 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/5/b/iv/solution

Computing the same expected value directly from the Poisson distribution instead gives

B (n) = E [X^{n}] = j = 0 \sum \infty j^{n} \frac{e ^{- 1}}{j !} = \frac{1}{e} j = 0 \sum \infty \frac{j ^{n}}{j !} .

n

, the ratio test shows that this sum converges. This identity is Dobiński's formula. The case

n = 0

also gives

B (0) = 1

0^{0}

is interpreted as one in this particular moment formula.

 New to topics? Read the docs here!