Mathematics Vs Music

congrui t ies . The b ib l iog raphy does not quite m a t c h wi th the text. And what do we m a k e o f the fo l lowing “sequence of p r imes” w h i c h a p p e a r s on page 80: 2, 3, 5, 7, 9, 11, 12, 13, 17, 19, 23, 29, 31, 37, 41, 43, 53, . . . As an ac t of pen i t ence , Seuil should re- vise and r ep r in t the edi t ion.

R E F E R E N C E S

[1 ] Michael Atiyah, The Geometry and Physics of Knots, Cambridge: Cambridge University

Press, 1990.

[2] Jim Hoste, Morwen Thistlethwaite, Jeff

Weeks, The First 1 701 936 Knots, Math- ematical Intelligencer 20 (1998), no. 4, 33-48.

Fermat’s Last Tancjo, a Musical MUSIC BY JOSHUA ROSENBLUM, BOOK BY JOANNE

SYDNEY LESSNER, LYRICS BY LESSNER & ROSENBLUMr

A STAGE PRODUCTION FOR THE YORK THEATRE

COMPANY; VIDEOCASSEqTE AND DVD PRODUCED

BY THE CLAY MATHEMATICS INSTITUTE, CAMBRIDGE,

MA, 2001.

REVIEWED BY MICHELE EMMER

I n D e c e m b e r 1996 J o s h u a Rosenb lum read a r ev i ew of Ami r Acz6l ‘ s b o o k Fermat’s Las t Theorem [1] in The New York Times. He s h o w e d it to Joanne Sydney Lessner . “Is t he re a mus ica l in here?” L e s s n e r ‘ s a n s w e r was a def ini te yes. In sp i te of be ing math- phobic , she i m m e d i a t e l y s aw a com- pel l ing de t ec t ive s to ry inheren t in this t h r ee -hundred -yea r -o ld s ea r ch for a proof .

This was the b i r th of the idea of making a mus i ca l on A n d r e w Wiles ‘s adven tu re and the p r o o f of F e r m a t ‘ s theorem. In 1996, S imon Singh had made a f i lm on this proof , and a few months l a te r he p u b l i s h e d a b o o k by the s ame t i t le [2], [3].

Singh’s idea, fo l lowed by John Lynch, the BBC p r o d u c e r of the ser ies “Horizon” [4], [5], w a s tha t the s to ry of Wiles and his d e m o n s t r a t i o n could be- c o m e grea t c inema. It was an adven- ture in h u m a n emot ion , l o a d e d wi th un- e x p e c t e d o u t c o m e s , a rea l drama.

The s u c c e s s o f Singh ‘s f i lm and

b o o k was a n e c e s s a r y p r e c o n d i t i o n to making a mus ica l on the Wiles story. Many peop l e knew the f i lm and the book, so the mus ica l ‘ s a u d i e n c e could eas i ly fol low the story.

Joanne Lessne r is the wife of pi- anist , compose r , and o r c h e s t r a d i rec- tor Jo shua Rosenblum. She he r se l f had wr i t t en sc r ip t s and songs for o the r mu- sicals. What d id they wan t out of a mu- s ical cal led Fermat ‘s Last Tango? In the au thor ‘ s no t e s in the v ideo on the show ( f inanced by The Clay Mathe- mat ics Inst i tute, an ins t i tu te d e d i c a t e d to increas ing and d i s semina t ing math- emat ica l knowledge) , Lessner wri tes , “Our a im with Fermat’s Last Tango is, above all, to enter ta in . We th ink i t ‘s a moving, t ime less s tory tha t benef i t s f rom being to ld in a whimsical , some- t imes i r r everen t fashion. If, a f te r see- ing it, you ‘ re i n sp i r ed to t ake a c r a c k at t rying to f ind Fer ina t ‘ s s imple , ele- gant p r o o f yourself , go r ight ahead.”

If even b iog raphy and fi lm are nec- essar i ly f i c t i o n – – i n that the a u t h o r de- c ides wha t to include, emphas ize or ne- g l e c t – – a mus ica l need not be j u d g e d by its h is tor ica l t ru th fu lness nor by the ac- cu racy of its r e f e r ences to so-ca l led real life.

Never theless , unl ike mos t mus i ca l authors , the a u tho r s r e s e a r c h e d the i r subject . They had a head-s tar t . S ingh ‘s fi lm gives all the e l emen t s of the s tory, f rom the Sh inmra-Tan iyama conjec- ture to ell iptic curves .

Lessner and Rosenb lum m a k e the good choice o f br inging in o t h e r fa- mous ma thema t i c i ans to form wi th F e r m a t a kind o f chorus . As in G r e e k t ragedy, they c o m m e n t and in tervene . They are the gua rd ians of a ma the – mat ic ians ‘ p a r a d i s e which Keane (Wiles ‘s name in the mus ica l ) w o u l d l ike to reach. They are Carl F r i e d r i c h Gauss ( in t e rp re ted by Gilles Chiasson) , Py thagoras (Mitchel l Kantor) , Euc l id (Chr is t ianne Tisdale , a woman) , Sir I saac Newton ( a n o t h e r woman , the tal- en ted Carie Wilshusen) . The main char- ac te r s are obv ious ly Daniel Keane (Chris Thompson) and Pier re de Fer – mat (a very convinc ing J o n a t h a n Rabb). Anna Keane, the wife, is p l a y e d by Edwardyne Cowan.

The s tage set is s imple, the d e c o r

down to a bare minimum: chairs , big numbers , a few books .

The mus ica l s ta r t s wi th an intro- duc t ion to the p rob lem, as if it were a scient i f ic foreword . Jou rna l i s t s a re m o b b i n g Keane. It is the ev ident ly muc h -a w a i t e d m o m e n t of glory. They canno t help but comment , “He m u s t be c r a z y ” – – a r ecu r r en t i dea in f i lms and b o o k s abou t ma themat i c i ans . A m o n g o t h e r things, the j ou rna l i s t s inquire a b o u t the mean ing of the theo rem. In Singh ‘s fi lm the s tuden t s r e p e a t a lim- e r i ck on Py thagoras ‘ s theorem. In the musica l , natural ly, they sing it.

In the song “The Beauty of Num- bers ,” Keane regre t s having f in ished the p r o o f – – a s in the fihn a P r ince ton m a t h e m a t i c i a n a sks who will ever pro- v ide a c o m p a r a b l e p rob lem. It is fine thea t re . Some of the l ines a re good, even f rom a ma the ma t i c a l s t andpoin t : “Per fec t ion l ike pi and the go lden sec- t ion.”

In an amusing song, F e r m a t m a k e s fun of the c omple x i t y of Wiles ‘ s proof . Keane-Wiles come back wi th “Show me you r mmazelous proof .” F e r m a t refuses: “My lips a re sealed.” Keane: “The wor ld dese rves to see.” Fe rmat : “But if the mys te ry we re re- vea l ed I wou ld lose m y immorta l i ty ; I t ake the a n s w e r to my death.” Keane has the last word: “You a re al- r e a d y dead.”

Along with some of the quar te t s sung by the f amous ma themat i c i ans , the K e a n e – F e n n a t due t s are a m o n g the be s t pa r t s of the musical . Keane ‘ s en- c o u n t e r wi th the famous four is a lso enter ta in ing. Each of the four pro- c la ims his own super ior i ty . Keane rec- ognizes a h a p p y and grateful Gauss who excla ims: “Finally!” Then E u c l i d – – yes, the E u c l i d – – i s i n t roduced to him. T o w a r d the end, when Keane th inks he has s e c u r e d a p lace among the f amous ma thema t i c i ans , Fermat , who is stil l keen on def la t ing him, te l ls him tha t in his p r o o f the re is a “big fat hole.” Fer – m a t is bri l l iant , ironic. The ma thema t i – c ians b u r s t out laughing. F e r m a t sings, “No one will share my fame. It is the on ly th ing I have.” In the mos t amus ing scene of the musical , Keane is then s c o r n e d because , at a lmos t 40, he is t oo o l d – – t o o old to be c ons id e r ed for the

VOLUME 25, NUMBER 1, 2003 77

F i e l d s m eda l . T h e c h o r u s s ings , “Math-

e m a t i c s is a y o u n g m a n ‘ s g a m e . ”

K e a n e is d e s p e r a t e . Like t h e rea l – l i fe

Wiles , h e t a k e s r e f u g e in h i s f ami ly un-

til t h e e n l i g h t e n m e n t f ina l ly a r r ives .

D r a m a t i c as in S ingh ‘ s f i lm, w h e r e t h e

t r u e Wiles is s h o w n w e e p i n g w i t h e m o –

t i o n e v e n s ix m o n t h s la ter .

“You wil l a l w a y s b e t h e r e / I t h i n k I

wi l l be a b e t t e r h u s b a n d / It is o f t e n

sa id , w i t h i n y o u r fa i lu re lie t h e s e e d s

o f y o u r s u c c e s s . ” Th is l a s t p h r a s e , by

K e a n e ‘ s wi fe , c u e s an a s s o c i a t i o n al-

l o w i n g h i m to c o m p l e t e t h e p r o o f . N o t

a v e r y c o n v i n c i n g p l o t g i m m i c k . T h e

p r e s s c o n f e r e n c e e n d i n g is a l s o p r e –

d i c t ab le . K e a n e c o m e s to F e r m a t ‘ s de- [4]

f en se , s ay i n g h e w a s a g r e a t m a t h e –

m a t i c i a n a n d s u r e l y h a d a p ro o f . T h a t ‘ s

n o t w h a t t h e r e a l Wiles s a y s in S ingh ‘ s

film. On b a l a n c e , t h o u g h , t h e g a m b l e o f [5]

t ry ing to p r o d u c e an e n t e r t a i n i n g a n d

m a t h e m a t i c a l l y c o r r e c t m u s i c a l t u r n e d

o u t a s u c c e s s .

REFERENCES

[1] Amir Aczel, Fermat’s Last Theorem, Four

Walls Eight Windows. New York, 1996.

[2] Simon Singh, Fermat’s Last Theorem,

1997.

[3] – – & John Lynch, Fermat’s Last Theo-

rem, Video, BBC (1996).

Simon Singh, L’ult imo teorema di Fermat: il

racconto di scienza del decennio, in M. Em-

mer, ed., Matematica e Cultura 2, Springer- Verlag Italia, Milano (1999), p. 40-43.

John Lynch, Alcune riflessioni sulla con-

struzione del film, in M. Emmer and M. Man-

aresi, eds., Matematica, arte, tecnologia,

cinema, Springer-Verlag Italia, Milano

(2002), p. 266-267; English edition to ap-

pear.

Dipartimento di Matematica

Universita Ca’ Foscari

Dorsoduro 3825/E, Ca’ Dolfin

30123 Venice, Italy

e-mail: emmermve@unive.it

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78 THE MATHEMATICAL INTELLIGENCER

Applications Of Derivatives

Optimization problems all look vastly different, but the steps to  solve them are actually quite similar. Explain the steps to solve an  Optimization problem, as well as the purpose behind each of those steps. 1-2 pages long with work cited page.

Literature Review Math ED + Literature Review -Summaries

Literature review -Summaries

STEM Teaching at the Elementary school

RQ: How do we prepare preservice teachers to engage in STEM teaching at the elementary?

Science, Technology, Engineering and Mathematics (STEM)

Author/Title Purpose Sample Results Implications
1

Supporting Elementary Pre-Service Teachers to Teach STEM Through Place-Based Teaching and Learning Experiences

Anne E. Adams

University of Idaho

Brant G. Miller

University of Idaho

Melissa Saul

University of Hawai’i – West O’ahu

Jerine Pegg

University of Alberta

Many elementary teachers feel less knowledgeable

about STEM content and less comfortable teaching STEM than other subjects.

Data were collected on elementary preservice teachers’ perceptions of their experiences as they participated in, planned, and enacted

integrated place-based STEM education lessons.

Findings indicate that experiences with STEM

learning and teaching through integrated, place-based activities had a positive impact on preservice teachers’ understanding of place-based approaches, their perceived ability, and projected

intent to design and implement place-based STEM learning activities.

In order to prepare teachers to meet the challenges of a rapidly changing human

landscape, we, as teacher educators, need to provide authentic, and meaningful experiences that

are situated in place, build community, and show pre-service teachers that they have resources

and partners eager to support the educational mission of community schools beyond the walls of

their school buildings.

Our findings suggest that within teacher education courses, methods instructors can

and should employ place-based pedagogy as a way to increase knowledge of STEM related

elementary teachers’ comfort with learning and teaching STEM content; and prepare pre-service

teachers for effectively using local spaces for inquiry instruction

2

Examining Elementary Pre-service Teachers’ Science, Technology,

Engineering, and Mathematics (STEM) Teaching Intention

Güney Hacıömeroğlu1

3

Taiwanese Preservice Teachers’ Science, Technology, Engineering, and Mathematics Teaching Intention

Kuen-Yi Lin

P. John Williams

International Journal of Science and Mathematics Education

August 2016,

4

Makerspace and reflective practice: Advancing pre-service teachers in STEM education

Blackley, Susan; Sheffield, Rachel; Maynard, Nicoleta; Koul, Rekha; Walker, Rebecca

Australian Journal of Teacher Education (Online)

(2017)

5

Using a Makerspace approach to engage Indonesian primary students with STEM

Susan Blackley Curtin University, Australia Yuli

Rahmawati, Ella Fitriani Universitas Negeri Jakarta, Indonesia

Rachel Sheffield and Rekha Koul Curtin University, Australia

Issues in Educational Research, 2018

6

Robotics to promote elementary education pre-service teachers’ STEM engagement, learning, and teaching

Kim, C., Kim, D., Yuan, J., Hill, R. B., Doshi, P., & Thai, C. N. (2015). Computers & Education, 91, 14-31.

7

An Integrated Model for STEM Teacher

Preparation: The Value of a Teaching Cooperative

Educational Experience

Ellen W. Eckman

Marquette University

Mary Allison Williams

Marquette University

This study aims at examining elementary pre-service teachers’ integrative STEM intentions.

This study applies the theory of planned behavior as a basis for exploring the impact of knowledge, values, subjective norms, perceived behavioral controls, and attitudes on the behavioral intention toward (STEM) education among Taiwanese preservice science teachers.

The research presented in this paper describes a type of Makerspace that is defined by its purpose: to improve the confidence and ability of primary education students in STEM education.

Examines the learning experiences integrated STEM project

purpose of helping teachers learn how to design and implement (STEM) lessons using robotics.

The purpose of this article is to evaluate an intensive, integrated model for teacher

preparation, specifically, a preservice STEM teacher education model

Quantitative study, data were gathered from 401 elementary pre-service teachers who were enrolled

in two public universities.

Questionnaires (N = 139) collected information on the behavioral intention of preservice science teachers engaging in STEM education. Data were analyzed using descriptive statistics, path analysis, and analysis of variance.

a large set of qualitative data was collected, this paper reports on the progress and reflections of the teacher education students, and shares insights into their personal learning and development as teachers.

9 female teacher education students and 71 schoolgirls in Years 5 and 6

Examines the learning experiences of 291 Year 5 and 6 Indonesian primary school students, across four schools in North Jakarta, who participated in an integrated STEM project

Data were collected from surveys, classroom observations, interviews, and lesson plans. Both quantitative and qualitative data analyses indicated that pre-service teachers engaged in robotics activities actively and mindfully.

STEM preservice teachers

participated in a cooperative teaching experience which placed them at the school site

for their university course work and field placements, thus ensuring a more seamless

connection between theory and practice.

Findings of this study showed that

there was no significant difference between pre-service teachers’ scores on knowledge, attitude,

value perceived behavioral control, and behavioral intention regarding gender.

However, there was

a significant difference between pre-service teachers’ scores on subjective norm regarding gender.

Results revealed that, in terms of direct effects, higher perceived behavioral control and subjective norms were associated with stronger STEM teaching intention. More positive attitude and greater knowledge were indirectly associated with higher subjective norms and perceived behavioral control, which resulted in stronger STEM teaching intention

The results indicated that a Makerspace approach was very effective in engaging students in the STEM space, and students were also challenged to work collaboratively in groups mentored by pre-service teachers.

analyses indicated that pre-service teachers engaged in robotics activities actively and mindfully. Their STEM engagement improved overall. Their emotional engagement (e.g., interest, enjoyment) in STEM significantly improved and in turn influenced their behavioral and cognitive engagement in STEM.

The findings from this comparative study of

the STEM preservice students in the teaching co-op and STEM preservice teachers in a

traditional preparation model indicates that the STEM preservice teachers in the teaching

cooperative model – – were more confident about their teaching skills, more comfortable with

their content knowledge, and prepared to work effectively with high-needs students

Using STEM activities in classroom environment would allow opportunities for pre-service teachers to become effective elementary teachers. Throughout the longitudinal

research studies,

observation notes can be taken, and interviews would be conducted to investigate

elementary pre-service teachers’ STEM teaching intentions deeply.

The most important factors appeared to include developing preservice teachers’ (a) positive appreciation regarding STEM outcomes and teaching and (b) competency in resolving difficulties related to STEM teaching

STEM teacher education programs must stress developing preservice teachers’ behavioral intention toward STEM teaching and related socioaffective factors and not just emphasize developing their knowledge in science, technology, engineering, and mathematics

With the application of STEM knowledge and skills, we also posit that the Makerspace approach is effective in the acquisition and demonstration of 21st century skills: problem-solving, critical and creative thinking, collaboration, and communication.

suggest that robotics can be used as a technology in activities designed to enhance teachers’ STEM engagement and teaching through improved attitudes toward STEM

Literature review

Summaries

STEM

Teaching at the Elementary school

RQ:

How do we prepare preservice teachers to engage in STEM

teaching at the elementary?

Science, Technology, Engineering and Mathematics (STEM)

Author/Title

Purpose

Sample

Results

Implications

1

Ordinary Differential Equation

In this mini project,you are expected to provide the following three required components:

(1)Problem Statement

(2) Background Information

(3)Required Work&Solution

The sections containing your problem statement and background information must be typed

using a font size of 12 in“Times New Roman”.Moreover the background information section

must contain“your own words”describing the method used in solving the given problem.

The last section can be hand-written but must be neat and easy to follow.

Discrete Math Problem

Discuss the concept of a fuzzy relation. How are fuzzy relations used?

Writing Assignments Requirements Each writing assignment is worth 20 points

should include the following sections:

Background (3 points): This is a discussion of how thenon-­‐mathematical and mathematical portions of your topic fit together. You might include a historical background of the topic, definitions of terms, the discrete mathematic sideas that are addressed(e.g. induction, logical fallacy, etc.), and some explanation about why these ideas were useful.

Examples (10 points): In most of your writing assignments you are asked to discuss and describe an aspect of discrete mathematics. Give two of three examples or techniques of the topic under discussion. Give generalinformation and also specific examples of the topic.

Bibliography(2 points): List the references you used to complete this report. Just list title and author for any books and articles you used. You should also include a list of people that you consultedor any other form of help that you received. For example, you might obtain some of your information from the internet; in this case, you could include the website.You’ll need at least one bookor articleas a reference, preferably two, and a total of at least two references.

You’ll notice that there are still 5 points unaccounted for. The remaining 5 points are for style: clarity, neatness, flow, design, organization and creativity-­‐-­‐it’s important to be able to communicate your ideas.Note:you don’t haveto put your report in the precise order given above. You may prefer to use the assigned problems to illustrate how the ideas of the subject fit together with the mathematical ideas that you will be using, in which case Background and Exampleswould be interwoven. Just make sure that these aspects appear in your report.

“Complex ANOVA, ANCOVA, Or MANOVA

T

his written assignment is based on the work conducted in the “Complex ANOVA, ANCOVA, or MANOVA” discussion forum. Based on this initial work, feedback received, and additional research, students should submit a basic research proposal that calls for the use of a complex ANOVA, ANCOVA, or MANOVA.

The paper should be APA formatted as a research proposal, and contain approximately 990-1320 words of content. Include a title page, and a reference page that includes any resources utilized.

Please include the following in the research proposal:

1. Introduction (1-2 paragraphs)

  • Present the research question of interest.
  • Explain how the chosen statistical test applies to this research question.
  • Provide the statistical notation and written explanations for the null and alternative hypotheses.

2. Methods (1 paragraph)

  • Participants
  • List how many participants will be selected.
  • Identify who will be the participants and their major demographic characteristics (e.g., sex, age, etc.).
  • Explain how participants will be selected for the study.

3. Procedures (1-2 paragraphs)

  • Identify the variables in the study.
  • Describe each variable’s scale of measurement (nominal, ordinal, interval, or ratio) and characteristics (i.e., discrete vs. continuous, qualitative vs. categorical, etc.).
  • Provide an operational definition for each variable, explaining how the variables will be measured.

4. Results (2-3 paragraphs)

  • Describe the statistical test that will be conducted. Be sure to include why the test was chosen and why it is appropriate for this study. Include in the discussion the necessary assumptions that should be met for the chosen test and how these will be addressed.
  • Identify the information that will be obtained from the results of this test and what will be needed to draw conclusions regarding the hypotheses. Be sure to include a discussion of applicable critical and calculated values, p levels, confidence intervals, effect sizes, post-hoc tests, and/or tables.

5. Discussion (1 paragraph)

  • Identify any expected biases, assumptions, or faults with the proposed study and the use of the identified statistical test.
  • Explain what conclusions can and cannot be made for this study, and using this statistical test.
  • Describe the practical significance or importance of the results.

Differential Equation

Actuarial Science

AS 3429/9429 Long Term Actuarial Math II

Practice Final Exam Solution

1. Solution:

(a) Under the Equivalence Principle

75, 000A40:10 + 25%P + 9%P ä40:10 + 20 + 5ä40:10 + 100A40:10 = P ä40:10

Hence, the gross premium is

P = 75, 100A40:10 + 20 + 5ä40:10

0.91ä40:10 − 0.25 = 6505.2

where from the LTAM standard ultimate life table

A40:10 = 0.61494, ä40:10 = 8.0863.

(b) The loss at issue random variable

Lg0 =75, 100v min(K40+1,10) + 0.25P + 20 + 5ämin(K40+1,10) − 0.91P ämin(K40+1,10)

=75, 100vmin(K40+1,10) + 0.25P + 20 + (5− 0.91P )1− v min(K40+1,10)

d

=(75, 100− 5− 0.91P d

)vmin(K40+1,10) + 0.25P + 20 + 5− 0.91P

d

The probability of making a loss

P(Lg0 > 0) =P((75, 100− 5− 0.91P

d )vmin(K40+1,10) + 0.25P + 20 +

5− 0.91P d

> 0)

=P(vmin(K40+1,10) > − 0.25P + 20 + 5−0.91Pd

75, 100− 5−0.91Pd )

=P(e−δmin(K40+1,10) > 0.61494)

=P(min(K40 + 1, 10) < − ln(0.61494)

δ )

=P(min(K40 + 1, 10) < 9.9658) =P(K40 + 1 < 9.9658)

=P(K40 < 8.9658) = 9q40 = 1− 9p40 = 1− l49 l40

= 0.006578

(c) • Gross premium policy value

5V g =75, 100A45:5 + 5ä45:5 − 0.91P ä45:5

=75, 100× 0.78387− (0.91× 6505.2− 5)× 4.53862 = 32, 023.9

where

ä45:5 = ä45 − 5E45 ä55 = 17.8162− 0.77991× 17.0245 = 4.53862

and A45:5 = 1− dä45:5 = 1− 0.04762× 4.53862 = 0.78387.

1

• Net premium policy value

5V n =75, 000A45:5 − Pn ä45:5

=75, 000× 0.78387− 5703.5356× 4.53862 = 32, 904.1

where

Pn = 75, 000A40:10

ä40:10 =

75, 000× 0.61494 8.0863

= 5703.5356

• FPT policy value

5V FPT =75, 000A45:5 − PFPT ä45:5

=75, 000× 0.78387− 6502.95× 4.53862 = 29275.8

where

ä41:9 = ä41 − 9E41 ä50 = 7.44458, A41:9 = 1− dä41:9 = 0.64549

PFPT = 75, 000A41:9

ä41:9 = 6502.95

*Numbers could be slightly different due to rounding.

2. Solution:

The loss at issue random variable for the portfolio is

L =

n∑ i=1

L0,i,

with mean E[L] = nE[L0,1] and variance V ar[L] = nV ar[L0,1], and n = 10, 000. For an individual policy,

L0,1 =150, 000v K50+1 + 20%P − 0.95P äK50+1

=150, 000vK50+1 + 0.2P − 0.95P 1− v K50+1

d

=

( 150, 000 +

0.95P

d

) vK50+1 + 0.2P − 0.95P

d

which gives E[L0,1] = 150, 000A50 + 0.2P − 0.95P ä50

and

V ar[L0,1] =

( 150, 000 +

0.95P

d

)2 (2A50 −A502)

Using normal approximation

P(L < 0) = P( L− E[L]√ V ar[L]

< 0− E[L]√ V ar[L]

) = 0.95 =⇒ − E[L]√ V ar[L]

= 1.645

i.e.,

− 10, 000(150, 000A50 + 0.2P − 0.95P ä50)( 150, 000 + 0.95Pd

)√ 10, 000(2A50 −A502)

= 1.645

which solves P = 1801.39.

2

3. Solution:

(a) For j = 0, …, 14,

Lj =

 1000vKx+j+1 − 3πäKx+j+1 , Kx+j = 0, 1, …, 14− j 1000vKx+j+1 − 3πä15−j − 2πv15−j äKx+j−(15−j)+1 , Kx+j = 15− j, …, 24− j 500vKx+j+1 − 3πä15−j − 2πv15−j äKx+j−(15−j)+1 , Kx+j = 25− j, …, 29− j 500vKx+j+1 − 3πä15−j − 2πv15−j ä15 , Kx+j = 30− j, 31− j, …

For j = 15, …, 24,

Lj =

 1000vKx+j+1 − 2πäKx+j+1 , Kx+j = 0, 1, …, 24− j 500vKx+j+1 − 2πäKx+j+1 , Kx+j = 25− j, …, 29− j 500vKx+j+1 − 2πä15−j , Kx+j = 30− j, 31− j, …

For j = 25, …, 29

Lj =

{ 500vKx+j+1 − 2πäKx+j+1 , Kx+j = j, j + 1, …, 29 500vKx+j+1 − 2πä15−j , Kx+j = 30− j, 31− j, …

For j = 30, 31, … Lj = 500v

Kx+j+1, Kx+j = 0, 1, …

(b) Under the equivalence principle,

500A40 + 500A 1 40:25 = 2πä40:30 + πä40:15 ,

which implies that

π = 500A40 + 500A

1 40:25

2ä40:30 + ä40:15 =

500 (0.202) + 500 (0.114)

2 (13.414) + 10.005 = 4.2896.

Policy values at times 10, 20 and 40 are respectively given by

10V = 500A50 + 500A 1 50:15 −

( 2πä50:20 + πä50:5

) = 500 (0.316) + 500 (0.149)− 4.2896 (2 (10.806) + 4.386) = 120.98,

20V = 500A60 + 500A 1 60:5 − 2πä60:10

= 500 (0.458) + 500 (0.112)− 2 (4.2896) (6.929) = 225.55,

and 40V = 500A80 = 500 (0.744) = 372.

3

4. Solution:

(a) The policy value at time 10 is

10V = 100, 200A70:10 = 100, 200× 0.63576 = 63703.152

Using the recursion formula

10V × (1 + 5%) = q70 × 100, 200 + p70 × 11V

which gives the policy value at time 11

11V = 66537.785

Hence the total profit or gain for this year (t = 10) is

nt(tV + Pt − eAt )× (1 + iAt )− [ dt(St+1 + E

A t+1) + nt+1 × t+1V

] =200(63703.152 + 0)(1 + 4%)− [2(100, 000 + 250) + 198(66537.785)] =− 124725.8 (loss)

(b) Expense (interest rate and mortality as expected)

200(10V )× (1 + 5%)− [200q70(100, 000 + 250) + 200p70 × 11V ] − {200(10V )× (1 + 5%)− [200q70(100, 000 + 200) + 200p70 × 11V ]}

=− 200q70 × 50 =− 104.13 (loss)

Interest (expenses as actual and mortality as expected)

200(10V )× (1 + 4%)− [200q70(100, 000 + 250) + 200p70 × 11V ] − {200(10V )× (1 + 5%)− [200q70(100, 000 + 250) + 200p70 × 11V ]}

=200(10V )(4%− 5%) =− 127406.304 (loss)

Mortality (expenses and interest as actual)

200(10V )× (1 + 4%)− [2(100, 000 + 250) + 198× 11V ] − {200(10V )× (1 + 4%)− [200q70(100, 000 + 250) + 200p70 × 11V ]}

=2784.629 (gain)

we can easily check −104.13− 127406.304 + 2784.629 = −124725.8.

5. Solution:

(a) From the CBD model, we know that logit(q(65, 2018)) follows a normal distribution with mean

µ = K (1) 2017 +c

(1) +(K (2) 2017 +c

(2))(x− x̄) = −3.2−0.02+(0.01+0.0006)(65−70) = −3.273

and variance

σ2 = σ2k1+(x−x̄) 2σ2k2+2(x−x̄)ρσk1σk2 = 0.03

2+(65−70)20.0052+2(65−70)×0.2×0.03×0.005 = 0.0352

hence the standard deviation is 0.035.

4

(b) Let X = logit(q(65, 2018)) and Y = p(65, 2018). Hence

X = log 1− Y Y

or Y = 1 1+eX

= f(X), we see that f(x) is a decreasing function of x, since

f ′(x) = − e X

(1 + eX)2 < 0

Therefore, the median of Y will be related to the median of X

πY0.5 = f(π X 0.5) =

1

1 + eπ X 0.5

= 1

1 + e−3.273 = 0.96349

and the 95th percentile of Y will be related to the 5th percentile of X

πY0.95 = f(π X 0.05) =

1

1 + eπ X 0.05

= 1

1 + e−3.330575 = 0.96546

Note that since X ∼ N(µ, σ2) the median of X πX0.5 = µ = −3.273 and the 5th percentile of X πX0.05 = µ+ σΦ

−1(0.05) = −3.273− 1.645× 0.035 = −3.330575.

6. Solution:

(a) Summarize the data

i yi si ri si/ri 1 1 1 20 1/20 2 2 1 19 1/19 3 4 2 17 2/17 4 5 1 13 1/13 5 8 3 11 3/11 6 9 4 8 4/8 7 12 2 3 2/3

ymax = 15

(b) Kaplan–Meier Estimate of survival function is

S20(y) =



1, 0 ≤ y < 1, 1− 1/20 = 0.950, 1 ≤ y < 2, 0.950(1− 1/19) = 0.900, 2 ≤ y < 4, 0.900(1− 2/17) = 0.794, 4 ≤ y < 5, 0.794(1− 1/13) = 0.733, 5 ≤ y < 8, 0.733(1− 3/11) = 0.533, 8 ≤ y < 9, 0.533(1− 4/8) = 0.267, 9 ≤ y < 12, 0.267(1− 2/3) = 0.089, 12 ≤ y < 15.

Tail correction:

• Efron’s method: S20(y) = 0 for y ≥ 15.

5

• Klein and Moeschberger’s method: S20(y) = 0.089 for 15 ≤ y < 22, and S20(y) = 0 for y ≥ 22. • Brown, Hollander and Korwar’s exponential tail correction: S20(y) = (0.089)

y/15 for y ≥ 15. Withe the change in value, we have ymax = y7 = 12 and S20(y) = 0 for y ≥ 12.

(c) Nelson–Aaloen estimates for cumulative hazard function and the corresponding esti- mated survival function are

Ĥ(y) =



0, 0 ≤ y < 1, 1/20 = 0.050, 1 ≤ y < 2, 0.050 + 1/19 = 0.103, 2 ≤ y < 4, 0.103 + 2/17 = 0.220, 4 ≤ y < 5, 0.220 + 1/13 = 0.297, 5 ≤ y < 8, 0.297 + 3/11 = 0.570, 8 ≤ y < 9, 0.570 + 4/8 = 1.070, 9 ≤ y < 12, 1.070 + 2/3 = 1.737, 12 ≤ y < 15.

Ŝ(y) =



e0 = 1, 0 ≤ y < 1, e−0.050 = 0.951, 1 ≤ y < 2, e−0.103 = 0.902, 2 ≤ y < 4, e−0.220 = 0.803, 4 ≤ y < 5, e−0.297 = 0.743, 5 ≤ y < 8, e−0.570 = 0.566, 8 ≤ y < 9, e−1.070 = 0.343, 9 ≤ y < 12, e−1.737 = 0.176, 12 ≤ y < 15.

Tail correction:

• Efron’s method: S20(y) = 0 for y ≥ 15. • Klein and Moeschberger’s method: S20(y) = 0.176 for 15 ≤ y < 22, and S20(y) = 0 for y ≥ 22. • Brown, Hollander and Korwar’s exponential tail correction: S20(y) = (0.176)

y/15 for y ≥ 15. (d) For the Kaplan–Meier estimator

V̂ ar[S20(2)] = [S20(2)] 2

[ 1

20(19) +

1

19(18)

] = 0.9002 × 0.00556 = 0.0045,

and for the Nelson–Aalen estimator,

V̂ ar[Ŝ(2)] = [Ŝ(2)]2 [

1(19)

203 +

1(18)

193

] = 0.9022 × 0.00500 = 0.004068.

(e) The 95% log-transform confidence interval for S(2) based on the Kaplan–Meier estima- tor is (

Ŝ(2) 1/U

, Ŝ(2) U )

= (0.65604, 0.97401)

with U = exp

( Φ−1( 1+α2 )×

√ V̂ ar[S(y)]

Ŝ(y)×ln Ŝ(y)

) = exp

( 1.96×

√ 0.0045

0.900×ln 0.900

) = 0.24994.

(f) The 95% log-transform confidence interval for S(2) based on the Nelson–Aalen estimator is (

exp ( −Ĥ(y)U

) , exp

( −Ĥ(y)/U

)) = (0.6733, 0.9735)

with U = exp

( Φ−1( 1+α2 )×

√ V̂ ar[H(y)]

Ĥ(y)

) = exp

( 1.96×

√ 0.00500

0.103

) = 3.84.

6