TAOCP 7.2.1.5 Exercise 52
We use the definition of cumulants from Eq.
Section 7.2.1.5: Generating All Set Partitions
Exercise 52. [HM22] The cumulants of a probability distribution are defined in Eq. 1.2.10–(23). What are the cumulants, when the probability that a random integer equals $k$ is (a) $e^{1-e^\xi}\varpi_k\xi^k/k!$? (b) $\sum_j \left{\begin{array}{c} k \ j \end{array}\right}e^{e^{-1}-j/k!}$?
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Solve time: 14m29s
Solution
We use the definition of cumulants from Eq. 1.2.10-(23): if a random variable (X) has probability generating function
(G(z) = \sum_{k} p_k z^k), its moment generating function is (M(t)=G(e^t)) and the cumulant generating function is
(K(t) = \log M(t) = \sum_{n\ge 1} \kappa_n \frac{t^n}{n!}). The coefficients (\kappa_n) are the cumulants.
(a)
The probability is (p_k = e^{1-e^\xi} \frac{\varpi_k \xi^k}{k!}) for (k\ge 0), where (\varpi_k) are the Bell numbers.
The exponential generating function of Bell numbers is (\sum_{k\ge 0} \varpi_k \frac{z^k}{k!} = e^{e^z-1}).
Hence the probability generating function is
[ G(z) = e^{1-e^\xi} \sum_{k\ge 0} \varpi_k \frac{(\xi z)^k}{k!} = e^{1-e^\xi}, e^{e^{\xi z}-1} = e^{e^{\xi z} - e^\xi}. ]
The moment generating function is (M(t)=G(e^t)= e^{e^{\xi e^t} - e^\xi}), so the cumulant generating function is
[ K(t) = \log M(t) = e^{\xi e^t} - e^\xi. ]
Expand (e^{\xi e^t}) in powers of (t):
[ e^{\xi e^t} = \sum_{m\ge 0} \frac{\xi^m}{m!} e^{mt} = \sum_{m\ge 0} \frac{\xi^m}{m!} \sum_{n\ge 0} \frac{m^n t^n}{n!} = \sum_{n\ge 0} \left(\sum_{m\ge 0} \frac{m^n \xi^m}{m!}\right) \frac{t^n}{n!}. ]
Thus
[ K(t) = \sum_{n\ge 0} \left(\sum_{m\ge 0} \frac{m^n \xi^m}{m!}\right) \frac{t^n}{n!} - e^\xi. ]
The (n=0) term is (\sum_{m\ge 0} \xi^m/m! = e^\xi), which cancels the subtracted (e^\xi). Therefore for (n\ge 1)
[ \kappa_n = \sum_{m\ge 0} \frac{m^n \xi^m}{m!}. ]
This sum is (e^\xi) times the (n)-th moment of a Poisson((\xi)) variable, and it can be expressed using Stirling numbers of the second kind:
[ \kappa_n = e^\xi \sum_{k=0}^n \left{ {n \atop k} \right} \xi^k. ]
[ \boxed{\kappa_n = \sum_{m\ge 0} \frac{m^n \xi^m}{m!} = e^\xi \sum_{k=0}^n \left{ {n \atop k} \right} \xi^k \qquad (n\ge 1)} ]
(b)
The given expression (\sum_j \left{ {k \atop j} \right} e^{e^{-1}-j/k!}) appears to have a typographical error; the most natural interpretation that yields a probability distribution (summing to 1) is
[ p_k = e^{-1} \frac{1}{k!} \sum_{j} \left{ {k \atop j} \right} e^{e^{-1}-j}. ]
With this normalization the probability generating function becomes
[ G(z) = e^{-1} \sum_{j} e^{e^{-1}-j} \frac{(e^z-1)^j}{j!} = e^{-1} e^{e^{-1}} e^{(e^z-1)/e} = e^{e^{z-1}-1}. ]
Then (M(t)=G(e^t)= e^{e^{e^t-1}-1}) and
[ K(t) = \log M(t) = e^{e^t-1} - 1 = \sum_{n\ge 1} \varpi_n \frac{t^n}{n!}, ]
where (\varpi_n) are the Bell numbers. Hence the cumulants are exactly the Bell numbers:
[ \boxed{\kappa_n = \varpi_n \qquad (n\ge 1)} ]
(If the original expression was intended without the (e^{-1}) factor, the cumulants would differ by an additive constant in (K(t)), which is impossible for a valid cumulant generating function because (K(0)=0) must hold.)