Notes:Mdm of a discrete distribution lemma

[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{\Floor}[1]{ \text{Floor}{\left({#1}\right)} } [/ilmath]

UPDATES

Caveats

This all requires that [ilmath]\lambda\ge 0[/ilmath] for a start, else floor isn't defined, and [ilmath]\alpha,\beta\ge 0[/ilmath] too. I am sure the result is right but PROOFS ARE NEEDED

New statement

We have the following:

• Let [ilmath]\alpha,\beta\in\mathbb{N}_0[/ilmath] be given with [ilmath]\alpha\le\beta[/ilmath] (Template:WLOG - swap as required)
• Let [ilmath]f:\{\alpha,\alpha+1,\ldots,\beta-1,\beta\}\rightarrow\mathbb{R} [/ilmath] be given
• Let [ilmath]\lambda\in\mathbb{R} [/ilmath] be given

The task is to evaluate the following summation:

• [math]S:\eq\sum_{k\eq\alpha}^\beta\vert k-\lambda\vert f(k) [/math]

We claim:

• [math]S\eq\sum_{k\eq\alpha}^{\text{Min}(\text{Floor}(\lambda),\beta)}(\lambda-k)f(k)+\sum_{k\eq\text{Min}(\text{Floor}(\lambda),\beta)+1}^\beta(k-\lambda)f(k)[/math]

Proof

We have (sort of) proved the case where [ilmath]\beta\ge\text{Min}(\text{Floor}(\lambda),\beta)+1[/ilmath] already. As in this case the second summation has at least one term. As is convention with summations the [ilmath]\sum^b_{k\eq a}c_k[/ilmath] for [ilmath]b<a[/ilmath] is [ilmath]0[/ilmath].

A complete proof will look something like this:

• Define [ilmath]\gamma:\eq\text{Min}(\text{Floor}(\lambda),\beta)[/ilmath]
  • So we claim that [math]S\eq\sum^\gamma_{k\eq\alpha}(\lambda-k)f(k)+\sum^\beta_{k\eq\gamma+1}(k-\lambda)f(k) [/math]

Original notes

Page notes

Let [ilmath]X[/ilmath] be a distribution defined on [ilmath]\mathbb{Z} [/ilmath] or a subset, which we will denote [ilmath]A[/ilmath], that is [ilmath]\P{X\eq k} [/ilmath] is defined for all [ilmath]k\in A[/ilmath] and there is no set of which [ilmath]A[/ilmath] is a strict subset with [ilmath]\P{X\eq k} [/ilmath] defined for all k within it.

• This definition might be too general.

Statement

Below we show

• for [ilmath]n\ge\Floor{\lambda}+1[/ilmath] then:
  • [math]\sum^n_{k\eq 0}\vert k-\lambda\vert f(k)\eq\sum^{\Floor{\lambda} }_{k\eq 0}(\lambda-k)f(k)+\sum^n_{k\eq\Floor{\lambda}+1}(k-\lambda)f(k)[/math]
• if [ilmath]n<\Floor{\lambda}+1[/ilmath] then we need to put like [ilmath]\text{Min}(\Floor{\lambda},n)[/ilmath] as the end value for the first sum, the modifications are trivial.

We need to modify this slightly for a sum between [ilmath]k\eq\alpha[/ilmath] to [ilmath]k\eq\beta[/ilmath] say - but this is certainly a good result for now.

(ORIGINAL STUFF)

Let's just go ahead and state it:

Original statement

Let:

• [math]S:\eq\sum_{k\eq 0}^\infty \vert k-\lambda\vert f(k)[/math]
  • We claim:
    • [math]S\eq \sum_{k\eq 0}^{\Floor{\lambda} }(\lambda-k)f(k)+\sum^\infty_{k\eq\Floor{\lambda}+1}(k-\lambda)f(k) [/math]

Proof

Let:

• [math]S:\eq\sum_{k\eq 0}^\infty \vert k-\lambda\vert f(k)[/math]
  • To evaluate this we must break the range [ilmath]\{0,1,2,\ldots,n-1,n,n+1,\ldots\} [/ilmath] into a region for which
    1. [ilmath]\vert k-\lambda\vert\eq k-\lambda[/ilmath] - meaning that [ilmath]k-\lambda\ge 0\ \implies k\ge \lambda[/ilmath] and
    2. [ilmath]\vert k-\lambda\vert\eq \lambda- k[/ilmath] meaning that [ilmath]k-\lambda\le 0\ \implies k\le \lambda[/ilmath]
  • When calculating such Mdms we often use the floor function to break the domain, breaking it around [ilmath]\Floor{\lambda} [/ilmath]

(Not sure where to take this)

Paper-style proof

Here's what I did on paper:

1. I used the "floor corollary" which states:
   • (It may have multiple forms, they all basically say the same thing)
     1. PRIMARY FORM: [math]\forall x\in\mathbb{R}_{\ge 0}\big[\Floor{x}\le x<\Floor{x}+1\big][/math]
     2. ... the other forms will be slightly more detailed, for example if [ilmath]x\in\mathbb{R}_{\ge 0} -\mathbb{N}_{\ge 0} [/ilmath] then [ilmath]\Floor{x}<x<\Floor{x}+1[/ilmath]
2. We look at the [ilmath]k\eq 0[/ilmath] forward case, and we see that if we sum over [ilmath]k[/ilmath] for [ilmath]k[/ilmath] from [ilmath]0[/ilmath] to [ilmath]\Floor{\lambda} [/ilmath] that:
   • [ilmath]0\le k\le\Floor{\lambda}\le\lambda<\Floor{\lambda}+1[/ilmath]
     • Specifically [ilmath]0\le k\le\Floor{\lambda} [/ilmath]
   • So for [ilmath]0\le k\le \Floor{\lambda}\le \lambda[/ilmath] we see that [ilmath]k\le \lambda\ \implies\ k-\lambda\le 0[/ilmath], so:
     • for [ilmath]0\le k\le\Floor{\lambda}\le\lambda[/ilmath] we have [ilmath]\lambda-k\ge 0[/ilmath] and thus [ilmath]\vert k-\lambda\vert\eq\vert\lambda-k\vert\eq \lambda-k[/ilmath] as [ilmath]\lambda-k\ge 0[/ilmath] in this range
       • Specifically: [ilmath]0\le k\le\Floor{\lambda} [/ilmath] means [ilmath]\vert k-\lambda\vert\eq\lambda-k[/ilmath]
   • So [math]\sum^{\Floor{\lambda} }_{k\eq 0}\vert k-\lambda\vert f(k)\eq\sum^{\Floor{\lambda} }_{k\eq 0}(\lambda-k)f(k) [/math] - the first part of our sum
3. Now suppose we are summing to [ilmath]n[/ilmath] for [ilmath]n>\Floor{\lambda} [/ilmath] then:
   • As [ilmath]\lambda<\Floor{\lambda}+1\le n[/ilmath] we see for the range [ilmath]k\in\{\Floor{\lambda}+1,\ \Floor{\lambda}+2,\ \ldots,\ n\} [/ilmath] we have [ilmath]\lambda<\Floor{\lambda}+1\le k\le n[/ilmath] - specifically [ilmath]\lambda<k[/ilmath]
     • So [ilmath]0<k-\lambda[/ilmath] which we shall write: [ilmath]k-\lambda>0[/ilmath]
   • Now if [ilmath]k-\lambda>0[/ilmath] then [ilmath]\vert k-\lambda\eq k-\lambda[/ilmath] when the [ilmath]\lambda<k[/ilmath] condition holds
     • Thus for [ilmath]\Floor{\lambda}+1\le k\le n[/ilmath] we have [ilmath]\vert k-\lambda\vert\eq k-\lambda[/ilmath]
   • So [math]\sum^n_{k\eq\Floor{\lambda}+1}\vert k-\lambda\vert f(k)\eq \sum^n_{k\eq\Floor{\lambda}+1}(k-\lambda)f(k)[/math]
4. Lastly we note that [ilmath]\{0,\ldots,\Floor{\lambda} \}\cup\{\Floor{\lambda}+1,\ldots,n\} [/ilmath] is the domain we intended to sum over

To get the stated result note that [ilmath]\sum^\infty_{k\eq 0}a_k:\eq \lim_{n\rightarrow\infty}(\sum_{k\eq 0}^n a_k) [/ilmath] (by definition of limit of a series

To conclude:

• for [ilmath]n\ge\Floor{\lambda}+1[/ilmath] then:
  • [math]\sum^n_{k\eq 0}\vert k-\lambda\vert f(k)\eq\sum^{\Floor{\lambda} }_{k\eq 0}(\lambda-k)f(k)+\sum^n_{k\eq\Floor{\lambda}+1}(k-\lambda)f(k)[/math]
• if [ilmath]n<\Floor{\lambda}+1[/ilmath] then we need to put like [ilmath]\text{Min}(\Floor{\lambda},n)[/ilmath] as the end value for the first sum, the modifications are trivial.