Mdm of the Poisson distribution

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TODO: Link with Poisson distribution page

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Statement

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

We will show that

• [math]\text{Mdm}(X)\eq 2\lambda e^{-\lambda}\frac{\lambda^u}{u!} [/math]
  • Where [ilmath]u:\eq[/ilmath][ilmath]\text{Floor}(\lambda)[/ilmath]

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

Recall the Mdm is defined as:

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

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

Calculation

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

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

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

    [math]\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]} [/math]

    [math]\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]} [/math] with the understanding that if [ilmath]u\eq 0[/ilmath] that the sum from [ilmath]k\eq 1[/ilmath] to [ilmath]u[/ilmath] evaluates to [ilmath]0[/ilmath], obviously

  • Notice now that:
    1. For the first sum, where [ilmath]1\le k\le u[/ilmath] (specifically that [ilmath]k\le u[/ilmath]) we have [ilmath]k\le \lambda[/ilmath]
       • and that from further above we noticed if [ilmath]k\le \lambda[/ilmath] then [ilmath]\big\vert k-\lambda\big\vert \eq \lambda-k[/ilmath]
    2. For the second sum, where [ilmath]k > u [/ilmath] that this meant [ilmath]k\ge \lambda[/ilmath]
       • and that from further above we noticed if [ilmath]k\ge\lambda[/ilmath] then [ilmath]\big\vert k-\lambda\big\vert\eq k-\lambda[/ilmath], so
  • [math]\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]} [/math]

    [math]\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]} [/math]

    • we now expand these sums:

    [math]\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]} [/math]

    [math]\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]} [/math], by grouping the terms and factorising where we can

    [math]\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]} [/math] by reindexing the latter two sums

    [math]\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]} [/math]

    [math]\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]} [/math]

  • For convienence let us assign the sums letters:
    • [math]\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]} [/math]
  • Now we combine the sums:
    • [math]\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]} [/math]

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

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

      [math]\eq 2\lambda e^{-\lambda}\frac{\lambda^u}{u!} [/math]

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