Difference between revisions of "Notes:Mdm of a discrete distribution lemma"

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(SAVING WORK)
 
(Adding in new statement)
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__TOC__{{ProbMacros}}{{M|\newcommand{\Floor}[1]{ \text{Floor}{\left({#1}\right)} } }}
 
__TOC__{{ProbMacros}}{{M|\newcommand{\Floor}[1]{ \text{Floor}{\left({#1}\right)} } }}
 +
==New statement==
 +
We have the following:
 +
* Let {{M|\alpha,\beta\in\mathbb{N}_0}} be given with {{M|\alpha\le\beta}} ({{WLOG}} - swap as required)
 +
* Let {{M|f:\{\alpha,\alpha+1,\ldots,\beta-1,\beta\}\rightarrow\mathbb{R} }} be given
 +
* Let {{M|\lambda\in\mathbb{R} }} be given
 +
 +
The task is to evaluate the following [[summation]]:
 +
* {{MM|S:\eq\sum_{k\eq\alpha}^\beta\vert k-\lambda\vert f(k) }}
 +
 +
We claim:
 +
* {{MM|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)}}
 
==Page notes==
 
==Page notes==
 
Let {{M|X}} be a distribution defined on {{M|\mathbb{Z} }} or a subset, which we will denote {{M|A}}, that is {{M|\P{X\eq k} }} is defined for all {{M|k\in A}} and there is no set of which {{M|A}} is a [[strict subset]] with {{M|\P{X\eq k} }} defined for all k within it.
 
Let {{M|X}} be a distribution defined on {{M|\mathbb{Z} }} or a subset, which we will denote {{M|A}}, that is {{M|\P{X\eq k} }} is defined for all {{M|k\in A}} and there is no set of which {{M|A}} is a [[strict subset]] with {{M|\P{X\eq k} }} defined for all k within it.

Revision as of 00:25, 22 January 2018

[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]

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]

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][Note 1]
    • 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"[Note 2] 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.

Notes

  1. For [ilmath]f(k):\eq\P{X\eq k} [/ilmath] and [ilmath]\lambda:\eq\E{X} [/ilmath] we get:
    • [math]\sum^\infty_{k\eq 0}\vert k-\E{X}\vert\P{X\eq k} [/math]
    Which if course is the Mdm of [ilmath]X[/ilmath] - provided [ilmath]X[/ilmath] is defined on [ilmath]\mathbb{N}_0[/ilmath] and no set of which [ilmath]\mathbb{N}_0[/ilmath] is a proper subset
  2. which should be on the floor function page but isn't there at time of writing (sig doesn't work? So: 21st of Jan 2018 ~ 9pm)