Problem 11537
(American Mathematical Monthly, Vol.117, December 2010) Proposed by L. Withers (USA).
Let p be a prime and a be a positive integer. Let X be a random variable having a Poisson distri- bution with mean a, and let M be the p-th moment of X. Prove that M ≡ 2a (mod p).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
For k = 1, . . . p − 1, by the Fermat’s little theorem, we have that
p k
= 1 k!
k−1
X
i=0
(−1)ik i
(k − i)p≡ 1 k!
k−1
X
i=0
(−1)ik i
(k − i) = 1 k
=
1 if k = 1,
0 if k > 1, (mod p)
where p k
denote the Stirling numbers of the second kind. Hence, since p p
= 1,
M = ea
∞
X
n=0
annp n! =
p
X
k=1
p k
ak =
p−1
X
k=1
p k
ak+ p p
ap≡ a + ap ≡ 2a (mod p).
