events in [ilmath]\Omega[/ilmath], thus:
• [ilmath]A\cap C[/ilmath][Note 2][ilmath]\eq\{0\} [/ilmath] so [ilmath]\mathbb{P}[A\wedge B]\eq\frac{1}{4} [/ilmath]
• [ilmath]A\cap D\eq\{1\} [/ilmath] so [ilmath]\mathbb{P}[A\wedge D]\eq\frac{1}{4} [/ilmath]
• [ilmath]B\cap C\eq\{\}\eq\emptyset [/ilmath] so [ilmath]\mathbb{P}[B\wedge C]\eq 0 [/ilmath]
• [ilmath]B\cap D\eq\{2,3\} [/ilmath] so [ilmath]\mathbb{P}[B\wedge D]\eq\frac{1}{2} [/ilmath]
• Notice as well that [ilmath]\mathbb{P}[A\cup B]\eq\mathbb{P}[A]+\mathbb{P}[B][/ilmath][Note 3][ilmath]\eq 1[/ilmath] and [ilmath]\mathbb{P}[C\cup D]\eq\mathbb{P}[C]+\mathbb{P}[D]\eq 1[/ilmath] for the same reasoning

Giving the following table:

A↓ B↓ [ilmath]\mathbf{\Sigma} [/ilmath]
C → [ilmath]\frac{1}{4} [/ilmath] [ilmath]0[/ilmath] [ilmath]\frac{1}{4} [/ilmath]
D → [ilmath]\frac{1}{4} [/ilmath] [ilmath]\frac{2}{4}\eq\frac{1}{2} [/ilmath] [ilmath]\frac{3}{4} [/ilmath]
[ilmath]\mathbf{\Sigma} [/ilmath] [ilmath]\frac{1}{2} [/ilmath] [ilmath]\frac{1}{2} [/ilmath] [ilmath]1[/ilmath]

But there are others, eg the following is if (or for) independent [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] where the individual distributions of [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] are the same, ie [ilmath]P[X\eq A]\eq\frac{1}{2} [/ilmath] still for example

A↓ B↓ [ilmath]\mathbf{\Sigma} [/ilmath]
C → [ilmath]\frac{1}{4}\times\frac{1}{2}\eq\frac{1}{8} [/ilmath] [ilmath]\frac{1}{4}\times\frac{1}{2}\eq\frac{1}{8} [/ilmath] [ilmath]\frac{2}{8}\eq\frac{1}{4} [/ilmath]
D → [ilmath]\frac{3}{4}\times\frac{1}{2}\eq\frac{3}{8} [/ilmath] [ilmath]\frac{3}{4}\times\frac{1}{2}\eq\frac{3}{8} [/ilmath] [ilmath]\frac{6}{8}\eq\frac{3}{4} [/ilmath]
[ilmath]\mathbf{\Sigma} [/ilmath] [ilmath]\frac{4}{8}\eq\frac{1}{2} [/ilmath] [ilmath]\frac{4}{8}\eq\frac{1}{2} [/ilmath] [ilmath]1[/ilmath]

In general:

A↓ B↓ [ilmath]\mathbf{\Sigma} [/ilmath]
C → [ilmath]\alpha:\eq\mathbb{P}[A\cap C] [/ilmath] [ilmath]\beta:\eq\mathbb{P}[B\cap C][/ilmath] [ilmath]\alpha+\beta\eq\mathbb{P}[C] [/ilmath]
D → [ilmath]\gamma:\eq\mathbb{P}[A\cap D][/ilmath] [ilmath]\delta:\eq\mathbb{P}[B\cap D][/ilmath] [ilmath]\gamma+\delta\eq\mathbb{P}[D] [/ilmath]
[ilmath]\mathbf{\Sigma} [/ilmath] [ilmath]\alpha+\gamma\eq\mathbb{P}[A] [/ilmath] [ilmath]\beta+\delta\eq\mathbb{P}[B] [/ilmath] [ilmath]1[/ilmath]

Solving the equations that arise and parameterising [ilmath]\alpha[/ilmath] as [ilmath]t[/ilmath] we get:

A↓ B↓ [ilmath]\mathbf{\Sigma} [/ilmath]
C → [ilmath]\frac{t}{4} [/ilmath] [ilmath]\frac{1-t}{4} [/ilmath] [ilmath]\frac{1}{4} [/ilmath]
D → [ilmath]\frac{2-t}{4} [/ilmath] [ilmath]\frac{1+t}{4} [/ilmath] [ilmath]\frac{3}{4} [/ilmath]
[ilmath]\mathbf{\Sigma} [/ilmath] [ilmath]\frac{1}{2} [/ilmath] [ilmath]\frac{1}{2} [/ilmath] [ilmath]1[/ilmath]
By using the constraints we see [ilmath]t\in[0,1][/ilmath] produces a "valid something" (see the
TODO: XXX note just below
)

This is our case for [ilmath]t\eq 1[/ilmath], for [ilmath]t\eq\frac{1}{2} [/ilmath] 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 [ilmath]t\eq\frac{\sqrt{2} }{2} [/ilmath] make sense?

So what have we parameterised I wonder?

### Final thoughts

Thinking about it.... really only [ilmath]t\eq 0[/ilmath] and [ilmath]t\eq 1[/ilmath] 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, [ilmath]Y[/ilmath] always maps 1 element of [ilmath]S[/ilmath] to either [ilmath]C[/ilmath] or [ilmath]D[/ilmath] and the remaining elements of [ilmath]S[/ilmath] to the other one, as [ilmath]X[/ilmath] maps 2 of [ilmath]S[/ilmath] to one and the other two to its other value, one of these will always never be from the single [ilmath]Y[/ilmath] one - so out of the 24 distributions of [ilmath](X,Y)[/ilmath] - there are actually only 2 "situations" (experiments?) described by them!

## Notes

1. Let [ilmath]Z:S\rightarrow T[/ilmath] where [ilmath]T[/ilmath] is any finite set, note that for [ilmath]U\in[/ilmath][ilmath]\mathcal{P}(T)[/ilmath][ilmath]\eq\sigma(T)[/ilmath] that [ilmath]Z^{-1}(U)\eq\bigcup_{V\in U}Z^{-1}(V) [/ilmath] - which is a finite union. As [ilmath]S[/ilmath] is a finite set in this example, we see that [ilmath]\sigma(S)\eq\mathcal{P}(S)[/ilmath], which gives us [ilmath]\forall s\in S[\{s\}\in \sigma(S)][/ilmath].
• Thus as [ilmath]Z[/ilmath] is a function, we see it must be measurable, both the codomains of [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] are finite, thus they're both measurable
2. Meaning "[ilmath]A[/ilmath] and [ilmath]C[/ilmath]"
3. As [ilmath]X[/ilmath] is a function [ilmath]A[/ilmath] and [ilmath]B[/ilmath] as events in [ilmath]\Omega[/ilmath] must be disjoint!