Mdm of the Poisson distribution

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TODO: Link with Poisson distribution page


Statement

Let XPoi(λ) for some λR>0. X may take any value in N0


We will show that

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

Recall the Mdm is defined as:

  • Mdm(X):=E[ |XE[X]| ]

Calculation

  • Mdm(X):=E[ |XE[X]| ]
    :=k=0|XE[X]|P[X=k]
    =eλ k=0λkk!|kE[X]|
    =eλ k=0λkk!|kλ|
    • Note that:
      1. if kλ  kλ0  |kλ|=kλ
      2. if kλ  kλ0 λk0 |kλ|=λk
    • Define the following two values:
      1. u:=RoundDownToInt(λ)[Note 1] (also known as the floor function[Note 2] and
      2. v:=u+1[Note 3]
      • This means we have uλ and vλ, specifically, we have the following two cases:
        1. if ku and as uλ we see kλ and
        2. if k>u then ku+1=vλ so kλ
    • Now, from above: Mdm(X)=eλ k=0λkk!|kλ|
      =eλ[λ00!|0λ| + k=1λkk!|kλ|]
      =eλ[λ + uk=1λkk!|kλ| + k=vλkk!|kλ|]
      with the understanding that if u=0 that the sum from k=1 to u evaluates to 0, obviously
    • Notice now that:
      1. For the first sum, where 1ku (specifically that ku) we have kλ
        • and that from further above we noticed if kλ then |kλ|=λk
      2. For the second sum, where k>u that this meant kλ
        • and that from further above we noticed if kλ then |kλ|=kλ, so
    • Mdm(X)=eλ[λ + uk=1λkk!|kλ| + k=vλkk!|kλ|]
      =eλ[λ + uk=1λkk!(λk) + k=vλkk!(kλ)]
      • we now expand these sums:
      =eλ[λ + first sumuk=1λk+1k!  uk=1kλkk! + second sumk=vkλkk!k=vλk+1k! ]
      =eλ[λ + λ(uk=1λkk!  k=vλkk!) + (k=vλk(k1)!  uk=1λk(k1)!)]
      , by grouping the terms and factorising where we can
      =eλ[λ + λ(uk=1λkk!  k=vλkk!) + (k=v1λk+1k!  u1k=0λk+1k!)]
      [Note 4] by reindexing the latter two sums
      =eλ[λ + λ(uk=1λkk!  k=vλkk!) + λ(k=v1λkk!  u1k=0λkk!)]
      =λeλ[1 + (uk=1λkk!  k=vλkk!) + (k=v1λkk!  u1k=0λkk!)]
    • For convienence let us assign the sums letters:
      • =λeλ[1 + uk=1λkk!A  k=vλkk!B + k=v1λkk!C  u1k=0λkk!D]
    • Now we combine the sums:
      • Mdm(X)=λeλ[1+u1k=1λkk!+λuu!A  k=vλkk!B + λv1(v1)! + k=vλkk!C  λ00!  u1k=1λkk!D]
        =λeλ[1  λ00! + λuu! + λv1(v1)!]
        - as the sum above in  A  cancels with the sum part of  D , and  B  cancels with the sum part of  C 
        =λeλ[11+2λuu!]
        - using that v1=u and tidying up
        =2λeλλuu!
  • Thus we see Mdm(X)=2λeλλuu!

Notes

  1. Jump up Recall λ>0, this means u0 and thus uN0
  2. Jump up Which is sometimes written:
    • TODO: It's [n] but with the bottom or top notches removed from the square brackets?
  3. Jump up Notice:
    • If λ is not N0 then u+1=RoundUpToInt(λ), so u+1=vλ
    • If λ is in N0 then u=λ and v=u+1>u=λ so v>λ
      • Notice (v>λ)(vλ)
    So either way, vλ
  4. Jump up Note that the third sum should have "1" as its upper index, however remember that when a sum is to this is actually a limit, it was in this case:
    • limn(nk=v)
    and became
    • limn(n1k=v1)

References

  1. Jump up Alec's own work, I actually kept muddling it up on paper so this page IS the reference!