Notes:Mdm of a discrete distribution lemma

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$\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } }$$\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } }$$\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } }$$\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} }$$\newcommand{\Floor}[1]{ \text{Floor}{\left({#1}\right)} }$

UPDATES

Caveats

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

New statement

We have the following:

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

The task is to evaluate the following summation:

• $$S:\eq\sum_{k\eq\alpha}^\beta\vert k-\lambda\vert f(k)$$

We claim:

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

Proof

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

A complete proof will look something like this:

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

Original notes

Page notes

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

• This definition might be too general.

Statement

Below we show

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

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

(ORIGINAL STUFF)

Let's just go ahead and state it:

Original statement

Let:

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

Proof

Let:

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

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

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

To conclude:

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