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AP/ADMS4504 final exam solutions Winter 2010 AP/ADMS4504 Fixed Income Securities and Risk Management Final Exam Solutions Winter 2010 Type A Exam Numerical Questions (2 points each) Key maturity Portfolio I Portfolio II Portfolio III 3-month 0.06 0.19 0.10 1-year 0.11 0.22 0.34 2-year 0.23 0.31 0.56 5-year 1.48 0.25 0.60 7-year 0.45 0.24 0.62 10-year 0.33 0.32 0.64 15-year 0.67 2.12 0.72 20-year 0.50 0.44 0.66 25-year 1.52 0.30 0.53 30-year 0.36 0.12 0.46 1. (Q. 1 in B) The above are the key rate durations for three portfolios of Treasury securities. One of these three portfolios is of ladder type; one is of barbell type; and another one is of bullet type. Consider the following nonparallel shift in the Treasury yield curve: the 5-year rate decreases by 35 basis points (bps), the 15- year rate increases by 20 bps, and the 25-year rate decreases by 40 bps. The rates for all other key maturities experience no change. What is the approximate change in the portfolio value for the ladder type portfolio? A) A 0.27% decrease. B) A 0.27% increase. C) A 0.28% decrease. D) A 0.28% increase. E) A 0.29% increase. Answer D Portfolio I is of the barbell type; portfolio II is of the bullet type; and portfolio III is of the ladder type. For portfolio III, the approximate change in the portfolio value = [-(0.60 × (-0.35%))] + [-(0.72 × (0.20%))] + [-(0.53 × (-0.40%))] = 0.28% increase. 1

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AP/ADMS4504 final exam solutions Winter 2010 2. (Q. 5 in B) Use the following daily yield observations to calculate a daily standard deviation of yields. The daily yield changes are assumed to be continuously compounded. Day Yield (%) 0 6.8112 1 6.7881 2 6.8203 3 6.8066 4 6.8122 5 6.8216 A) 0.3166% B) 0.3206% C) 0.3126% D) 0.3246% E) 0.3086% Answer A %. 3166 . 0 1002 . 0 1002 . 0 ) 1 5 ( ) 0305 . 0 1379 . 0 ( ) 0305 . 0 0822 . 0 ( ) 1 5 ( ) 0305 . 0 2011 . 0 ( ) 0305 . 0 4732 . 0 ( ) 0305 . 0 3397 . 0 ( . 0305 . 0 X . 1379 . 0 )] % 8122 . 6 % 8216 . 6 [ln( 100 X , 0822 . 0 )] % 8066 . 6 % 8122 . 6 [ln( 100 X , 2011 . 0 )] % 8203 . 6 % 8066 . 6 [ln( 100 X , 4732 . 0 )] % 7881 . 6 % 8203 . 6 [ln( 100 X , 3397 . 0 )] % 8112 . 6 % 7881 . 6 [ln( 100 X 2 2 2 2 2 5 4 3 2 1 = = = + + + + = = = = = = = = = = = = deviation standard squared, s percentage Variance and 3. (Q. 6 in B) Use the daily yields in the following table to construct a 3-day moving average equal weight volatility (i.e., standard deviation) forecast for Day 5, assuming that the expected value of the daily change in yield is zero. Day Yield (%) 0 3.750 1 3.752 2 3.749 3 3.746 2
AP/ADMS4504 final exam solutions Winter 2010 4 3.744 5 3.748 A) 0.0962% B) 0.0985% C) 0.1016% D) 0.1020% E) 0.1026% Answer C 4. (Q. 2 in B) Assume normally distributed yield changes. The annual standard deviation of yields is 22%. The current level of yield is 5.6%. Which one of the following is the most likely range for future yields with a probability of 99.7%? A) [1.842%; 8.969%]

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