W13Tute 1001 - 03 with probability 0 . 4 with probability 0...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Examples 1. Jade is 18 years old and has current salary of $20000. The continuous compounding salary growth rates for each year are independent and identically distributed with distribution δ i = ( 0 . 05 with probability 0 . 4 0 . 02 with probability 0 . 6 . 1.1 Calculate E [ S 19 ] 1.2 Explain why the growth rate can be expressed as δ i = 0 . 02 + 0 . 03 X , where X Binomial(1 , 0 . 4) 1.3 Hence, or otherwise, show that E [ S 18+ t ] = 20000 e 0 . 02 t (0 . 4 e 0 . 03 + 0 . 6) t You may use the fact that if X and Y are independent random variables, then E [ f ( X ) g ( Y )] = E [ f ( X )] E [ g ( Y )] for arbitrary functions f and g . 1.4 Contributions of 1% of Jade’s salary are made into the fund each year. The effective rate of interest is 10% p.a. and t ( ap ) 18 = e - 0 . 02 t . Calculate the expected present value of the next 10 contributions, if the first contribution is made in one year’s time. UNSW Week 13 ACTL1001 Tutorial
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Examples 1.1, 1.2 E [ S 19 ] = 20000 e 0 . 05 × 0 . 4 + 20000 e 0 . 02 × 0 . 6 = 20652 . 58485 Since X is Binomial(1, 0.4) it can be expressed as X = ( 1 with probability 0 . 4 0 with probability 0 . 6 Hence 0 . 03 X = ( 0 .
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 03 with probability 0 . 4 with probability 0 . 6 and . 02 + 0 . 03 X = ( . 05 with probability 0 . 4 . 02 with probability 0 . 6 UNSW Week 13 ACTL1001 Tutorial Examples 1.3 E [ S 18+ t ] = E [ S 18 e 1 e 2 . . . e t ] = 20000 E [ e 1 ] E [ e 2 ] . . . E [ e t ] due to independence = 20000( E [ e 1 ]) t since the deltas are identically distributed = 20000(0 . 4 e . 05 + 0 . 6 e . 02 ) t = 20000 e . 02 t (0 . 4 e . 03 + 0 . 6) t UNSW Week 13 ACTL1001 Tutorial Examples 1.4 We want . 01 10 X i =1 E [ S 18+ t ] t ( ap ) 18 1 + i t = 0 . 01 10 X i =1 20000 e . 02 t (0 . 4 e . 03 + 0 . 6) t e-. 02 t (1 + i ) t = 200 10 X i =1 . 4 e . 03 + 0 . 6 1 . 1 t = 200 10-1 -1 = 1302 . 0336 where = . 4 e . 03 + 0 . 6 1 . 1 UNSW Week 13 ACTL1001 Tutorial...
View Full Document

This note was uploaded on 06/12/2011 for the course ASB 1001,2522, taught by Professor Nicole during the One '09 term at University of New South Wales.

Page1 / 4

W13Tute 1001 - 03 with probability 0 . 4 with probability 0...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online