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Write

A_{i} = Y_{i - 1} Y_{i}

, so

S_{n} = \sum_{i = 1}^{n} A_{i}

. Each product has mean one, and its variance is

Var (A_{i}) = E (Y_{i - 1}^{2}) E (Y_{i}^{2}) - 1 = 2 \cdot 2 - 1 = 3.

(1)

Adjacent products share one factor, giving covariance one by part (a)(iii). Products whose indices differ by at least two involve disjoint independent random variables, so their covariance is zero. There are

n - 1

adjacent pairs. By the variance of a sum,

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

(2)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / b / ii / Solution by

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For random variables with finite second moments, the variance of a sum 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}) .

(1)

No independence assumption is needed.

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / b / i / Solution by

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Assume finite second moments and write

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

. Expanding the square in the definition of variance 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}) .

(1)

Recognizing the two variances and the covariance, we obtain

Var (X_{1} + X_{2}) = Var (X_{1}) + Var (X_{2}) + 2 Cov (X_{1}, X_{2}) .

(2)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / a / iii / Solution by

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The products

T U

and

U V

U

, so they are not independent. Using the definition of covariance and the independence of

T, U, V

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

(1)

By contrast,

T U

and

VW

depend on disjoint sets of independent random variables, so they are independent and have zero covariance. Hence

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

(2)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / a / ii / Solution by

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By independence, the expected value of the product factors:

E (T U) = E T E U = 1.

(1)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / a / i / Solution by

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For the rate-one exponential distribution, the probability density function is

e^{- t}

for

t \geq 0

. By integration by parts,

E T = \int_{0}^{\infty} t e^{- t} d t = 1, E (T^{2}) = \int_{0}^{\infty} t^{2} e^{- t} d t = 2.

(1)

Thus

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

(2)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 5 / b / iv / Solution by

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Starting from

X

at time

n

, the markov property and the time-independent transition rules give

P (M_{2 n} = X ∣ M_{n} = X) = p_{n}

. Therefore

P (M_{n} = X ∣ M_{2 n} = X) = \frac{P ( M _{n} = X , M _{2 n} = X )}{p _{2 n}} = \frac{p _{n}^{2}}{p _{2 n}} .

(1)

Write

p_{n} = \frac{3}{4} (\frac{3}{4})^{n} (1 + \frac{1}{3} (- \frac{1}{3})^{n})

. The conditional probability becomes

\frac{3}{4} \frac{( 1 + \frac{1}{3} ( - \frac{1}{3} ) ^{n} ) ^{2}}{1 + \frac{1}{3} ( \frac{1}{9} ) ^{n}} ⟶ \frac{3}{4} .

(2)

The conditioning selects paths which have avoided the absorbing state up to time

2 n

; it is therefore quite different from the unconditioned probability

p_{n}

, which tends to zero.

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 5 / b / iii / Solution by

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The characteristic equation of a linear recurrence is

16 r^{2} - 8 r - 3 = (4 r - 3) (4 r + 1) = 0.

(1)

Its distinct roots are

3/4

and

- 1/4

, so

p_{n} = A (3/4)^{n} + B (- 1/4)^{n}

. The initial conditions are

p_{0} = 1

and

p_{1} = 1/2

, giving

A + B = 1

and

3 A - B = 2

. Hence

A = 3/4

B = 1/4

, and

p_{n} = \frac{3}{4} (\frac{3}{4})^{n} + \frac{1}{4} (- \frac{1}{4})^{n} = \frac{3 ^{n + 1} + ( - 1 ) ^{n}}{4 ^{n + 1}}, n \geq 0.

(2)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 5 / b / ii / Solution by

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Let

q_{n} = P (M_{n} = Y)

. The law of total probability applied to the current markov chain state gives

p_{n} = \frac{1}{2} p_{n - 1} + \frac{3}{8} q_{n - 1}, q_{n - 1} = \frac{1}{2} p_{n - 2} (n \geq 2) .

(1)

The second equality holds because

Y

can only be reached from

X

. Eliminating

q_{n - 1}

gives the linear recurrence relation

16 p_{n} = 8 p_{n - 1} + 3 p_{n - 2}, n \geq 2.

(2)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 5 / b / i / Solution by

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There are two paths from

X

back to

X

in two steps:

X \to X \to X

and

X \to Y \to X

. By the markov property, their probabilities multiply along each path. Therefore

p_{2} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{3}{8} = \frac{1}{4} + \frac{3}{16} = \frac{7}{16} .

(1)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 5 / a / ii / Solution by

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For a finite or countable partition of a set into events

B_{i}

, the law of total probability states

P (A) = i : P (B_{i}) > 0 \sum P (A ∣ B_{i}) P (B_{i}) .

(1)

The events

A \cap B_{i}

are pairwise disjoint and have union

A

, so countable additivity proves the formula. Terms with

P (B_{i}) = 0

contribute zero and are omitted to avoid undefined conditional probabilities.

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 5 / a / i / Solution by

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A partition of a set

Ω

is a collection of nonempty subsets whose union is

Ω

and which are pairwise disjoint:

⋃_{i \in I} B_{i} = Ω, B_{i} \cap B_{j} = \emptyset (i \neq = j) .

(1)

For use with probabilities, the

B_{i}

must also be events in the sample space.

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / b / iii / Solution by

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For the geometric distribution,

μ = 1/ p

, so the probability generating function of the size-biased distribution is

G_{Y} (s) = \frac{p ^{2} s}{( 1 - q s ) ^{2}} .

(1)

X_{1}, X_{2}

are independent and identically distributed with geometric distribution of parameter

p

, the probability generating function of an independent sum gives

G_{X_{1} + X_{2} - 1} (s) = s^{- 1} (\frac{p s}{1 - q s})^{2} = \frac{p ^{2} s}{( 1 - q s ) ^{2}} .

(2)

At zero, interpret the expression by its continuous extension. The probability generating functions agree, hence

Y = d X_{1} + X_{2} - 1, X_{1}, X_{2} \sim Geom (p) independently .

(3)

One can also check the probability mass functions directly: for

k \geq 1

, there are

k

positive integer pairs with sum

k + 1

, each having probability

p^{2} q^{k - 1}

. Thus both sides assign probability

k p^{2} q^{k - 1}

k

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / b / ii / Solution by

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By the definition of a probability generating function,

G_{Y} (s) = \sum_{k \geq 1} \frac{k p _{k}}{μ} s^{k} = \frac{s}{μ} G_{Y}^{'} (s) .

(1)

Termwise differentiation of the power series is valid for

∣ s ∣ < 1

. At

s = 1

, the equality extends by the finite mean assumption, and both sides equal one.

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / b / i / Solution by

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Since

Y \geq 1

, its finite mean satisfies

1 \leq μ < \infty

. Every

q_{k} = k p_{k} / μ

is nonnegative, and

\sum_{k \geq 1} q_{k} = \frac{1}{μ} \sum_{k \geq 1} k p_{k} = \frac{E Y}{μ} = 1.

(1)

Thus

(q_{k})

is a probability mass function. It defines the size-biased distribution of

Y

: an outcome of size

k

receives

k

times its original relative weight.

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / a / iv / Solution by

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The event

W > k

occurs exactly when both

X_{1} > k

and

X_{2} > k

. By independence,

P (W > k) = q^{k} q^{k} = q^{2 k} = (1 - (2 p - p^{2}))^{k} .

(1)

Thus

W \sim Geom (2 p - p^{2}) .

(2)

The new success probability

2 p - p^{2}

is the chance that at least one of the two trials succeeds.

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / a / iii / Solution by

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The mean is the derivative at one of the probability generating function. Therefore

E X = G_{X}^{'} (1) = \frac{p}{p ^{2}} = \frac{1}{p} .

(1)

Evaluation at one is valid here because the probability generating function has radius of convergence

1/ (1 - p) > 1

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / a / ii / Solution by

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Using the geometric series, the probability generating function is

G_{X} (s) = E (s^{X}) = \sum_{k \geq 1} p q^{k - 1} s^{k} = \frac{p s}{1 - q s}, ∣ q s ∣ < 1.

(1)

Differentiation gives

G_{X}^{'} (s) = \frac{p ( 1 - q s ) + pq s}{( 1 - q s ) ^{2}} = \frac{p}{( 1 - s + p s ) ^{2}} .

(2)

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / a / i / Solution by

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Put

q = 1 - p

. Since

X

takes positive integer values,

P (X > 0) = 1 = q^{0}

. Taking differences of the given tail probabilities therefore yields, also for

k = 1

P (X = k) = P (X > k - 1) - P (X > k) = p q^{k - 1}, k \geq 1.

(1)

This is the probability mass function of the geometric distribution on

{1, 2, \dots}

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 5 / c / ii / Solution by

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The finite geometric series identity gives, for each

∣ x ∣ < 1

f_{n} (x) = \frac{x ( 1 - x ^{n} )}{1 - x} ⟶ f (x) = \frac{x}{1 - x}, f (x) - f_{n} (x) = \frac{x ^{n + 1}}{1 - x} .

(1)

For every fixed

n

, the absolute value of this error tends to infinity as

x \to 1^{-}

. Thus

- 1 < x < 1 sup ∣ f (x) - f_{n} (x) ∣ = \infty,

(2)

and there is no uniform convergence on

(- 1, 1)

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 Pinned article: Introduction to the OurBigBook Project

Welcome to the OurBigBook Project! Our goal is to create the perfect publishing platform for STEM subjects, and get university-level students to write the best free STEM tutorials ever.

Everyone is welcome to create an account and play with the site: ourbigbook.com/go/register. We belive that students themselves can write amazing tutorials, but teachers are welcome too. You can write about anything you want, it doesn't have to be STEM or even educational. Silly test content is very welcome and you won't be penalized in any way. Just keep it legal!

Video 1.

Intro to OurBigBook

. Source.

We have two killer features:

topics: topics group articles by different users with the same title, e.g. here is the topic for the "Fundamental Theorem of Calculus" ourbigbook.com/go/topic/fundamental-theorem-of-calculus
Articles of different users are sorted by upvote within each article page. This feature is a bit like:
- a Wikipedia where each user can have their own version of each article
- a Q&A website like Stack Overflow, where multiple people can give their views on a given topic, and the best ones are sorted by upvote. Except you don't need to wait for someone to ask first, and any topic goes, no matter how narrow or broad
This feature makes it possible for readers to find better explanations of any topic created by other writers. And it allows writers to create an explanation in a place that readers might actually find it.
Figure 1.
Screenshot of the "Derivative" topic page
. View it live at: ourbigbook.com/go/topic/derivative
Video 2.
OurBigBook Web topics demo
. Source.
local editing: you can store all your personal knowledge base content locally in a plaintext markup format that can be edited locally and published either:
- to OurBigBook.com to get awesome multi-user features like topics and likes
- as HTML files to a static website, which you can host yourself for free on many external providers like GitHub Pages, and remain in full control
This way you can be sure that even if OurBigBook.com were to go down one day (which we have no plans to do as it is quite cheap to host!), your content will still be perfectly readable as a static site.
Figure 2.
You can publish local OurBigBook lightweight markup files to either OurBigBook.com or as a static website
.
Figure 3.
Visual Studio Code extension installation
.
Figure 4.
Visual Studio Code extension tree navigation
.
Figure 5.
Web editor
. You can also edit articles on the Web editor without installing anything locally.
Video 3.
Edit locally and publish demo
. Source. This shows editing OurBigBook Markup and publishing it using the Visual Studio Code extension.
Video 4.
OurBigBook Visual Studio Code extension editing and navigation demo
. Source.
Internal cross file references done right:
Infinitely deep tables of contents:
Figure 6.
Dynamic article tree with infinitely deep table of contents
.
Live URL: ourbigbook.com/cirosantilli/chordate
Descendant pages can also show up as toplevel e.g.: ourbigbook.com/cirosantilli/chordate-subclade

All our software is open source and hosted at: github.com/ourbigbook/ourbigbook

Further documentation can be found at: docs.ourbigbook.com

Feel free to reach our to us for any help or suggestions: docs.ourbigbook.com/#contact

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