Mdm of the Poisson distribution

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TODO: Link with Poisson distribution page

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Statement

Let $X\sim$$\text{Poi}$$(\lambda)$ for some $\lambda\in\mathbb{R}_{>0}$. $X$ may take any value in $\mathbb{N}_0$

We will show that

• $$\text{Mdm}(X)\eq 2\lambda e^{-\lambda}\frac{\lambda^u}{u!}$$
  • Where $u:\eq$$\text{Floor}(\lambda)$

I have confirmed this experimentally in Experimental evidence for the Mdm of the Poisson distribution

Recall the Mdm is defined as:

• $$\text{Mdm}(X):\eq\mathbb{E}\Big[\ \big\vert X-\mathbb{E}[X]\big\vert\ \Big]$$

$\newcommand{\LHS}[0]{ {\text{Mdm}(X)} }$

Calculation

• $$\text{Mdm}(X):\eq\mathbb{E}\Big[\ \big\vert X-\mathbb{E}[X]\big\vert\ \Big]$$ $$:\eq\sum^\infty_{k\eq 0}\big\vert X-\mathbb{E}[X]\big\vert\cdot\P{X\eq k}$$

  $$\eq e^{-\lambda}\ \sum^\infty_{k\eq 0}\frac{\lambda^k}{k!}\big\vert k-\mathbb{E}[X]\big\vert$$ $$\eq e^{-\lambda}\ \sum^\infty_{k\eq 0}\frac{\lambda^k}{k!}\big\vert k-\lambda\big\vert$$

  • Note that:
    1. if $k\ge \lambda\ \implies\ k-\lambda\ge 0\ \implies\ \vert k-\lambda\vert \eq k-\lambda$
    2. if $k\le \lambda\ \implies\ k-\lambda\le 0\ \implies \lambda-k\ge 0\ \implies \vert k-\lambda\vert \eq \lambda-k$
  • Define the following two values:
    1. $u:\eq\text{RoundDownToInt}(\lambda)$ (also known as the floor function and
    2. $v:\eq u+1$
    • This means we have $u\le \lambda$ and $v\ge \lambda$, specifically, we have the following two cases:
      1. if $k\le u$ and as $u\le \lambda$ we see $k\le \lambda$ and
      2. if $k> u$ then $k \ge u+1\eq v \ge\lambda$ so $k\ge \lambda$
  • Now, from above: $$\LHS{}\eq e^{-\lambda}\ \sum^\infty_{k\eq 0}\frac{\lambda^k}{k!}\big\vert k-\lambda\big\vert$$

    $$\eq e^{-\lambda}{\left[\frac{\lambda^0}{0!}\big\vert 0-\lambda\vert \ +\ \sum^\infty_{k\eq 1}\frac{\lambda^k}{k!}\big\vert k-\lambda\big\vert\right]}$$

    $$\eq e^{-\lambda}{\left[\lambda\ +\ \sum^u_{k\eq 1}\frac{\lambda^k}{k!}\big\vert k-\lambda\big\vert\ +\ \sum^\infty_{k\eq v}\frac{\lambda^k}{k!}\big\vert k-\lambda\big\vert \right]}$$ with the understanding that if $u\eq 0$ that the sum from $k\eq 1$ to $u$ evaluates to $0$, obviously

  • Notice now that:
    1. For the first sum, where $1\le k\le u$ (specifically that $k\le u$) we have $k\le \lambda$
       • and that from further above we noticed if $k\le \lambda$ then $\big\vert k-\lambda\big\vert \eq \lambda-k$
    2. For the second sum, where $k > u$ that this meant $k\ge \lambda$
       • and that from further above we noticed if $k\ge\lambda$ then $\big\vert k-\lambda\big\vert\eq k-\lambda$, so
  • $$\LHS{}\eq e^{-\lambda}{\left[\lambda\ +\ \sum^u_{k\eq 1}\frac{\lambda^k}{k!}\big\vert k-\lambda\big\vert\ +\ \sum^\infty_{k\eq v}\frac{\lambda^k}{k!}\big\vert k-\lambda\big\vert \right]}$$

    $$\eq e^{-\lambda}{\left[\lambda\ +\ \sum^u_{k\eq 1}\frac{\lambda^k}{k!}(\lambda-k)\ +\ \sum^\infty_{k\eq v}\frac{\lambda^k}{k!}( k-\lambda)\right]}$$

    • we now expand these sums:

    $$\eq e^{-\lambda}{\Bigg[\lambda\ +\ \overbrace{\sum^u_{k\eq 1}\frac{\lambda^{k+1} }{k!}\ -\ \sum^u_{k\eq 1}\frac{k\lambda^k}{k!} }^\text{first sum}\ +\ \overbrace{\sum^\infty_{k\eq v}\frac{k\lambda^k}{k!}-\sum^\infty_{k\eq v}\frac{\lambda^{k+1} }{k!} }^\text{second sum}\ \Bigg]}$$

    $$\eq e^{-\lambda}{\left[\lambda\ +\ \lambda{\left(\sum^u_{k\eq 1}\frac{\lambda^k}{k!}\ -\ \sum^\infty_{k\eq v}\frac{\lambda^k}{k!}\right)}\ +\ {\left( \sum^\infty_{k\eq v}\frac{\lambda^k}{(k-1)!}\ -\ \sum^u_{k\eq 1}\frac{\lambda^k}{(k-1)!}\right)}\right]}$$ , by grouping the terms and factorising where we can

    $$\eq e^{-\lambda}{\left[\lambda\ +\ \lambda{\left(\sum^u_{k\eq 1}\frac{\lambda^k}{k!}\ -\ \sum^\infty_{k\eq v}\frac{\lambda^k}{k!}\right)}\ +\ {\left( \sum^\infty_{k\eq v-1}\frac{\lambda^{k+1} }{k!}\ -\ \sum^{u-1}_{k\eq 0}\frac{\lambda^{k+1} }{k!}\right)}\right]}$$ by reindexing the latter two sums

    $$\eq e^{-\lambda}{\left[\lambda\ +\ \lambda{\left(\sum^u_{k\eq 1}\frac{\lambda^k}{k!}\ -\ \sum^\infty_{k\eq v}\frac{\lambda^k}{k!}\right)}\ +\ \lambda{\left( \sum^\infty_{k\eq v-1}\frac{\lambda^{k} }{k!}\ -\ \sum^{u-1}_{k\eq 0}\frac{\lambda^{k} }{k!}\right)}\right]}$$

    $$\eq \lambda e^{-\lambda}{\left[1\ +\ {\left(\sum^u_{k\eq 1}\frac{\lambda^k}{k!}\ -\ \sum^\infty_{k\eq v}\frac{\lambda^k}{k!}\right)}\ +\ {\left( \sum^\infty_{k\eq v-1}\frac{\lambda^{k} }{k!}\ -\ \sum^{u-1}_{k\eq 0}\frac{\lambda^{k} }{k!}\right)}\right]}$$

  • For convienence let us assign the sums letters:
    • $$\eq \lambda e^{-\lambda}{\Bigg[1\ +\ \underbrace{\sum^u_{k\eq 1}\frac{\lambda^k}{k!} }_\text{A}\ -\ \underbrace{\sum^\infty_{k\eq v}\frac{\lambda^k}{k!} }_\text{B}\ +\ \underbrace{\sum^\infty_{k\eq v-1}\frac{\lambda^{k} }{k!} }_\text{C}\ -\ \underbrace{\sum^{u-1}_{k\eq 0}\frac{\lambda^{k} }{k!} }_\text{D} \Bigg]}$$
  • Now we combine the sums:
    • $$\LHS{}\eq\lambda e^{-\lambda}{\Bigg[1+\underbrace{\sum^{u-1}_{k\eq 1}\frac{\lambda^k}{k!}+\frac{\lambda^u}{u!} }_\text{A}\ -\ \underbrace{\sum_{k\eq v}^\infty \frac{\lambda^k}{k!} }_\text{B}\ +\ \underbrace{ \frac{\lambda^{v-1} }{(v-1)!}\ +\ \sum^\infty_{k\eq v}\frac{\lambda^k}{k!} }_\text{C}\ -\ \underbrace{ \frac{\lambda^0}{0!}\ -\ \sum^{u-1}_{k\eq 1}\frac{\lambda^k}{k!} }_\text{D} \Bigg]}$$

      $$\eq \lambda e^{-\lambda}{\Bigg[ 1\ -\ \frac{\lambda^0}{0!} \ +\ \frac{\lambda^u }{u!} \ +\ \frac{\lambda^{v-1} }{(v-1)!} \Bigg]}$$ - as the sum above in $\text{ A }$ cancels with the sum part of $\text{ D }$, and $\text{ B }$ cancels with the sum part of $\text{ C }$

      $$\eq \lambda e^{-\lambda}{\left[ 1-1+2\frac{\lambda^u}{u!} \right]}$$ - using that $v-1\eq u$ and tidying up

      $$\eq 2\lambda e^{-\lambda}\frac{\lambda^u}{u!}$$

• Thus we see $$\text{Mdm}(X)\eq 2\lambda e^{-\lambda}\frac{\lambda^u}{u!}$$