View source for Notes:Mdm of a discrete distribution lemma

__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==
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.
* This definition might be too general. 

===Statement===
Below we show
* for {{M|n\ge\Floor{\lambda}+1}} then:
** {{MM|\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)}}
* if {{M|n<\Floor{\lambda}+1}} then we need to put like {{M|\text{Min}(\Floor{\lambda},n)}} as the end value for the first sum, the modifications are trivial.

We need to modify this slightly for a sum between {{M|k\eq\alpha}} to {{M|k\eq\beta}} say - but this is certainly a good result for now.

: ''(ORIGINAL STUFF)''
Let's just go ahead and state it:
===Original statement===
Let:
* {{MM|S:\eq\sum_{k\eq 0}^\infty \vert k-\lambda\vert f(k)}}<ref group="Note">For {{M|f(k):\eq\P{X\eq k} }} and {{M|\lambda:\eq\E{X} }} we get:
* {{MM|\sum^\infty_{k\eq 0}\vert k-\E{X}\vert\P{X\eq k} }} 
Which if course ''is'' the [[Mdm]] of {{M|X}} - provided {{M|X}} is defined on {{M|\mathbb{N}_0}} and no set of which {{M|\mathbb{N}_0}} is a [[proper subset]]</ref>
** We claim:
*** {{MM|S\eq \sum_{k\eq 0}^{\Floor{\lambda} }(\lambda-k)f(k)+\sum^\infty_{k\eq\Floor{\lambda}+1}(k-\lambda)f(k) }}
==Proof==
Let:
* {{MM|S:\eq\sum_{k\eq 0}^\infty \vert k-\lambda\vert f(k)}}
** To evaluate this we must break the range {{M|\{0,1,2,\ldots,n-1,n,n+1,\ldots\} }} into a region for which
**# {{M|\vert k-\lambda\vert\eq k-\lambda}} - meaning that {{M|k-\lambda\ge 0\ \implies k\ge \lambda}} and
**# {{M|\vert k-\lambda\vert\eq \lambda- k}} meaning that {{M|k-\lambda\le 0\ \implies k\le \lambda}} 
** When calculating such [[Mdm|Mdms]] we often use the [[floor function]] to break the domain, breaking it around {{M|\Floor{\lambda} }}
(Not sure where to take this)

===Paper-style proof===
Here's what I did on paper:
# I used the "floor corollary"<ref group="Note">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) </ref> which states:
#* (It may have multiple forms, they all basically say the same thing)
#*# '''PRIMARY FORM: ''' {{MM|\forall x\in\mathbb{R}_{\ge 0}\big[\Floor{x}\le x<\Floor{x}+1\big]}}
#*# ... the other forms will be slightly more detailed, for example if {{M|x\in\mathbb{R}_{\ge 0} -\mathbb{N}_{\ge 0} }} then {{M|\Floor{x}<x<\Floor{x}+1}}
#  We look at the {{M|k\eq 0}} forward case, and we see that if we sum over {{M|k}} for {{M|k}} from {{M|0}} to {{M|\Floor{\lambda} }} that:
#* {{M|0\le k\le\Floor{\lambda}\le\lambda<\Floor{\lambda}+1}}
#** Specifically {{M|0\le k\le\Floor{\lambda} }}
#* So for {{M|0\le k\le \Floor{\lambda}\le \lambda}} we see that {{M|k\le \lambda\ \implies\ k-\lambda\le 0}}, so:
#** for {{M|0\le k\le\Floor{\lambda}\le\lambda}} we have {{M|\lambda-k\ge 0}} and thus {{M|\vert k-\lambda\vert\eq\vert\lambda-k\vert\eq \lambda-k}} as {{M|\lambda-k\ge 0}} in this range
#*** Specifically: {{M|0\le k\le\Floor{\lambda} }} means {{M|\vert k-\lambda\vert\eq\lambda-k}}
#* So {{MM|\sum^{\Floor{\lambda} }_{k\eq 0}\vert k-\lambda\vert f(k)\eq\sum^{\Floor{\lambda} }_{k\eq 0}(\lambda-k)f(k) }} - the first part of our sum
# Now suppose we are summing to {{M|n}} for {{M|n>\Floor{\lambda} }} then:
#* As {{M|\lambda<\Floor{\lambda}+1\le n}} we see for the range {{M|k\in\{\Floor{\lambda}+1,\ \Floor{\lambda}+2,\ \ldots,\ n\} }} we have {{M|\lambda<\Floor{\lambda}+1\le k\le n}} - specifically {{M|\lambda<k}}
#** So {{M|0<k-\lambda}} which we shall write: {{M|k-\lambda>0}} 
#* Now if {{M|k-\lambda>0}} then {{M|\vert k-\lambda\eq k-\lambda}} when the {{M|\lambda<k}} condition holds
#** Thus for {{M|\Floor{\lambda}+1\le k\le n}} we have {{M|\vert k-\lambda\vert\eq k-\lambda}}
#* So {{MM|\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)}}
# Lastly we note that {{M|\{0,\ldots,\Floor{\lambda} \}\cup\{\Floor{\lambda}+1,\ldots,n\} }} is the domain we intended to sum over
To get the stated result note that {{M|\sum^\infty_{k\eq 0}a_k:\eq \lim_{n\rightarrow\infty}(\sum_{k\eq 0}^n a_k) }} (by definition of [[limit of a series]]


To conclude:
* for {{M|n\ge\Floor{\lambda}+1}} then:
** {{MM|\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)}}
* if {{M|n<\Floor{\lambda}+1}} then we need to put like {{M|\text{Min}(\Floor{\lambda},n)}} as the end value for the first sum, the modifications are trivial.
==Notes==
<references group="Note"/>
{{Notes|Probability|Statistics|Elementary Probability}}