# 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][ilmath]\newcommand{\Min}[1]{\text{Min}{\left({#1}\right)} } [/ilmath][ilmath]\newcommand{\Floor}[1]{\text{Floor}{\left({#1}\right)} } [/ilmath][ilmath]\newcommand{\l}[0]{\lambda} [/ilmath]

## Document

I claim:

• $\forall\lambda\in\mathbb{R}_{\ge 0}\forall\alpha,\beta\in\mathbb{N}_0\forall f\in\mathcal{F}\big(\{\alpha,\alpha+1,\ldots,\beta-1,\beta\}\subseteq\mathbb{N}_0,\mathbb{R}_{\ge 0}\big)\left[\big(\alpha\le\beta\big)\implies\left(\sum^\beta_{k\eq\alpha}\big\vert k-\lambda\big\vert f(k)\eq \sum_{k\eq\alpha}^{\Min{\Floor{\lambda,\beta} } }\big(\lambda-k\big)f(k)+\sum^\beta_{k\eq\Min{\Floor{\l},\beta}+1}\big(k-\lambda\big)f(k)\right)\right]$

## Lemmas

1. $\forall k\in\mathbb{N}_0\forall\lambda\in\mathbb{R}_{\ge 0}\big[\big(k\le\Floor{\l}\big)\implies\big(\vert k-\l\vert\eq \l-k\big)\big]$ - Proved on paper
2. $\forall k\in\mathbb{N}_0\forall\lambda\in\mathbb{R}_{\ge 0}\big[\big(k\ge\Floor{\l}\big)\implies\big(\vert k-\l\vert\eq k-\l\big)\big]$ - Proved on paper

## Process

We partition [ilmath]\{\alpha,\ldots,\beta\} [/ilmath] so that:

1. on one part [ilmath]k\le\Floor{\l} [/ilmath] holds, and on the other part [ilmath]k>\Floor{\l} [/ilmath] holds - OR -
2. on one part [ilmath]k<\Floor{\l} [/ilmath] holds, and on the other part [ilmath]k\ge\Floor{\l} [/ilmath] holds

We then sum over these parts, paying special attention to what happens if either part of the partition is empty.

## Key workings

There are 2 key statements:

1. Given [ilmath]k\le \l[/ilmath]:
• It is easy to see that this implies the following two statements:
1. [ilmath]\vert k-\lambda\vert\eq\lambda-k[/ilmath]
2. [ilmath]k\le\Floor{\l} [/ilmath][Note 1]
• But also:
• [ilmath]k\le\Floor{\l} \implies k\le \lambda[/ilmath][Note 2]
2. Given [ilmath]k\ge \l[/ilmath]
• It is easy to see that this implies the following two statements:
1. [ilmath]\vert k-\lambda\vert\eq k-\lambda[/ilmath]
2. [ilmath]\Floor{\l}\le k[/ilmath][Note 3] - which we shall write [ilmath]k\ge\Floor{\l} [/ilmath]
• But also:
• [ilmath]k\ge\Floor{\l} \implies k\ge \l[/ilmath][Note 4]

## Notes

1. As:
• [ilmath]k\le \l[/ilmath]
[ilmath]\implies k\le \l<\Floor{\l}+1[/ilmath] - by corollary on the page Floor function
[ilmath]\implies k<\Floor{\l}+1[/ilmath]
[ilmath]\implies k\le\Floor{\l} [/ilmath] by the nature of strict orderings on the natural numbers,