Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 3F / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

Independence gives

M_{S} (t) = \prod_{j = 1}^{n} M_{X_{j}} (t) = [exp (e^{t} - 1)]^{n} = exp {n (e^{t} - 1)} .

This is the moment generating function of a Poisson distribution with mean

n

. Hence

S \sim Poisson (n) .

ii

Solution

The exponential Markov bound gives, for every

t > 0

,

Pr (S \geq 2 n) \leq exp {- 2 n t + n (e^{t} - 1)} .

The exponent is minimized when

e^{t} = 2

, so take

t = lo g 2

. Then

Pr (S \geq 2 n) \leq exp {n (1 - 2 lo g 2)} = (\frac{e}{4})^{n} .

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