University of Hong Kong: MFIN7012 Fixed Income Assignment Solution

This document presents a comprehensive solution to an assignment on Fixed Income Securities and Interest Rate Modelling, likely for a Master's level finance course at the University of Hong Kong. The solution addresses key concepts such as one-step interest rate trees, the calculation of expected rates, market price of risk, and risk-neutral probabilities. It includes detailed calculations for option pricing, replicating portfolios using ZCBs, and the construction of risk-neutral interest rate trees. The assignment also covers interest rate swaps and the application of the Ho-Lee and Vasicek interest rate models, including parameter estimation and term structure analysis. Furthermore, the solution explores the impact of varying parameters on the term structure of interest rates and discusses derivative markets, including forwards, futures, swaps, and options. Finally, it touches upon the binomial model and its application in pricing and risk management.

Fixed Income Securities and Interest Rate Modelling 1
FIXED INCOME SECURITIES AND INTEREST RATE MODELLING
by Student’s Name
Code + Course Name
Professor’s Name
University Name
City, State
Date

Fixed Income Securities and Interest Rate Modelling 2
Fixed Income Securities and Interest Rate Modelling
1.
a. Expected 6 month treasury rate
Po=97.4845, Pu = 0.5 Pd = 0.5
When r1=r1 ,u P1 , u=100 e−0.04 × 0.5=98.0199
The return then is 98.0199
97.4845 −1=0.00549
When r1=r1 ,d P1 ,d =100 e−0.01 ×0.5 =99.5012
The return here is 99.5012
97.4845 −1=0.0207
E [ r1 ]= ( 0.00549 × 0.5+0.0207 ×0.5 )
¿ 0.013095=1.31%
b. Market price of risk
λ= e−r0
E [ P1 ] −P0
P1 , u−P1 ,d
E [ P1 ]=99.5012 ×0.5+98.0199× 0.5
¿ 98.76055
λ= e−0.02× 0.5 98.76055−97.4845
98.0199−99.5012
¿−0.1980

Fixed Income Securities and Interest Rate Modelling 3
c. Risk neutral probability for an upward movement in the interest rate
λ¿= e−ro E¿ [ P1 ]−Po
P1 ,u −P1 ,d
=0
E¿=P u¿ P1 ,u−P d¿ P1 , d∧P d¿+P u¿=1∧0<Pu¿<1 ,
P d¿=1−P u¿
Hence Po=e−ro E¿ [ P1 ]
Solving, 98.76055=e−0.02 ×0.5 ( 98.0199+99.5012−99.5012 )
P u¿=98.76055 e−0.02 × 0.5−99.5012
98.0199−99.5012
¿ 0.1667
P d¿=1−0.1667=0.8333
d. Option
i. Price of option
r1=0.04 , thus0.04−0.02=0.02
max {0.02 , 0 }=0.02
payoff of A=100× 0.02
¿ 2
ii. Replicating portfolio
Using replication
Payoff ¿ 2 with price=99.005
the 2 period has 98.0199∧99.5012 with price=97.4845
We consider a portfolio A:

Fixed Income Securities and Interest Rate Modelling 4
Investing 25.89 and -26.677 of A in the above, we have
¿ 2 ×99.005−2× 97.4845
¿ 3.041
2. Risk-neutral interest rate tree
r0 ¿ 0.05
r1=0.055× 0.5+0.047 × 0.5=0.051
r2=0.06 ×0.5+0.05 × 0.5+0.045 ×0.5=0.1025
a. Interest rate swap at t=0
The fixed rate is 100 ×5 %=5
Float ¿ 5 ×100 ×r2 ( t−1)
The floating leg
t rates float fixed
0 5% 5
1 5.1% 5.0 5
2 10.25% 5.1 5
The swap rate c ¿ 0.5 ( 1−c ) =1
c=1 %
b. Price at t=0

Fixed Income Securities and Interest Rate Modelling 5
K = 100
R = 5%
T = 3 years
P0 ¿ 100 e−0.05 ×3=86.0708
c. Value of the a 3 year callable bond
Par value = P
Coupon = c
Maturity = T
Vt= Ex-coupon price
V t=E¿ [ e−rt ( c +Vt +1 ) ]
VT=P
3. 6 month ZCB FV= 100 price = 100
1 year ZCB FV= 100 price =100
A two step Ho-Lee
r0 =100 e−r0 ×0.5 =98 →r 0=0.04
ru =0.04+∅ 0.5+0.01 √ 0.5
rd =0.04 +∅ 0.5−0.01 √ 0.5=ru −2× 0.01 √ 0.5
96=100 e−r0 × 0.5 [ 0.5 e−r u ×0.5+0.5 e− ( ru ×0.5 ) ×0.5 ]
e−ru × 0.5=
96
100 e−r 0 ×0.5
0.5+0.5 e−2 × 0.01 √0.5 × 0.5 =0.9829
ru =0.0345

Fixed Income Securities and Interest Rate Modelling 6
erd
=0.9829−2× 0.01 √ 0.5=0.9687
rd =0.0636
∅ = 0.0636−.04+0.01 √0.5
0.5 =0.0613
So, ru =0.0345, rd =0.0636, ∅ =0.0613
4. Vasicek interest rate model
With parameters γ= 0.3262, r= 0.0509, σ= 0.0221
risk-neutral process has parameters γ*= 0.4653, r¿= 0.0634
ro= 0.055
a. d rt =γ [ r−rt ] dt+σdWι
Ε [ r1 ]=Ε [ r0 e−γ +r (1−e−γ ) +σ ( γ∗+r¿ ) e− γ
]
¿ 0.055 e−0.3262+ 0.0509 ( 1−e−0.3262 )+ 0.0221 ( 0.4653+ 0.0634 ) e−0.3262
¿ 0.0655
r10=0.055 e−0.3262 ×10 +0.0509 ( 1−e−0.3262 ×10 ) +0.0221 ( 0.4653 e−10+ 0.0634 e−10 ) e−0.3262
¿ 0.050167
The slope shows a decreasing curve, the expectation is that the rates should increase over time
but the observation made show a decreasing curve which may be due to market segmentation
b. Market price
At t = 0;
The market price for 100 nominal will be

Fixed Income Securities and Interest Rate Modelling 7
100 ( e−0.0655 )
¿ 93.6599
c. Plot of the term structure of interest rates
r5 =0.055 e−0.3262×5 +0.0509 ( 1−e−0.3262× 5 ) + 0.0221 ( 0.4653 e−5 +0.0634 e−5 ) e−0.3262=0.052
r15=0.055e−0.3262 ×15 +0.0509 ( 1−e−0.3262 ×15 ) +0.0221 ( 0.4653 e−15+0.0634 e−15 ) e−0.3262=0.051
r20=0.055 e−0.3262 ×20 +0.0509 ( 1−e−0.3262 ×20 ) +0.0221 ( 0.4653 e−20+0.0634 e−20 ) e−0.3262=0.05091
r30=0.055 e−0.3262 ×30 +0.0509 ( 1−e−0.3262 ×30 ) +0.0221 ( 0.4653 e−30+ 0.0634 e−30 ) e−0.3262=0.05090
d. Varying γ
Say, γ1> γ
r5 =0.055 e−0.1234× 5+ 0.0509 ( 1−e−0.1234 ×5 ) +0.0221 ( 0.4653 e−5 +0.0634 e−5 ) e−0.1234=0.0532
r15=0.055e−0.1234 ×15 +0.0509 ( 1−e−0.1234× 15 ) + 0.0221 ( 0.4653 e−15 +0.0634 e−15 ) e−0.1234=0.0515

Fixed Income Securities and Interest Rate Modelling 8
The rates are increasing with an increasing γ.
e. Varying r
Taking a greater r say 0.065>0.05091
r5 =0.055 e−0.3262×5 +0.065 ( 1−e−0.3262× 5 )+ 0.0221 ( 0.4653 e−5 +0.0634 e−5 ) e−0.3262=0.0596
r15=0.055e−0.3262 ×15 +0.065 ( 1−e−0.3262 ×15 ) +0.0221 ( 0.4653 e−15+0.0634 e−15 ) e−0.3262=0.06343
The rates are increasing with an increasing r
f. Varying σ
Taking a greater σ, say 0.035>0.0221
r5 =0.055 e−0.3262×5 +0.0509 ( 1−e−0.3262× 5 ) + 0.035 ( 0.4653 e−5+ 0.0634 e−5 ) e−0.3262=0.05322
r15=0.055e−0.3262 ×15 +0.0509 ( 1−e−0.3262 ×15 ) +0.035 ( 0.4653 e−15+0.0634 e−15 ) e−0.3262=0.0725
An increasing σ, leads to increased return rates.