Mdm of a discrete distribution lemma

From Maths
Jump to: navigation, search

I may have found a useful transformation for calculating Mdm's of distributions defined on Z or a subset. I document my work so far below:

  • Notes:Mdm of a discrete distribution lemma
  • Notes:Mdm of a discrete distribution lemma - round 2
    \newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }
    \newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } } \newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } } \newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } } \newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} }

Statement

Notice: - there are plans to generalise this lemma:- specifically to allow \lambda to take any real value (currently only non-negative allowed) and possibly also allow \alpha,\beta to be negative too

Let \lambda\in\mathbb{R}_{\ge 0} and let \alpha,\beta\in\mathbb{N}_0 such that \alpha\le\beta, let f:\{\alpha,\alpha+1,\ldots,\beta-1,\beta\}\subseteq\mathbb{N}_0\rightarrow\mathbb{R} be a function, then we claim:

  • \sum^\beta_{k\eq\alpha}\big\vert k-\lambda\big\vert f(k) \eq\sum^\gamma_{k\eq\alpha}(\lambda-k)f(k) +\sum_{k\eq\gamma+1}^\beta (k-\lambda)f(k) where:
    • \gamma:\eq\text{Min}(\text{Floor}(\lambda),\beta)

Note that \beta\eq\infty is valid for this expression (standard limits stuff, see sum to infinity)

Applications to computing Mdm

Let X be a discrete random variable defined on \{\alpha,\alpha+1,\ldots,\beta-1,\beta\}\subseteq\mathbb{N}_0 (remember that \beta\eq\infty is valid and just turns the second sum into a limit), then:

  • define \lambda:\eq\E{X}
  • define f:k\mapsto \P{X\eq k}

Recall the mdm of x is defined to be:

  • \Mdm{X}:\eq \sum^\beta_{k\eq\alpha}\big\vert k-\E{X}\big\vert\ \P{X\eq k}

It is easy to see that with the definitions substituted that:

  • \sum^\beta_{k\eq\alpha}\big\vert k-\lambda\big\vert f(k)\eq\Mdm{X}

Proof

Grade: A**
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
Note follow
  • Initial notes Alec (talk) 01:24, 22 January 2018 (UTC)
    1. A lot of work has been done in Notes:Mdm of a discrete distribution lemma and I've done each of the 4 cases individually (\alpha\eq\beta, \beta<\text{Floor}(\lambda), \beta>\text{Floor}(\lambda) and \beta\eq\text{Floor}(\lambda) - but they need to be put together.
    2. There is a 5th case where \lambda<0 is introduced
    3. I'd like to generalise this to \alpha,\beta\in\mathbb{Z} - generalising beyond \alpha,\beta being non-negative

Notes

References