Mdm of a discrete distribution lemma

I may have found a useful transformation for calculating Mdm's of distributions defined on [ilmath]\mathbb{Z} [/ilmath] 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
  [ilmath]\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} }[/ilmath]
  [ilmath]\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } } [/ilmath][ilmath]\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} } [/ilmath]

Statement

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

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

• [math]\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) [/math] where:
  • [ilmath]\gamma:\eq\text{Min}(\text{Floor}(\lambda),\beta)[/ilmath]

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

Applications to computing Mdm

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

• define [ilmath]\lambda:\eq\E{X} [/ilmath]
• define [ilmath]f:k\mapsto \P{X\eq k} [/ilmath]

Recall the mdm of [ilmath]x[/ilmath] is defined to be:

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

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

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

Proof

Grade: A**

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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 ([ilmath]\alpha\eq\beta[/ilmath], [ilmath]\beta<\text{Floor}(\lambda)[/ilmath], [ilmath]\beta>\text{Floor}(\lambda)[/ilmath] and [ilmath]\beta\eq\text{Floor}(\lambda)[/ilmath] - but they need to be put together.
  2. There is a 5th case where [ilmath]\lambda<0[/ilmath] is introduced
  3. I'd like to generalise this to [ilmath]\alpha,\beta\in\mathbb{Z} [/ilmath] - generalising beyond [ilmath]\alpha,\beta[/ilmath] being non-negative