Mdm of the Poisson distribution
From Maths
TODO: Link with Poisson distribution page
Contents
[hide]Statement
Let X∼Poi(λ) for some λ∈R>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[ |X−E[X]| ]
Calculation
- Mdm(X):=E[ |X−E[X]| ]:=∞∑k=0|X−E[X]|⋅P[X=k]
- =e−λ ∞∑k=0λkk!|k−E[X]|=e−λ ∞∑k=0λkk!|k−λ|
- Note that:
- if k≥λ ⟹ k−λ≥0 ⟹ |k−λ|=k−λ
- if k≤λ ⟹ k−λ≤0 ⟹λ−k≥0 ⟹|k−λ|=λ−k
- Define the following two values:
- u:=RoundDownToInt(λ)[Note 1] (also known as the floor function[Note 2] and
- v:=u+1[Note 3]
- This means we have u≤λ and v≥λ, specifically, we have the following two cases:
- if k≤u and as u≤λ we see k≤λ and
- if k>u then k≥u+1=v≥λ so k≥λ
- Now, from above: Mdm(X)=e−λ ∞∑k=0λkk!|k−λ|
- =e−λ[λ00!|0−λ| + ∞∑k=1λkk!|k−λ|]
- =e−λ[λ + u∑k=1λkk!|k−λ| + ∞∑k=vλkk!|k−λ|]with the understanding that if u=0 that the sum from k=1 to u evaluates to 0, obviously
- =e−λ[λ00!|0−λ| + ∞∑k=1λkk!|k−λ|]
- Notice now that:
- For the first sum, where 1≤k≤u (specifically that k≤u) we have k≤λ
- and that from further above we noticed if k≤λ then |k−λ|=λ−k
- For the second sum, where k>u that this meant k≥λ
- and that from further above we noticed if k≥λ then |k−λ|=k−λ, so
- For the first sum, where 1≤k≤u (specifically that k≤u) we have k≤λ
- Mdm(X)=e−λ[λ + u∑k=1λkk!|k−λ| + ∞∑k=vλkk!|k−λ|]
- =e−λ[λ + u∑k=1λkk!(λ−k) + ∞∑k=vλkk!(k−λ)]
- we now expand these sums:
- =e−λ[λ + first sum⏞u∑k=1λk+1k! − u∑k=1kλkk! + second sum⏞∞∑k=vkλkk!−∞∑k=vλk+1k! ]
- =e−λ[λ + λ(u∑k=1λkk! − ∞∑k=vλkk!) + (∞∑k=vλk(k−1)! − u∑k=1λk(k−1)!)], by grouping the terms and factorising where we can
- =e−λ[λ + λ(u∑k=1λkk! − ∞∑k=vλkk!) + (∞∑k=v−1λk+1k! − u−1∑k=0λk+1k!)][Note 4] by reindexing the latter two sums
- =e−λ[λ + λ(u∑k=1λkk! − ∞∑k=vλkk!) + λ(∞∑k=v−1λkk! − u−1∑k=0λkk!)]
- =λe−λ[1 + (u∑k=1λkk! − ∞∑k=vλkk!) + (∞∑k=v−1λkk! − u−1∑k=0λkk!)]
- =e−λ[λ + u∑k=1λkk!(λ−k) + ∞∑k=vλkk!(k−λ)]
- For convienence let us assign the sums letters:
- =λe−λ[1 + u∑k=1λkk!⏟A − ∞∑k=vλkk!⏟B + ∞∑k=v−1λkk!⏟C − u−1∑k=0λkk!⏟D]
- =λe−λ[1 + u∑k=1λkk!⏟A − ∞∑k=vλkk!⏟B + ∞∑k=v−1λkk!⏟C − u−1∑k=0λkk!⏟D]
- Now we combine the sums:
- Mdm(X)=λe−λ[1+u−1∑k=1λkk!+λuu!⏟A − ∞∑k=vλkk!⏟B + λv−1(v−1)! + ∞∑k=vλkk!⏟C − λ00! − u−1∑k=1λkk!⏟D]
- =λe−λ[1 − λ00! + λuu! + λv−1(v−1)!]- as the sum above in A cancels with the sum part of D , and B cancels with the sum part of C
- =λe−λ[1−1+2λuu!]- using that v−1=u and tidying up
- =2λe−λλuu!
- =λe−λ[1 − λ00! + λuu! + λv−1(v−1)!]
- Mdm(X)=λe−λ[1+u−1∑k=1λkk!+λuu!⏟A − ∞∑k=vλkk!⏟B + λv−1(v−1)! + ∞∑k=vλkk!⏟C − λ00! − u−1∑k=1λkk!⏟D]
- =e−λ ∞∑k=0λkk!|k−E[X]|
- Thus we see Mdm(X)=2λe−λλuu!
Notes
- Jump up ↑ Recall λ>0, this means u≥0 and thus u∈N0
- Jump up ↑ Which is sometimes written:
- TODO: It's [n] but with the bottom or top notches removed from the square brackets?
-
- Jump up ↑ Notice:
- If λ is not ∈N≥0 then u+1=RoundUpToInt(λ), so u+1=v≥λ
- If λ is in N≥0 then u=λ and v=u+1>u=λ so v>λ
- Notice (v>λ)⟹(v≥λ)
- 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→∞(n∑k=v⋯)
- limn→∞(n−1∑k=v−1⋯)
- limn→∞(n∑k=v⋯)
References
- Jump up ↑ Alec's own work, I actually kept muddling it up on paper so this page IS the reference!
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