Difference between revisions of "Mdm of the Binomial distribution"
From Maths
(Saving work for restart) |
(Saving work) |
||
| Line 6: | Line 6: | ||
==Proof== | ==Proof== | ||
| − | * {{ | + | We begin: |
| + | * {{MM|\Mdm{X}:\eq\sum^n_{k\eq 0}{\big\vert k-\E{X}\big\vert\cdot\P{X\eq k} } }} | ||
| + | *: {{MM|\eq\sum^n_{k\eq 0}\big\vert k-np\big\vert\cdot\ncr{n}{k}\cdot p^k\cdot (1-p)^{n-k} }} - by substitution of the [[expectation]] of the [[binomial distribution]], as well as the expression for [[probability]] of {{M|X\eq k}} | ||
| + | |||
| + | We now operate on the {{link|abs|function}} part, there are 2 cases (and we will introduce the [[floor (function)|{{M|\lfloor\cdot\rfloor}}]] or "''[[floor function]]''") | ||
| + | * Note that as {{M|np\ge 0}} we will have {{M|\lfloor np\rfloor\le np}}<ref group="Note">Defining {{link|floor|function}} for negative values can be "controversial"</ref> | ||
| + | Case analysis: | ||
| + | # Suppose {{M|k\ge np}} | ||
| + | #* now {{M|k-np\ge 0}} so by the definition of [[absolute value]] (specifically that if {{M|x\ge 0}} then {{M|\vert x\vert\eq x}}) we see: | ||
| + | #** if {{M|k\ge np}} then {{M|k-np\ge 0}} and if {{M|k-np\ge 0}} then {{M|\vert k-np\vert\eq k-np}} | ||
| + | #*** Conclusion: for this case we have {{M|\vert k-np\vert\eq k-np}} | ||
| + | #* Note additionally that {{M|\lfloor np\rfloor\le np}} so we have: {{M|k\ge np\ge \lfloor np\rfloor}} {{link|implies}} {{M|k\ge \lfloor np\rfloor}} | ||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
Latest revision as of 01:32, 12 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]
Statement
Let [ilmath]n\in\mathbb{N}_{\ge 1} [/ilmath] and [ilmath]p\in[0,1]\subseteq[/ilmath][ilmath]\mathbb{R} [/ilmath] and:
- [ilmath]X\sim[/ilmath][ilmath]\text{Bin} [/ilmath][ilmath](n,p)[/ilmath] - so [ilmath]X[/ilmath] is a Binomial random variable
We will calculate the Mdm of [ilmath]X[/ilmath], [ilmath]\E{\big\vert X-\E{X}\big\vert} [/ilmath]
Proof
We begin:
- [math]\Mdm{X}:\eq\sum^n_{k\eq 0}{\big\vert k-\E{X}\big\vert\cdot\P{X\eq k} } [/math]
- [math]\eq\sum^n_{k\eq 0}\big\vert k-np\big\vert\cdot\ncr{n}{k}\cdot p^k\cdot (1-p)^{n-k} [/math] - by substitution of the expectation of the binomial distribution, as well as the expression for probability of [ilmath]X\eq k[/ilmath]
We now operate on the [ilmath]\vert\cdot\vert[/ilmath] part, there are 2 cases (and we will introduce the [ilmath]\lfloor\cdot\rfloor[/ilmath] or "floor function")
- Note that as [ilmath]np\ge 0[/ilmath] we will have [ilmath]\lfloor np\rfloor\le np[/ilmath][Note 1]
Case analysis:
- Suppose [ilmath]k\ge np[/ilmath]
- now [ilmath]k-np\ge 0[/ilmath] so by the definition of absolute value (specifically that if [ilmath]x\ge 0[/ilmath] then [ilmath]\vert x\vert\eq x[/ilmath]) we see:
- if [ilmath]k\ge np[/ilmath] then [ilmath]k-np\ge 0[/ilmath] and if [ilmath]k-np\ge 0[/ilmath] then [ilmath]\vert k-np\vert\eq k-np[/ilmath]
- Conclusion: for this case we have [ilmath]\vert k-np\vert\eq k-np[/ilmath]
- if [ilmath]k\ge np[/ilmath] then [ilmath]k-np\ge 0[/ilmath] and if [ilmath]k-np\ge 0[/ilmath] then [ilmath]\vert k-np\vert\eq k-np[/ilmath]
- Note additionally that [ilmath]\lfloor np\rfloor\le np[/ilmath] so we have: [ilmath]k\ge np\ge \lfloor np\rfloor[/ilmath] [ilmath]\implies[/ilmath] [ilmath]k\ge \lfloor np\rfloor[/ilmath]
- now [ilmath]k-np\ge 0[/ilmath] so by the definition of absolute value (specifically that if [ilmath]x\ge 0[/ilmath] then [ilmath]\vert x\vert\eq x[/ilmath]) we see:
