7. Continuously Varying Interest
Rates
7.1 The Continuous Varying Interest Rate
Formula.
Suppose that interest is continuously compounded with a rate which is changing in time.
Let the present time be time 0, and let r(s)
denote the interest rate per unit at time s
units, s ≥ 0.
The quantity r(s) is called the spot or the
instantaneous interest rate at time s.
Let D(t) be the amount that you will have
in your account at time t if you deposit P at
time 0. In order to determine D(t) in terms
of the interest rates r(s), 0 ≤ s ≤ t, note that
by the Simple Interest Formula 1.8, for small
h, we have
D(s + h) ≈ D(s)(1 + r(s) · h),
(≈ means “is approximately equal to”)
D(s + h) ≈ D(s) + D(s) · r(s) · h,
D(s + h) − D(s) ≈ D(s) · r(s) · h,
D(s + h) − D(s)
≈ D(s) · r(s).
h
By the definition of the derivative 4.7, taking
the limit as h → 0, we have
D0(s) = D(s) · r(s)
or
D0(s)
= r(s).
D(s)
To solve this differential equation, integrate
both sides:
Z t 0
D (s)
0 D(s)
ds =
Z t
0
r(s)ds.
Thus
[ln D(s)]t0 =
Z t
0
r(s)ds
or
Z t
ln D(t) − ln D(0) =
0
r(s)ds.
Since D(0) = P, we obtain from the preceding
equation that
ln D(t) − ln P =
Z t
0
r(s)ds
exp (ln D(t) − ln P) = exp
Z t
0
exp (ln D(t)) · exp ln P−1 = exp
D(t) · P−1 = exp
Z t
0
r(s)ds
Z t
0
r(s)ds
and
D(t) = P · exp
Z t
0
r(s)ds .
r(s)ds
7.2 Example. Suppose that interest is continuously compounded with a rate which is
changing in time. Let the present time be
time 0, and let
1
1+s
be the interest rate per unit at time s units,
s ≥ 0.
r(s) = 0.10 − 0.05
(i) Draw a graph of the function r(s).
(ii) Find the amount D(t) that you will have
in your account at time t, t ≥ 0, if you
deposit £1 , 000 at time 0.
(iii) Suppose the unit is equal to one year.
How much money do you have in your account after 10 years?
Solution.
(i) The graph of the function
1 is
y1(s) = 0.05 1+s
1
The graph of the function y2(s) = −0.05 1+s
is
1
The graph of the function r(s) = 0.10−0.05 1+s
is
(ii) By the Continuous Varying Interest Rate
Formula 7.1,
D(t) = £1 , 000 exp
Z t
0
!
r (s )ds .
Note that
Z t
Z t
1
ds
r(s)ds =
0.10 − 0.05
1+s
0
0
(by the linearity of the integrals)
=
Z t
0
0.10ds − 0.05
Z t
1
ds
0 1+s
= 0.10 · t − 0.05 · [ln |1 + s|]t0
= 0.10 · t − 0.05 · (ln |1 + t| − ln |1|)
= 0.10 · t − 0.05 · ln |1 + t|.
Hence
D(t) = £1 , 000 · exp (0 .10t − 0 .05 · ln |1 + t |)
D(t) = £1 , 000 · e
0 .10t +ln |1 +t |−0 .05
= £1 , 000 · e 0 .10t · e ln |1 +t |
−0 .05
= £1 , 000 · e 0 .10t · |1 + t |−0 .05 .
(iii) For t = 10,
D(10) = £1 , 000 · e 0 .10 ·10 · (1 + 10 )−0 .05
= £1 , 000 · e · 11 −0 .05
(Recall that e ≈ 2.71828182. )
= £1 , 000 · 2 .71828182 · 0 .88701378 = £2 , 411 .15 .
7.3 The Continuous Compounded Interest Formula. The Continuous Compounded
Interest Formula 4.10 is a special case of the
Continuous Varying Interest Rate Formula 7.1,
arising when the interest rate r(s), s ≥ 0, is
r . Let a principal P be boralways equal to 100
rowed at an annual rate of interest r% compounded continuously for t years.
Then, according to 7.1, for D(0) = P and
r , s ≥ 0, the amount A owed on
r(s) = 100
the principal P at time t is
A = D(t) = P · exp
Z t
0
r(s)ds
= P · exp
Z t
0
r
ds
100
r
= P · exp
·t
100
=P
r ×t
e 100 .
7.4 The Present Value.
The Continuous Varying Interest Rate Formula 7.1 states
that a principal P earning a continuously compounded interest r(s) per unit at time s units
will be worth
D(t) = P · exp
Z t
0
r(s)ds
at time t. Therefore
P = D(t) · exp
Z t
= D(t) · exp −
0
−1
r(s)ds
Z t
0
r(s)ds
.
Definition. The present value P (t) corresponding to an amount A and to a variable
interest rate r(s), s ≥ 0, is given by
P (t) = A · exp −
Z t
0
, t ≥ 0,
r(s)ds
and is equal to the amount to be invested at
time 0 to accumulate A at time t by continuous compounding at the interest rate r(s),
s ≥ 0.
7.5 Definition. The function
P (t) = exp −
Z t
0
r(s)ds
, t ≥ 0,
which is at each point t equal to the present
value P (t) corresponding to the amount 1 and
to a variable interest rate r(s), s ≥ 0, is called
the present value function.
In other words, the present value function at t
is equal to the amount to be invested at time
0 to accumulate 1 at time t by continuous
compounding at the interest rate r(s), s ≥ 0.
7.6 Example. How much money should be
invested at time 0 at a continuously compounded interest rate
1
1+s
per unit at time s units, s ≥ 0, in order to
accumulate £5 , 000 , 000 at time t?
r(s) = 0.10 − 0.05
Solution. By 7.4, the present value P (t) of
£5 , 000 , 000 is
"
P (t) = £5 , 000 , 000 · exp −
Z t
0
!#
r (s )ds
.
In Example 7.2 we showed that
Z t
0
r(s)ds =
Z t
0
0.10 − 0.05
1
ds
1+s
= 0.10 · t − 0.05 · ln |1 + t|.
Hence the present value P (t) is
£5 , 000 , 000 · exp [− (0 .10 · t − 0 .05 · ln |1 + t |)]
= £5 , 000 , 000 · exp [−0 .10 · t + 0 .05 · ln |1 + t |]
= £5 , 000 , 000 ·exp (−0 .10 · t )·exp ln |1
+ t |0 .05
= £5 , 000 , 000 · exp (−0 .10 · t ) · |1 + t |0 .05 .
Suppose t = 10: then
P (10) = £5 , 000 , 000 ·exp (−0 .10 · 10 )·|1 +10 |0 .05
1
= £5 , 000 , 000 · · 11 0 .05
e
= £2 , 073 , 696 .33 .
7.7 Definition.
The average of the spot
interest rates up to time t
t
1
r̄(t) = ·
r(s)ds, t ≥ 0,
t 0
is called the yield curve.
Z
7.8 Example. Find the yield curve and the
present value function if
r(s) = r1 ·
1
s
+ r2 ·
.
1+s
1+s
Solution. Note that we can rewrite r(s) as
r1 + r2 · s
r(s) =
1+s
r − r2 + r2 · (1 + s)
= 1
1+s
r − r2
= 1
+ r2 .
1+s
Hence, for t ≥ 0,
Z t
Z t
r1 − r2
r(s)ds =
+ r2ds
0
0 1+s
Z t
Z t
r1 − r2
=
ds +
r2ds
0 1+s
0
Z t
t
1
= (r1 − r2)
ds + r2
ds
0 1+s
0
Z
= (r1 − r2)[ln |1 + s|]t0 + r2[s]t0
= (r1 − r2) · ln(1 + t) + r2 · t.
Therefore, for t ≥ 0, the yield curve is
t
1
r̄(t) = ·
r(s)ds
t 0
Z
=
1
· [(r1 − r2) · ln(1 + t) + r2 · t]
t
r − r2
= 1
· ln(1 + t) + r2.
t
Consequently, for t ≥ 0, the present value
function is
P (t) = exp −
Z t
0
r(s)ds
= exp [− ((r1 − r2) · ln(1 + t) + r2 · t)]
(r
−r
)
= exp ln[(1 + t) 2 1 ] · exp (−r2 · t)
= (1 + t)(r2−r1) · exp (−r2 · t).