10F (Probability)

G_{X} (s) = \sum_{m = 1}^{\infty} P (X = m) s^{m} .

(1)

Differentiating

n

times,

G_{X}^{(n)} (s) = \sum_{m = n}^{\infty} \frac{m !}{( m - n )!} P (X = m) s^{m - n} .

(2)

s = 0

, only the term

m = n

remains. Hence

P (X = n) = \frac{G _{X}^{(n)} ( 0 )}{n !} .

(3)

b

 0  0

Solution

 0  0

For

0 \leq s \leq 1

, independence gives

G_{X + Y} (s) = E [s^{X + Y}] = E [s^{X} s^{Y}] = E [s^{X}] E [s^{Y}] .

(1)

Therefore

G_{X + Y} (s) = G_{X} (s) G_{Y} (s) .

(2)

c

 0  0

Solution

 0  0

For the geometric distribution on

{1, 2, \dots}

P (X = m) = p (1 - p)^{m - 1} .

(1)

Summing the geometric series,

G_{X} (s) = \sum_{m = 1}^{\infty} p (1 - p)^{m - 1} s^{m} = \frac{p s}{1 - ( 1 - p ) s} .

(2)

d

 0  0

Solution

 0  0

When

r

red marbles remain, the next draw reduces their number with probability

r / n

. The waiting time

W_{r}

for that reduction has a geometric distribution on

{1, 2, \dots}

and therefore

G_{W_{r}} (s) = \frac{( r / n ) s}{1 - ( 1 - r / n ) s} = \frac{rs}{n - ( n - r ) s} .

(1)

The successive waiting times are independent, and

T = W_{n} + W_{n - 1} + \dots + W_{1} .

(2)

Using the product rule for probability-generating functions,

G_{T} (s) = r = 1 \prod n \frac{rs}{n - ( n - r ) s} .

(3)

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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 10F (Probability)

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