Mdm of the Binomial distribution

$\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }$

$\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

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 $\lfloor np\rfloor\le np$

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: $k\ge np\ge \lfloor np\rfloor$ $\implies$ $k\ge \lfloor np\rfloor$