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CONTENTS PAGE MATERIAL SUMMARY
Essential Topic: Stochastic interest ratesChapter 12
Mathematics of Finance: A Deterministic Approachby S. J. Garrett
CONTENTS PAGE MATERIAL SUMMARY
CONTENTS PAGE
MATERIAL
MotivationVarying interest rate modelAccumulation of a single investmentExampleAccumulation of annual investmentsExampleLog-normal distributionExample
SUMMARY
CONTENTS PAGE MATERIAL SUMMARY
MOTIVATION
I We have so far been concerned with fixed, piecewiseconstant, or term-structured interest rates.
I In each case, the rates have been known in advance and allcash-flow analyses have been deterministic.
I However, in reality interest rates are not known inadvance. This motivates the use of stochastic models.
I A simplification is to assume the varying interest rate modelwith i.i.d. rates across the years.
CONTENTS PAGE MATERIAL SUMMARY
VARYING INTEREST RATE MODEL
I Under the varying interest rate model, the single rate thatapplies in each year is unknown in advance. However,once it has been determined (at the start of the year) it isfixed throughout that year.
I Furthermore, we assume that the interest rates that applyin each year are independent and identically distributed (i.i.d.)across all years.
I Essentially we are assuming that the interest rate in eachyear takes an unknown realization of a known distribution.
I Properties of the distribution of the random rate I areknown in advance and
E[I] = j var(I) = s2.
CONTENTS PAGE MATERIAL SUMMARY
ACCUMULATION OF A SINGLE INVESTMENT
I Consider a unit investment made at time t = 0 which willaccumulate under the i.i.d. varying interest rate model forn years.
I The accumulation after n years is denoted Sn.I As the interest rate that applies in each year is random {it},
the accumulation Sn will also be random.I We seek to determine the mean and variance of Sn in terms
of properties of the interest rate distribution.
CONTENTS PAGE MATERIAL SUMMARY
ACCUMULATION OF A SINGLE INVESTMENT
I The expected value of Sn is calculated as
E[Sn] =E [(1 + I1)× (1 + I2)× · · · × (1 + In)]
=E [1 + I1]× E [1 + I2]× · · · × E [1 + In] (ind.)
=
n∏t=1
(1 + E[It])
=⇒ E[Sn] =(1 + j)n (if identically distributed)
I This gives the expected value of the accumulation underthe i.i.d. varying interest rate model.
CONTENTS PAGE MATERIAL SUMMARY
ACCUMULATION OF A SINGLE INVESTMENTI The variance of Sn is calculated from the standard
expression
var(Sn) = E[S2
n]− E [Sn]
2
I We know that E[Sn]2 = (1 + j)2n
I Further
E[S2n] =E
[(1 + I1)
2 × (1 + I2)2 × · · · × (1 + In)
2]=E[(1 + I1)
2]× E[(1 + I2)
2]× · · · × E[(1 + In)
2] (ind.)
=
n∏t=1
(1 + 2E[It] + E[I2
t ])
=(1 + 2j + s2 + j2)n (if identically distributed)
I The expression for the variance of the accumulation underthe model is then
var(Sn) =((1 + j)2 + s2)n − (1 + j)2n.
CONTENTS PAGE MATERIAL SUMMARY
EXAMPLE
The annual interest earned on an investment each year is acollection of i.i.d. random variables with mean 0.05 andstandard deviation 0.15. Calculate the expected value andstandard deviation of the accumulation of £10 after 5 years.
Answer
We have that j = 0.05 and s2 = 0.152. The expected value andstandard deviation of the accumulation are then
E [10S5] =10× 1.055 = £12.76
var (10S5) =102 ×((1.052 + 0.152)5 − 1.0510
)= (4.16)2
=⇒ s.d. =£4.16
CONTENTS PAGE MATERIAL SUMMARY
ACCUMULATION OF ANNUAL INVESTMENTS
I We now consider the n-year accumulation of a unitinvestment made at the start of each year for n years underthe i.i.d. varying interest rate model.
I This accumulation is a random variable, An, such that
An = (1 + I1)(1 + I2)(1 + I3) . . . (1 + In)+(1 + I2)(1 + I3) . . . (1 + In)
+(1 + I3) . . . (1 + In)...
+(1 + In).
I To make progress towards expressing properties of An interms of properties of I, we note the recursive relationship
An = (1 + In)(1 + An−1).
CONTENTS PAGE MATERIAL SUMMARY
ACCUMULATION OF ANNUAL INVESTMENTSI We take the expectation of the recursive relationship to
find a recursive relationship for the expected value of then-year accumulation, µn
µn = E[An] = E[(1 + In)(1 + An−1)] (ind.)
I For identically distributed rates, this reduces to
µn =(1 + j)(1 + µn−1)
=s̈n at rate j
I Similarly the second moment, mn = E[A2n], can be formed in
terms of a recursive relationship
mn =[var(1 + In) + E[1 + In]
2]E[1 + 2An−1 + A2
n−1]
(ind.)
=[s2 + (1 + j)2] [1 + 2µn−1 + mn−1] (iden. dist.)
I The variance of the n-year accumulation is then
var(An) = mn − µ2n.
CONTENTS PAGE MATERIAL SUMMARY
EXAMPLEThe annual rate of interest earned on a particular fund is arandom variable with i.i.d. annual rates. In all years theexpected value of interest is 5% and the standard deviation 8%.Calculate the mean and standard deviation of the totalaccumulated amount at the end of 2 years if £1 is invested atthe start of the first and second years.
Answer
We have that j = 0.05 and s2 = 0.082 and require E[A2] and√var(A2).
E [A2] =µ2 = s̈2 = £2.15
var (A2) =m2 − µ22 =
(s2 + (1 + j)2) (1 + 2µ1 + m1)− 2.152
with m1 =(1 + j)2 + s2 = 1.1089 and µ1 = 1.05
=⇒ var (A2) =(0.082 + 1.052) (1 + 2× 1.05 + 1.1089)− 2.152
=(£0.21)2.
CONTENTS PAGE MATERIAL SUMMARY
LOG-NORMAL DISTRIBUTIONI An important distribution for the i.i.d. varying interest
rate model is the log-normal distribution
1 + I ∼ ln N(µ, σ2)
where µ and σ are parameters of the distribution.I The properties of the distribution are such that
ln Sn =
n∑t=1
ln(1 + It)
∼N(nµ,nσ2)
I The distribution parameters, µ and σ2, are linked to j ands2 via
1 + j = eµ+σ22 and s2 = e2µ+σ2 ×
(eσ
2 − 1).
I These form two simultaneous equations for thedistribution parameters in terms of j and s.
CONTENTS PAGE MATERIAL SUMMARY
EXAMPLE
A single investment of £2000 is made into a fund that paysinterest under the i.i.d. varying interest rate model with alog-normal distribution. The annual rates have mean 5% andstandard deviation 10%.a.) Calculate the expected value of the accumulation after 10
years.b.) Calculate the probability that the accumulated value after
10 years will be less than 90% of the expected value froma.).
CONTENTS PAGE MATERIAL SUMMARY
EXAMPLEAnswer
a.) We have j = 5% and so
E[2000S10] = 2000× (1.05)10 = £3257.79
b.) We can determine that µ = 0.0443 and σ2 = 0.0952.Therefore
P(
S10 < 0.90× 1.0510)=P (S10 < 1.4660)
=P (ln S10 < ln 1.4660)
=P(
ln S10 − 10µ√10σ2
<ln 1.4660− 10µ√
10σ2
)=P(
Z <ln 1.4660− 10× 0.0433√
10× 0.0952
)=P (Z < −0.201) ≈ 42%
CONTENTS PAGE MATERIAL SUMMARY
SUMMARY
I Under the i.i.d. varying interest rate model the interest rate in anyyear is unknown in advance but is a realization of the randomvariable I with known distribution. We have E[I] = j andvar(I) = s2.
I Sn denotes the n-year accumulation of a unit investment under thei.i.d. varying interest rate model. We have
E[Sn] = (1 + j)n and var(Sn) =((1 + j)2 + s2)2 − (1 + j)2n.
I An denotes the n-year accumulation of an annual unit investmentmade in advance. We have
E[An] =µn = s̈n and var(An) = mn − µ2n
where mn =[s2 + (1 + j)2
][1 + 2µn−1 + mn−1].
I The log-normal distribution is an important example of adistribution for I
1 + I ∼ ln N(µ, σ2) =⇒ ln Sn ∼ N(nµ,nσ2).