Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 12F (Probability)

Table of contents
- a 12F
  - Solution a
- b 12F
  - Solution b
- c 12F
  - Solution c
- d 12F
  - Solution d

a

Solution

By Tonelli theorem for the nonnegative indicators,

E [N] = E [\sum_{n \geq 1} 1_{A_{n}}] = \sum_{n \geq 1} P (A_{n}) .

If this is finite, then

N

is finite almost surely: alternatively, Markov inequality gives

P (N \geq m) \leq \frac{E [ N ]}{m} ⟶ 0.

Since

{N = \infty} \subseteq {N \geq m}

for every

m

,

P (N = \infty) = 0 .

This is the first Borel-Cantelli lemma.

b

Solution

Write

p_{n} = P (A_{n})

. Independence gives, first for finite partial counts and then by monotone convergence,

E [2^{- N}] = \prod_{n \geq 1} E [2^{- 1_{A_{n}}}] = \prod_{n \geq 1} (1 - \frac{p _{n}}{2}) .

Using

1 - x \leq e^{- x}

factor by factor,

E [2^{- N}] \leq exp (- \frac{1}{2} n \geq 1 \sum p_{n}) = e^{- E [N] /2} .

c

Solution

If

\sum_{n} P (A_{n}) = \infty

, part (b) gives

0 \leq E [2^{- N}] \leq e^{- \infty} = 0.

Thus

2^{- N} = 0

almost surely. By the convention in the question, this occurs exactly when

N = \infty

. Therefore

P (N = \infty) = 1 .

This is the second Borel-Cantelli lemma for independent events.

d

Solution

Divide the keystrokes into disjoint blocks of five, and let

A_{n}

be the event that block

n

is exactly HELLO. The events are independent events, and

P (A_{n}) = 2 6^{- 5} .

Consequently

\sum_{n = 1}^{\infty} P (A_{n}) = \infty.

Part (c) shows that infinitely many of these block events occur almost surely. Each occurrence is an occurrence of HELLO in the full typed sequence, so

P (HELLO appears infinitely often) = 1 .

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version