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The Arbitrage-Free Valuation Framework Level 2 Q bank with Ans
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Set 1 Questions 1. The arbitrage opportunity which is based on the idea that the value of the whole should equal the sum of the parts is best known as: A. dominance. B. value additivity. C. law of one price. 2. The arbitrage-free value of option-free bonds is the: A. sum of present ...

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  • January 30, 2025
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  • 2024/2025
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The Arbitrage-Free Valuation Framework Q Bank


Set 1 Questions

1. The arbitrage opportunity which is based on the idea that the value of the whole should equal
the sum of the parts is best known as: A. dominance.
B. value additivity.
C. law of one price.

2. The arbitrage-free value of option-free bonds is the:
A. sum of present values of the future values using par rates.
B. sum of present values of the expected future values using the benchmark spot rates.
C. sum of the future values of the bond based on yield to maturity.

3. The yield for a 3.5% coupon 5-year annual pay bond in Karachi (Bond X) is 2.8%. The same
bond sells for PKR 101.98 in Lahore. Is there an arbitrage opportunity and if so, how can it
be exploited?
A. There is no arbitrage opportunity.
B. There is an arbitrage opportunity which can be exploited by buying the bond in Karachi
and selling in Lahore.
C. There is an arbitrage opportunity which can be exploited by buying the bond in Lahore
and selling in Karachi.

The ffollowing finformation frelates fto fquestions f4 f- f6. f
f
Benchmark fPar fCurve f
Maturity f(Years) f Par fRate f Bond fPrice f
1f 1.00% f 100 f
2f 2.00% f 100 f
3f 3.00% f 100 f
Bond fA fis f3-year f4% fcoupon fannual-pay fbond. fIt fhas fthe fsame frisk fand fliquidity fas fthe
fbenchmark fand fsells ffor f$102.8286 ftoday fto fyield f3%. f
f
4. Calculate fthe fone-year fspot frates ffrom fthe fgiven fterm fstructure? fThe fspot frates ffor feach
fyear’s fcash fflow fare: f
A. 1.00%, f2.00%, f3.00%. f
B. 2.10%, f3.20%, f4.50%. f
C. 1.00%, f2.01%, f3.04% f
f
5. Which fof fthe ffollowing fstatements fis fmost flikely fcorrect fregarding fthe farbitrage-free
fprice fof fBond fA fgiven fthe fterm fstructure fabove? f
A. Bond fA’s fcash fflows fmust fbe fdiscounted fby fits fyield fto fmaturity fto fdetermine fthe
farbitrage-free fprice. f f
B. Bond fA’s fcash fflows fmust fbe fdiscounted fby fthe fspot frates fto fobtain fthe farbitrage-free
fprice. f




Copyright © IFT. All rights reserved. Page 1

, The Arbitrage-Free Valuation Framework Q Bank


C. Bond fA fmust fbe fdiscounted fby fthe fyield fto fmaturity fof fa fthree-year fbenchmark fbond
fto ffind fthe farbitrage-free fprice. f
ff
6. Using fthe fanswers fof fquestions f4 fand f5, fthe farbitrage-free fprice fof fBond fA fis fclosest fto:
f
A. $102.8286 f
B. $100.0000 f
C. $100.8682 f
f
7. An finterest frate ftree frepresents finterest frates fbased fon: f
A. an finterest frate fmodel fand fan fassumption fabout fvolatility fof finterest frates. f
B. both fpositive fand fnegative finterest frates. f
C. higher fand flower fforward frates fdetermined fby fchanging fvolatility fat feach fnode. f
f
8. The finterest frate fmodel fis fbased fupon: fA. fpathwise fvaluation. f
B. Pascal fTriangle. f
C. a flognormal fmodel fof finterest frates. f
f
9. The fmethod(s) fmost flikely fused fto festimate finterest frate fvolatility fis f(are): fA. fthe
fhistorical fvolatility fmethod fonly. f
B. the fimplied fvolatility fapproach fonly. f
C. the fhistorical fvolatility fmethod for fthe fimplied fvolatility fapproach. f
f
10. A flognormal fmodel fof finterest frates finsures fwhich fof fthe ffollowing? f fA. fHigher
fvolatility fat fhigher frates. f
B. Constant fvolatility facross fhigh for flow frates. f
C. Lower fvolatility fat fhigher frates. f
f
The ffollowing finformation frelates fto fquestions f11 f- f13. f
f
Three-Year fBinomial fInterest fRate fTree f




f
f
Implied fValues f(in f$) ffor fBond fZ: fA f4% fcoupon, fthree-year, fannual fpay fbond f
fbased fon fthe fabove finterest frate ftree f
Time f0 f Time f1 f Time f2 f Time f3 f
V0 f Node f1–1=102.1942 f Node f2-1=101.1963 f Node f3-1=104.0 f
f Node f1-2 f=105.9699 f Node f2-2 f= f? f Node f3-2=104.0 f
f



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