Difference between revisions of "Poisson distribution/RV"

From Maths
Jump to: navigation, search
(Saving work)
 
(PROBLEM)
Line 17: Line 17:
 
*** Specifically consider {{M|\big(\mathbb{N}_0,\ }}[[power set|{{M|\mathcal{P}(\mathbb{N}_0)}}]]{{M|\big)}} as a [[sigma-algebra]] and {{M|X:[0,1]\rightarrow\mathbb{N}_0}} by:
 
*** Specifically consider {{M|\big(\mathbb{N}_0,\ }}[[power set|{{M|\mathcal{P}(\mathbb{N}_0)}}]]{{M|\big)}} as a [[sigma-algebra]] and {{M|X:[0,1]\rightarrow\mathbb{N}_0}} by:
 
**** {{M|X:x\mapsto\left\{\begin{array}{lr}0&\text{if }x\in[0,p_1)\\1 & \text{if }x\in[p_1,p_2)\\ \vdots & \vdots \\ k & \text{if }x\in[p_k,p_{k+1})\\ \vdots & \vdots \end{array}\right.}} for {{MM|p_1:\eq e^{-\lambda} \frac{\lambda^1}{1!} }} and {{M|p_k:\eq p_{k-1}+e^{-\lambda}\frac{\lambda^k}{k!} }}
 
**** {{M|X:x\mapsto\left\{\begin{array}{lr}0&\text{if }x\in[0,p_1)\\1 & \text{if }x\in[p_1,p_2)\\ \vdots & \vdots \\ k & \text{if }x\in[p_k,p_{k+1})\\ \vdots & \vdots \end{array}\right.}} for {{MM|p_1:\eq e^{-\lambda} \frac{\lambda^1}{1!} }} and {{M|p_k:\eq p_{k-1}+e^{-\lambda}\frac{\lambda^k}{k!} }}
Giving the setup shown on the left.
+
Giving the setup shown on the left.<noinclude>
 +
=={{XXX|TODO:}}==
 +
* Surely it should be {{M|[0,1)}} and {{M|\mathcal{B}\big([0,1)\big)}} for this to work? [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 20:49, 26 February 2018 (UTC)
 +
</noinclude>

Revision as of 20:49, 26 February 2018

Definition

As a formal random variable

[ilmath]\xymatrix{ & [0,1] \ar[r]^X & \mathbb{N}_0 \\ & \mathcal{B}([0,1]) \ar[dl]_-{\lambda} & \mathcal{P}(\mathbb{N}_0) \ar[l]_-{X^{-1} } \ar@{-->}@/^1em/[dll]^-{\mathbb{P}:\eq \lambda\circ X^{-1} } \\ \mathbb{R} & & } [/ilmath]
Situation for our RV

There is no unique way to define a random variable, here is one way.

  • Let [ilmath]\big([/ilmath][ilmath][0,1][/ilmath][ilmath],\ [/ilmath][ilmath]\mathcal{B}([0,1])[/ilmath][ilmath],\ [/ilmath][ilmath]\lambda[/ilmath][ilmath]\big)[/ilmath] be a probability space - which itself could be viewed as a rectangular distribution's random variable
    • Let [ilmath]\lambda\in\mathbb{R}_{>0} [/ilmath] be given, and let [ilmath]X\sim\text{Poi}(\lambda)[/ilmath]
      • Specifically consider [ilmath]\big(\mathbb{N}_0,\ [/ilmath][ilmath]\mathcal{P}(\mathbb{N}_0)[/ilmath][ilmath]\big)[/ilmath] as a sigma-algebra and [ilmath]X:[0,1]\rightarrow\mathbb{N}_0[/ilmath] by:
        • [ilmath]X:x\mapsto\left\{\begin{array}{lr}0&\text{if }x\in[0,p_1)\\1 & \text{if }x\in[p_1,p_2)\\ \vdots & \vdots \\ k & \text{if }x\in[p_k,p_{k+1})\\ \vdots & \vdots \end{array}\right.[/ilmath] for [math]p_1:\eq e^{-\lambda} \frac{\lambda^1}{1!} [/math] and [ilmath]p_k:\eq p_{k-1}+e^{-\lambda}\frac{\lambda^k}{k!} [/ilmath]

Giving the setup shown on the left.

TODO: TODO:

  • Surely it should be [ilmath][0,1)[/ilmath] and [ilmath]\mathcal{B}\big([0,1)\big)[/ilmath] for this to work? Alec (talk) 20:49, 26 February 2018 (UTC)