Notes:Mdm of a discrete distribution lemma - round 2

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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 \{\alpha,\ldots,\beta\} so that:

  1. on one part k\le\Floor{\l} holds, and on the other part k>\Floor{\l} holds - OR -
  2. on one part k<\Floor{\l} holds, and on the other part k\ge\Floor{\l} 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 k\le \l:
    • It is easy to see that this implies the following two statements:
      1. \vert k-\lambda\vert\eq\lambda-k
      2. k\le\Floor{\l} [Note 1]
    • But also:
      • k\le\Floor{\l} \implies k\le \lambda[Note 2]
  2. Given k\ge \l
    • It is easy to see that this implies the following two statements:
      1. \vert k-\lambda\vert\eq k-\lambda
      2. \Floor{\l}\le k[Note 3] - which we shall write k\ge\Floor{\l}
    • But also:
      • k\ge\Floor{\l} \implies k\ge \l[Note 4]

Notes

  1. Jump up ↑ As:
    • k\le \l
      \implies k\le \l<\Floor{\l}+1 - by corollary on the page Floor function
      \implies k<\Floor{\l}+1
      \implies k\le\Floor{\l} by the nature of strict orderings on the natural numbers,
      TODO: Link somewhere!
  2. Jump up ↑ Simply because \Floor{\l}\le \lambda - as described on the floor function page
  3. Jump up ↑ As: \Floor{\lambda}\le\lambda\le k
  4. Jump up ↑ Suppose \Floor{\l}\le k
    • From the floor function page we know that \Floor{\l}\le \l <\Floor{\l}+1
      • So using only the 2nd inequality there: \l-1<\Floor{\l}
    • Thus \l-1<\Floor{\l}\le k
    • or just: \l-1<k
    Which is the same as:
    • \l\le k - by the strict orderings related to orderings on the natural numbers
      TODO: Also link this somewhere!
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