 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/7/a/ii/solution

G_{X} (s) = \sum_{k = 0}^{n} (k n) p^{k} (1 - p)^{n - k} s^{k} = (1 - p + p s)^{n} .

G_{X}^{'} (1) = n p, G_{X}^{''} (1) = n (n - 1) p^{2} = E [X (X - 1)] .

Since

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

, the variance is

G_{X}^{''} (1) + G_{X}^{'} (1) - G_{X}^{'} (1)^{2} = n (n - 1) p^{2} + n p - n^{2} p^{2} = n p (1 - p) .

Thus

G_{X} (s) = (1 - p + p s)^{n}, E [X] = n p, var (X) = n p (1 - p) .

The formulas also hold for

n = 0, 1

and the degenerate choices

p = 0, 1

, with derivatives read directly from the resulting polynomial.

 New to topics? Read the docs here!