Dobiński's formula

In the setting of probability theory, Dobiński's formula represents the nth moment of the Poisson distribution with mean 1. Sometimes Dobiński's formula is stated as saying that the number of partitions of a set of size n equals the nth moment of that distribution.

The computation of the sum of Dobiński's series can be reduced to a finite sum of n+o(n) terms, taking into account the information that ${\displaystyle B_{n}}$ is an integer. Precisely one has, for any integer K > 1

${\displaystyle B_{n}=\left\lceil {1 \over e}\sum _{k=0}^{K-1}{\frac {k^{n}}{k!}}\right\rceil }$

provided ${\displaystyle {\frac {K^{n}}{K!}}\leq 1}$ (a condition that implies K > n, but that is satisfied by some K of size n+o(n)). Indeed, one has ${\displaystyle {\frac {(K+j)^{n}}{(K+j)!}}\leq {\frac {K^{n}}{K!}}{1 \over j!}\leq {1 \over j!}}$ for all j ≥ 0 so that the tail ${\displaystyle \sum _{k\geq K}{\frac {k^{n}}{k!}}=\sum _{j\geq 0}{\frac {(K+j)^{n}}{(K+j)!}}}$ is dominated by the series ${\displaystyle \sum _{j\geq 0}{\frac {1}{j!}}=e}$, which implies ${\displaystyle 0<B_{n}-{1 \over e}\sum _{k=0}^{K-1}{\frac {k^{n}}{k!}}<1}$, whence the reduced formula.

Dobiński's formula can be seen as a particular case, for ${\displaystyle x=0}$, of the more general relation:

${\displaystyle {1 \over e}\sum _{k=x}^{\infty }{\frac {k^{n}}{(k-x)!}}=\sum _{k=0}^{n}{\binom {n}{k}}B_{k}x^{n-k}.}$

One proof[2] relies on a formula for the generating function for Bell numbers,

${\displaystyle e^{e^{x}-1}=\sum _{n=0}^{\infty }{\frac {B_{n}}{n!}}x^{n}.}$

The power series for the exponential gives

${\displaystyle e^{e^{x}}=\sum _{k=0}^{\infty }{\frac {e^{kx}}{k!}}=\sum _{k=0}^{\infty }{\frac {1}{k!}}\sum _{n=0}^{\infty }{\frac {(kx)^{n}}{n!}}}$

so

${\displaystyle e^{e^{x}-1}={\frac {1}{e}}\sum _{k=0}^{\infty }{\frac {1}{k!}}\sum _{n=0}^{\infty }{\frac {(kx)^{n}}{n!}}}$

The coefficient of ${\displaystyle x^{n}}$ in this power series must be ${\displaystyle B_{n}/n!}$, so

${\displaystyle B_{n}={\frac {1}{e}}\sum _{k=0}^{\infty }{\frac {k^{n}}{k!}}.}$

Another style of proof was given by Rota.[3] Recall that if x and n are nonnegative integers then the number of one-to-one functions that map a size-n set into a size-x set is the falling factorial

${\displaystyle (x)_{n}=x(x-1)(x-2)\cdots (x-n+1)}$

Let ƒ be any function from a size-n set A into a size-x set B. For any b ∈ B, let ƒ ⁻¹(b) = {a ∈ A : ƒ(a) = b}. Then {ƒ ⁻¹(b) : b ∈ B} is a partition of A. Rota calls this partition the "kernel" of the function ƒ. Any function from A into B factors into

• one function that maps a member of A to the element of the kernel to which it belongs, and
• another function, which is necessarily one-to-one, that maps the kernel into B.

The first of these two factors is completely determined by the partition π that is the kernel. The number of one-to-one functions from π into B is (x)_|π|, where |π| is the number of parts in the partition π. Thus the total number of functions from a size-n set A into a size-x set B is

${\displaystyle \sum _{\pi }(x)_{|\pi |},}$

the index π running through the set of all partitions of A. On the other hand, the number of functions from A into B is clearly xⁿ. Therefore, we have

${\displaystyle x^{n}=\sum _{\pi }(x)_{|\pi |}.}$

Rota continues the proof using linear algebra, but it is enlightening to introduce a Poisson-distributed random variable X with mean 1. The equation above implies that the nth moment of this random variable is

${\displaystyle E(X^{n})=\sum _{\pi }E((X)_{|\pi |})}$

where E stands for expected value. But we shall show that all the quantities E((X)_k) equal 1. It follows that

${\displaystyle E(X^{n})=\sum _{\pi }1,}$

and this is just the number of partitions of the set A.

The quantity E((X)_k) is called the kth factorial moment of the random variable X. To show that this equals 1 for all k when X is a Poisson-distributed random variable with mean 1, recall that this random variable assumes each value integer value ${\displaystyle j\geq 0}$ with probability ${\displaystyle 1/(ej!)}$. Thus

${\displaystyle E((X)_{k})=\sum _{j=0}^{\infty }{\frac {(j)_{k}}{ej!}}={\frac {1}{e}}\sum _{j=0}^{\infty }{\frac {j(j-1)\cdots (j-k+1)}{j(j-1)\cdots 1}}={\frac {1}{e}}\sum _{j=i}^{\infty }{\frac {1}{(j-i)!}}=1.}$