 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/8/a/iv/solution

1

G (s) = \sum_{j = 0}^{\infty} \frac{e ^{- 1} s ^{j}}{j !} = e^{s - 1}, s \in R .

G^{(k)} (s) = e^{s - 1}

E [X (X - 1) \dots (X - k + 1)] = 1 (k \geq 0) .

In particular

G^{'} (1) = G^{''} (1) = 1

, and the variance is

var (X) = 1 + 1 - 1^{2} = 1.

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