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

Latest revision as of 17:53, 24 January 2018

I may have found a useful transformation for calculating Mdm's of distributions defined on $\mathbb{Z}$ or a subset. I document my work so far below:

• Notes:Mdm of a discrete distribution lemma
• Notes:Mdm of a discrete distribution lemma - round 2
  $\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }$
  $\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} }$

Statement

Notice: - there are plans to generalise this lemma:- specifically to allow $\lambda$ to take any real value (currently only non-negative allowed) and possibly also allow $\alpha,\beta$ to be negative too

Let $\lambda\in\mathbb{R}_{\ge 0}$ and let $\alpha,\beta\in\mathbb{N}_0$ such that $\alpha\le\beta$, let $f:\{\alpha,\alpha+1,\ldots,\beta-1,\beta\}\subseteq\mathbb{N}_0\rightarrow\mathbb{R}$ be a function, then we claim:

• $$\sum^\beta_{k\eq\alpha}\big\vert k-\lambda\big\vert f(k) \eq\sum^\gamma_{k\eq\alpha}(\lambda-k)f(k) +\sum_{k\eq\gamma+1}^\beta (k-\lambda)f(k)$$ where:
  • $\gamma:\eq\text{Min}(\text{Floor}(\lambda),\beta)$

Note that $\beta\eq\infty$ is valid for this expression (standard limits stuff, see sum to infinity)

Applications to computing Mdm

Let $X$ be a discrete random variable defined on $\{\alpha,\alpha+1,\ldots,\beta-1,\beta\}\subseteq\mathbb{N}_0$ (remember that $\beta\eq\infty$ is valid and just turns the second sum into a limit), then:

• define $\lambda:\eq\E{X}$
• define $f:k\mapsto \P{X\eq k}$

Recall the mdm of $x$ is defined to be:

• $$\Mdm{X}:\eq \sum^\beta_{k\eq\alpha}\big\vert k-\E{X}\big\vert\ \P{X\eq k}$$

It is easy to see that with the definitions substituted that:

• $$\sum^\beta_{k\eq\alpha}\big\vert k-\lambda\big\vert f(k)\eq\Mdm{X}$$

Proof

Notes

References