Assignment 3 ANSWER KEY Assigned February 11, Due February 18 PubH 8462 Advanced Survival Analysis Instructor: David M. Vock

1. Verify that the optional variation process is an unbiased estimators of the variance. Let m(t) be a martingale with respect to the filtration Ft . The optional variation process is: X [m](t) = = [4m(s)]2 s≤t

=

X {m(s) − m(s− )}2

=

X

s≤t

m2 (s) − 2m(s)m(s− ) + m2 (s− )

s≤t

note E{[m](t)}

X = E{ m2 (s) − 2m(s)m(s− ) + m2 (s− )} s≤t

=

X

=

X

=

X

=

X

E[E{m2 (s) − 2m(s)m(s− ) + m2 (s− )|Fs− }]

s≤t

E{m2 (s)} − E[2E{m(s)|Fs− }m(s− )] + E{m2 (s− )}

s≤t

E{m2 (s)} − 2E[m2 (s− )] + E{m2 (s− )}

s≤t

E{m2 (s)} − E{m2 (s− )}

s≤t

Telescoping sum if we think of the above as limit

Pn

i=1

where n → ∞ and m(s0 ) = 0

2

Thus, E{m (t)} = var(m(t)). Alternatively if we think of m(t) as a counting process martingale in continuous time then X [m](t) = [4m(s)]2 s≤t

X = [4N (s)]2 s≤t

= N (t)

1

Note that E{N (t)}

= = = =

Z t E{ dN (u)} 0 Z t E{dN (u)} 0 Z t E[E{dN (u)|Fu− }] 0 Z t E{yi (u)λi (u)du} 0

= E[< m > (t)] = var{m(t)} 2. Using any result given in lecture, prove that the Doob-Meyer decomposition must be unique (as). Let m(t) ¯ be a submargingle. Assume that there exists martingales m1 (t) and m2 (t) wrt Ft− and predictable process C1 (t) and C2 (t) wrt Ft− so that C1 (0) = 0 and C2 (0) = 0 and m(t) ¯

=

m1 (t) + C1 (t) = m2 (t) + C2 (t)

So m1 (t) − m2 (t)

=

C2 (t) − C1 (t)

Now the difference of two martingales wrt the same filtration is a martingale. Similarly the difference of 2 predictable processes is a predictable process. Thus m1 (t)−m2 (t) is a martingale and C2 (t)−C1 (t) is a predictable process. Thus X(t) = m1 (t) − m2 (t) = C2 (t) − C1 (t) is both a predictable process and a martingale. We showed in class that if X(t) is both a martingale and predictable process it must be a constant a.s. Thus m2 (t) = m1 (t) + k C2 (t) = C1 (t) − k for some constant k. If C2 (t) = 0 and C1 (t) = 0 then k=0. Thus the decomposition is unique (a.s.). 3. K&P 5.2: Note that for part (b) we are looking for the unconditional mean. Also for part (c) plot five of the realizations of N· (t) and the corresponding martingale before plotting the average of the 30 realizations. Also, part(c) does not specify a value for β so please use β = 1. (a)Intensity function for counting process N. (t)

= E{dN. (t)|Ft− } n X = E{ dN. (t)|Ft− } i=1

= =

n X i=1 n X i=1

2

E{dN. (t)|Ft− } yi (u) exp{xi (t)β}α

(b) Note that over the interval where xi (t) is constant, the counting process is a homogeneous poisson process. That is if xt (t) is constant between t1 and t2 then E{Ni (t2 ) − Ni (t1 )} = exp(β)α(t2 − t1 ). For any t ≤ τi E{Ni (t)|Ai } = note E{I(t > Ai )}

=

E{I(t > Ai )|Ai }

=

E[E{Ni (t)|Ai }]

I(t > Ai ){exp(β)αAi + α(t − Ai )} + I(t ≤ Ai ) exp(β)αt τ −t t , E{I(t ≤ Ai )} = τ τ Z t 1 Ai dAi τ 0

=

Ai 2 t t2 |0 = 2τ 2τ

=

{exp(β) − α}

= =

t2 α τ − t t2 + + exp(β)αt 2τ 2τ τ t2 exp(β)α t2 α t2 α τ exp(β)αt t2 exp(β)α − + + − 2τ 2τ τ τ τ 2 2 t α t exp(β)α − + t exp(β)α 2τ 2τ

for t > τi , E{Ni (t)} = E{Ni (τi )} thus E{N. (t)} =

n X t∗2 α i=1





t∗2 i exp(β)α + t∗i exp(β)α 2τ

where t∗i = min(t, τi ) 4. K&P 5.3

Z < u > (t)

t

var{du(t)|Ft− }

= 0

Z

t

E{du(t)du(t)T |Ft− }

= 0

Z

t

=

n n X X E[{ Gi (u)dmi (u)}{ Gi (u)dmi (u)}T |Ft− ]

0

= =

i=1

Z tX n 0 i=1 Z tX n

i=1

E[{Gi (u)dmi (u)}{Gi (u)dmi (u)}T |Ft− ], m1 , ...mn are orthogonal Gi (u)Gi (u)T E[{dmi (u)}2 |Ft− ], Gi (u) is predictable

0 i=1

= =

Z tX n 0 i=1 n XZ t i=1

Gi (u)Gi (u)T d < mi > (u) Gi (u)Gi (u)T d < mi > (u)

0

5. Let M (t), t ∈ [0, τ ] be a martingale with respect to the filtration {Ft , t ∈ [0, τ ]}. Suppose M (0) = 0 (wp1). For 0 < s < t ≤ u < v < τ show that cov{M (t) − M (s), M (v) − M (u)} = 0. Because M (t) has zero mean for all t, cov(M (t)−M (s), M (v)−M (u))=E[(M (t)−M (s))(M (v)−M (u))]. Now condition on Fu . 3

E{E[(M (t) − M (s))(M (v) − M (u))|Fu ]} = E{[M (t) − M (s)]E[M (v) − M (u)|Fu ]} But E[(m(v) − M (u))|Fu ] = 0 .s. by the martingale property. By taking expectations, we see the covariance is 0. 6. Let Y and X1 , X2 , . . . be random variables defined on (Ω, F, P ) with EY 2 < ∞. Let Fn = σ(X1 , X2 , . . . , Xn ). Let Yn = E(Y |Fn ). Prove that {Yn } is a margingale sequence with respect to the filtration {Fn , n ≥ 1}. Since this is a discrete time martingale we do not need to worry about the cadlag property. For each n, Yn is adapted to Fn by its definition. The key martingale property is easy to establish. E(Yn |Fn−1 )

=

E(Y |Fn |Fn−1 ), definition of Yn

= E(Y |Fn−1 ), Tower property of conditional expectation = Yn−1 , definition of Yn−1

4

Answer Key

E[E1m2(s) - 2m(s)m(s−) + m2(s−)|Fs− l]. = ∑ ... n i=1 where n → с and m(s0)=0. Thus, E1m2(t)l = var(m(t)). Alternatively if we think of m(t) as a counting process ...

112KB Sizes 7 Downloads 445 Views

Recommend Documents

Answer Key
Answer Key: 1. D. 11. A. 21. D. 31. A. 2. E. 12. C. 22. D. 32. B. 3. C. 13. D. 23. E. 33. E. 4. B. 14. A. 24. A. 34. B. 5. D. 15. C. 25. B. 35. C. 6. D. 16. E. 26. C. 36. E. 7.

Answer Key
2. } 3. 23. all real numbers 24. all real numbers. 25. f(x) 5 x 2 200 26. g(x) 5 0.8x. 27. g( f(x)) 5 0.8x 2 160. 28. f(g(x)) 5 0.8x 2 200 29. pay less with discount first.

Answer Key
Jan 7, 2017 - PROVISIONAL ANSWER KEY. Advt. No. : 5/2016-17 - Vistaran Adhikari (Sahakar) (VA(S)-2016-17). QUESTION PAPER SERIES - A.

Final Answer Key
71. D. 96. C. 22. C. 47. C. 72. B. 97. B. 23. D. 48. B. 73. D. 98. D. 24. C. 49. C. 74. D. 99. A. 25. B. 50. A. 75. D. 100. A. QUESTION PAPER SERIES - A. GUJARAT PANCHAYAT SERVICES SELECTION BOARD. FINAL ANSWER KEY. Advt. No. : 4/2016-17 - Vistaran A

answer key -
CENTERS : MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA. IIT-JEE 2014. PRACTICE TEST - 3(MAIN). ANSWER KEY. Que. 1. 2. 3. 4. 5. 6.

Answer Key
of wrongs that others do. 1 Corinthians 13:5 (CeV). Session Five. Love isn't selfish. 1 Corinthians 13:5 (CeV). Session Six. [Love] always protects, always trusts,.

answer key -
ANSWER KEY. 1. C. 31. B. 2. D. 32. B. 3. D. 33. A. 4. C. 34. D. 5. A. 35. B. 6. B. 36. D. 7. D. 37. B. 8. B. 38. A. 9. D. 39. A. 10. C. 40. D. 11. D. 41. B. 12. C. 42. D. 13.

Answer Key Hindi.pdf
Whoops! There was a problem previewing this document. Retrying... Download. Connect more apps... Answer Key Hindi.pdf. Answer Key Hindi.pdf. Open.

answer key copy.indt
MN 7. radius is 3 units 8.Yes, 62 1 82 5 102. 9. 7.2 10. 12 11. 18 12. 180° 13. 20°. 14. 340° 15. 90° 16. 270° 17. 90°. 18. 270° 19. x 5 58 20. SY 5 8. 21. L. M. K.

Answer Key Hindi.pdf
Page 1 of 1. ifj"dkj oYM Z. REET Level-II Exam-2018 (Answer Key). fganh. iz-l a- iz'u mÙkj. 31- or Zuh dh fof/kor~ f'k{kk dk Kku------ (B) mPp izkFkfed Lrj. 32- Hkkf"kd lkexzh gS& (D) e wy fo"k;& oLr q ,oa izR;;k sa. 33- cqySfVu ckMs Z gS& (B) n `';

DFSL Recruitment 2017 Answer Key [email protected] ...
20 D0219 INGLE AMOL VINAYAK OPEN 64.18. 21 D0513 VAIRAL MAHADEV BHIKAJIRAO NT-C 64.14. 22 D0297 KAYASTHA JAYANT KANTILAL OBC 64.07. 23 D0001 DESHMUKH SWAPNIL DADARAO OPEN 64.04. 24 D0269 SANAP GANESH RAGHUNATH NT-D 63.80. 25 D0107 MUNDE RAJNANDINI DH

answer key _10 hindi
At¸m /font>. BsI kvtIm 40. D¯-c-sa-gp-tX− tNmZy-§ fpsS F®w b>18. G.Somasekharan,G.H.S.S.Sadanandapuram,Kottarakkara,hindisabhaktr.blogspot.com.

DFSL Recruitment 2017 Answer Key [email protected] ...
DFSL Recruitment 2017 Answer Key [email protected]. DFSL Recruitment 2017 Answer Key [email protected]. Open. Extract. Open with.

TNPSC Group 4 Answer Key 2018_English_Language.pdf ...
Adanya jaminan Undang-Un- dang terkait besaran dana pen- didikan membuat pembayaran .... TNPSC Group 4 Answer Key 2018_English_Language.pdf.

iift-answer-key-TIME.pdf
Page 1 of 1. iift-answer-key-TIME.pdf. iift-answer-key-TIME.pdf. Open. Extract. Open with. Sign In. Main menu. Displaying iift-answer-key-TIME.pdf. Page 1 of 1.

KCET-2016-ANSWER-KEY-BIOLOGY.pdf
There was a problem previewing this document. Retrying... Download. Connect more apps... Try one of the apps below to open or edit this item.

Answer key Urdu10.pdf
Page 3 of 5. Answer key Urdu10.pdf. Answer key Urdu10.pdf. Open. Extract. Open with. Sign In. Main menu. Displaying Answer key Urdu10.pdf. Page 1 of 5.

TNPSC Group 4 Answer Key 2018_Tamil_Language_question ...
https://docs.google.com/file/d/0Bx6Qe6ibTFN1bHF2SWEyT0RLZFU/. - Star warsavicomplete. Page 1. Whoops! There was a problem loading this page. Retrying... Whoops! There was a problem loading this page. Retrying... TNPSC Group 4 Answer Key 2018_Tamil_La

xat 2017 answer key ed.pdf
Page. 1. /. 2. Loading… Page 1 of 2. OWNER'S. GUIDE. NV751. MANUEL DU. PROPRIÉTAIRE. NV751. MANUAL DEL. USUARIO. NV751. www.PoweredLiftAway.com 800.798.7398. ®. Page 1 of 2. Page 2 of 2. Main menu. Displaying xat 2017 answer key ed.pdf.

vistaran adhikaree answer key wsatermark.pdf
a PRIMARY AGRICULTURAL COOPERATIVE SOCIETY b PRIMARY AGRICULTURE COOPERATIVE SOCIETY. c PRIME AGRICULTURE COOPERATIVE ...

ANSWER KEY 2018 ANNUAL.pdf
]mTw 11. t]Pv 49,50. ... Hcn-°¬am{Xw kzoI-cn-°m-hp-∂Xpw H∂n-e-[nIw {]mhiyw kzoI-cn-°m-hp-∂-hbpw ̨-Hmtcm IqZm-i- ..... ANSWER KEY 2018 ANNUAL.pdf.

Mauryan Empire Answer Key Handout.pdf
Page 1. Whoops! There was a problem loading more pages. Retrying... Mauryan Empire Answer Key Handout.pdf. Mauryan Empire Answer Key Handout.pdf.

Answer Key 4.4 Practice B.pdf
Loading… Page 1. Whoops! There was a problem loading more pages. Retrying... Answer Key 4.4 Practice B.pdf. Answer Key 4.4 Practice B.pdf. Open. Extract.