Difference between revisions of "Mdm of the Binomial distribution"

Revision as of 12:11, 8 January 2018 (view source) Alec (Talk | contribs) (Saving work for restart)		Latest revision as of 01:32, 12 January 2018 (view source) Alec (Talk | contribs) (Saving work)
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	==Proof==		==Proof==
−	* {{M|\Mdm(X)}}	+	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"/>

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Latest revision as of 01:32, 12 January 2018

$\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

Let $n\in\mathbb{N}_{\ge 1}$ and $p\in[0,1]\subseteq$$\mathbb{R}$ and:

• $X\sim$$\text{Bin}$$(n,p)$ - so $X$ is a Binomial random variable

We will calculate the Mdm of $X$, $\E{\big\vert X-\E{X}\big\vert}$

Proof

We begin:

• $$\Mdm{X}:\eq\sum^n_{k\eq 0}{\big\vert k-\E{X}\big\vert\cdot\P{X\eq k} }$$

  $$\eq\sum^n_{k\eq 0}\big\vert k-np\big\vert\cdot\ncr{n}{k}\cdot p^k\cdot (1-p)^{n-k}$$ - by substitution of the expectation of the binomial distribution, as well as the expression for probability of $X\eq k$

We now operate on the $\vert\cdot\vert$ part, there are 2 cases (and we will introduce the $\lfloor\cdot\rfloor$ or "floor function")

• Note that as $np\ge 0$ we will have $\lfloor np\rfloor\le np$^{[Note 1]}

Case analysis:

1. Suppose $k\ge np$
   • now $k-np\ge 0$ so by the definition of absolute value (specifically that if $x\ge 0$ then $\vert x\vert\eq x$) we see:
     • if $k\ge np$ then $k-np\ge 0$ and if $k-np\ge 0$ then $\vert k-np\vert\eq k-np$
       • Conclusion: for this case we have $\vert k-np\vert\eq k-np$
   • Note additionally that $\lfloor np\rfloor\le np$ so we have: $k\ge np\ge \lfloor np\rfloor$ $\implies$ $k\ge \lfloor np\rfloor$

Notes

1. Jump up ↑ Defining floor for negative values can be "controversial"