Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 9

Table of contents
- a 9
  - i a
    - Solution i
  - ii a
    - Solution ii
  - iii a
    - Solution iii
- b 9
  - i b
    - Solution i
  - ii b
    - Solution ii
- c 9
  - i c
    - Solution i
  - ii c
    - Solution ii

a

i

Solution

The rate-one exponential distribution has density

e^{- t}

for

t \geq 0

. Integration by parts gives

E T = 1

and

E [T^{2}] = 2

, so

E T = 1, Var (T) = 2 - 1^{2} = 1.

ii

Solution

Because

T

and

U

are independent random variables, their product's expected value factors:

E [T U] = E T E U = 1.

The original independence assumption is essential to this factorization.

iii

Solution

For the overlapping products, factor using the independence of

T, U, V

:

Cov (T U, U V) = E [T U^{2} V] - E [T U] E [U V] = 1 \cdot 2 \cdot 1 - 1 \cdot 1 = 1.

This is the covariance of overlapping independent products: the shared

U

contributes its second moment. In contrast,

T U

and

VW

use disjoint independent factors and are themselves independent, giving

Cov (T U, U V) = 1, Cov (T U, VW) = 0.

b

i

Solution

Assume finite second moments and write

Z_{i} = X_{i} - E X_{i}

. Expanding the squared centered sum gives

Var (X_{1} + X_{2}) = E [(Z_{1} + Z_{2})^{2}] = E [Z_{1}^{2}] + E [Z_{2}^{2}] + 2 E [Z_{1} Z_{2}] = Var (X_{1}) + Var (X_{2}) + 2 Cov (X_{1}, X_{2}) .

This variance of a sum identity does not require independence; the covariance term records the contribution that would otherwise be lost.

ii

Solution

For finite second moments, the general variance of a sum formula is

Var (i = 1 \sum n X_{i}) = i = 1 \sum n Var (X_{i}) + 2 1 \leq i < j \leq n \sum Cov (X_{i}, X_{j}) .

Each distinct unordered pair appears once in the second sum and twice when the full square is expanded. No independence assumption is needed.

c

i

Solution

Put

Z_{i} = Y_{i - 1} Y_{i}

for

1 \leq i \leq n

. Independence of the underlying exponentials gives

E Z_{i} = 1

and

Var (Z_{i}) = E [Y_{i - 1}^{2}] E [Y_{i}^{2}] - 1 = 4 - 1 = 3.

By the covariance of overlapping independent products, adjacent terms have covariance one. Terms separated by at least two indices use disjoint independent factors, so their covariance is zero. There are

n - 1

adjacent pairs. Linearity of expectation and the variance of a sum formula therefore yield

E S_{n} = n, Var (S_{n}) = 3 n + 2 (n - 1) = 5 n - 2, n \geq 1.

In particular, the products themselves are not independent merely because the original

Y_{i}

are.

ii

Solution

The normalized sum has mean one and variance

(5 n - 2) / n^{2}

. For every fixed

ϵ > 0

, the Chebyshev inequality gives

0 \leq P (\frac{S _{n}}{n} > 1 + ϵ) \leq P (\frac{S _{n}}{n} - 1 > ϵ) \leq \frac{5 n - 2}{n ^{2} ϵ ^{2}} ⟶ 0.

This proves the requested limit, and in fact proves convergence in probability of

S_{n} / n

to one. It is a finite covariance range weak law: adjacent dependence is harmless here because only linearly many covariance terms survive.

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