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| Ft = E e | Ft .
B(t, T ) = ОІ(t)E t
ОІ(T )
LetвЂ™s derive the SDE satisп¬Ѓed by B(t, T ). Let Nt = E [1/ОІ(T ) | Ft ]. This is a
martingale. By the martingale representation theorem,
t
Nt = E [1/ОІ(T )] + Hs dWs
0

for some adapted integrand Hs . So B(t, T ) = ОІ(t)Nt . Here T is п¬Ѓxed. By ItoвЂ™s product
formula,
dB(t, T ) = ОІ(t)dNt + Nt dОІ(t)
= ОІ(t)Ht dWt + Nt r(t)ОІ(t)dt
= ОІ(t)Ht dWt + B(t, T )r(t)dt,

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and we thus have
dB(t, T ) = ОІ(t)Ht dWt + B(t, T )r(t)dt. (25.1)

Forward rates. We now discuss forward rates. If one holds T п¬Ѓxed and graphs B(t, T ) as
a function of t, the graph will not clearly show the behavior of r. One sometimes speciп¬Ѓes
interest rates by what are known as forward rates.
Suppose we want to borrow \$1 at time T and repay it with interest at time T + Оµ.
At the present time we are at time t в‰¤ T . Let us try to accomplish this by buying a zero
coupon bond with maturity date T and shorting (i.e., selling) N zero coupon bonds with
maturity date T + Оµ. Our outlay of money at time t is

B(t, T ) в€’ N B(t, T + Оµ) = 0.

If we set
N = B(t, T )/B(t, T + Оµ),

our outlay at time t is 0. At time T we receive \$1. At time T +Оµ we pay B(t, T )/B(t, T +Оµ).
The eп¬Ђective rate of interest R over the time period T to T + Оµ is

B(t, T )
eОµR = .
B(t, T + Оµ)

Solving for R, we have
log B(t, T ) в€’ log B(t, T + Оµ)
R= .
Оµ
We now let Оµ в†’ 0. We deп¬Ѓne the forward rate by

в€‚
f (t, T ) = в€’ log B(t, T ). (25.2)
в€‚T

Sometimes interest rates are speciп¬Ѓed by giving f (t, T ) instead of B(t, T ) or r(t).
Recovering B from f . Let us see how to recover B(t, T ) from f (t, T ). Integrating, we have

T T
в€‚
log B(t, u)du = в€’ log B(t, u) |u=T
f (t, u)du = в€’ u=t
в€‚u
t t
= в€’ log B(t, T ) + log B(t, t).

Since B(t, t) is the value of a zero coupon bond at time t which expires at time t, it is
equal to 1, and its log is 0. Solving for B(t, T ), we have
T
в€’ f (t,u)du
B(t, T ) = e . (25.3)
t

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Recovering r from f . Next, let us show how to recover r(t) from the forward rates. We
have T
в€’ r(u)du
| Ft .
B(t, T ) = E e t

Diп¬Ђerentiating,
в€‚ T
в€’ r(u)du
B(t, T ) = E в€’ r(T )e | Ft .
t
в€‚T
Evaluating this when T = t, we obtain

E [в€’r(t) | Ft ] = в€’r(t). (25.4)

On the other hand, from (25.3) we have

в€‚ T
в€’ f (t,u)du
B(t, T ) = в€’f (t, T )e .
t
в€‚T

Setting T = t we obtain в€’f (t, t). Comparing with (25.4) yields

r(t) = f (t, t). (25.5)

100
26. Some interest rate models.
Heath-Jarrow-Morton model
Instead of specifying r, the Heath-Jarrow-Morton model (HJM) speciп¬Ѓes the forward
rates:
df (t, T ) = Пѓ(t, T )dWt + О±(t, T )dt. (26.1)

Let us derive the SDE that B(t, T ) satisп¬Ѓes. Let
T T
в€— в€—
О± (t, T ) = О±(t, u)du, Пѓ (t, T ) = Пѓ(t, u)du.
t t

T
Since B(t, T ) = exp(в€’ t f (t, u)du), we derive the SDE for B by using ItoвЂ™s formula with
T
the function ex and Xt = в€’ t f (t, u)du. We have
T
dXt = f (t, t)dt в€’ df (t, u)du
t
T
= r(t)dt в€’ [О±(t, u)dt + Пѓ(t, u)dWt ] du
t
T T
= r(t)dt в€’ О±(t, u)du dt в€’ Пѓ(t, u)du dWt
t t
в€— в€—
= r(t)dt в€’ О± (t, T )dt в€’ Пѓ (t, T )dWt .

Therefore, using ItoвЂ™s formula,

dB(t, T ) = B(t, T )dXt + 2 B(t, T )(Пѓ в€— (t, T ))2 dt
1

= B(t, T ) r(t) в€’ О±в€— + 2 (Пѓ в€— )2 dt в€’ Пѓ в€— B(t, T )dWt .
1

From (25.1) we know the dt term must be B(t, T )r(t)dt, hence

dB(t, T ) = B(t, T )r(t)dt в€’ Пѓ в€— B(t, T )dWt .

Comparing with (26.1), we see that if P is the risk-neutral measure, we have О±в€— = 2 (Пѓ в€— )2 .
1

See Note 1 for more on this.
Hull and White model
In this model, the interest rate r is speciп¬Ѓed as the solution to the SDE

dr(t) = Пѓ(t)dWt + (a(t) в€’ b(t)r(t))dt. (26.2)

Here Пѓ, a, b are deterministic functions. The stochastic integral term introduces random-
ness, while the a в€’ br term causes a drift toward a(t)/b(t). (Note that if Пѓ(t) = Пѓ, a(t) =
a, b(t) = b are constants and Пѓ = 0, then the solution to (26.2) becomes r(t) = a/b.)

101
t
(26.2) is one of those SDEвЂ™s that can be solved explicitly. Let K(t) = b(u)du.
0
Then

d eK(t) r(t) = eK(t) r(t)b(t)dt + eK(t) a(t) в€’ b(t)r(t) dt + eK(t) [Пѓ(t)dWt ]
= eK(t) a(t)dt + eK(t) [Пѓ(t)dWt ].

Integrating both sides,
t t
K(t) K(u)
eK(u) Пѓ(u)dWu .
e r(t) = r(0) + e a(u)du +
0 0

Multiplying both sides by eв€’K(t) , we have the explicit solution
t t
в€’K(t) K(u)
eK(u) Пѓ(u)dWu .
r(t) = e r(0) + e a(u)du +
0 0

If F (u) is deterministic, then
t
F (ui )(Wui+1 в€’ Wui ).
F (u)dWu = lim
0

From undergraduate probability, linear combinations of Gaussian r.v.вЂ™s (Gaussian = nor-
mal) are Gaussian, and also limits of Gaussian r.v.вЂ™s are Gaussian, so we conclude that the
t
r.v. 0 F (u)dWu is Gaussian. We see that the mean at time t is
t
в€’K(t)
eK(u) a(u)du .
E r(t) = e r(0) +
0

We know how to calculate the second moment of a stochastic integral, so
t
в€’2K(t)
e2K(u) Пѓ(u)2 du.
Var r(t) = e
0

(One can similarly calculate the covariance of r(s) and r(t).) Limits of linear combinations
T
of Gaussians are Gaussian, so we can calculate the mean and variance of 0 r(t)dt and get
an explicit expression for
T
в€’ r(u)du
B(0, T ) = E e .
0

Cox-Ingersoll-Ross model
One drawback of the Hull and White model is that since r(t) is Gaussian, it can take
negative values with positive probability, which doesnвЂ™t make sense. The Cox-Ingersoll-
Ross model avoids this by modeling r by the SDE

dr(t) = (a в€’ br(t))dt + Пѓ r(t)dWt .

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The diп¬Ђerence from the Hull and White model is the square root of r in the stochastic
integral term. This square root term implies that when r(t) is small, the п¬‚uctuations in
1
r(t) are larger than they are in the Hull and White model. Provided a в‰Ґ 2 Пѓ 2 , it can be
shown that r(t) will never hit 0 and will always be positive. Although one cannot solve
for r explicitly, one can calculate the distribution of r. It turns out to be related to the
square of what are known in probability theory as Bessel processes. (The density of r(t),
for example, will be given in terms of Bessel functions.)

Note 1. If P is not the risk-neutral measure, it is still possible that one exists. Let Оё(t) be a
t 1t
function of t, let Mt = exp(в€’ 0 Оё(u)dWu в€’ 2 0 Оё(u)2 du) and deп¬Ѓne P(A) = E [MT ; A] for
A в€€ FT . By the Girsanov theorem,

dB(t, T ) = B(t, T ) r(t) в€’ О±в€— + 2 (Пѓ в€— )2 + Пѓ в€— Оё]dt в€’ Пѓ в€— B(t, T )dWt ,
1

where Wt is a Brownian motion under P. Again, comparing this with (25.1) we must have

О±в€— = 1 (Пѓ в€— )2 + Пѓ в€— Оё.
2

Diп¬Ђerentiating with respect to T , we obtain

О±(t, T ) = Пѓ(t, T )Пѓ в€— (t, T ) + Пѓ(t, T )Оё(t).

If we try to solve this equation for Оё, there is no reason oп¬Ђ-hand that Оё depends only on t and
not T . However, if Оё does not depend on T , P will be the risk-neutral measure.

103
Problems

1. Show E [XE [Y | G] ] = E [Y E [X | G] ].
2. Prove that E [aX1 + bX2 | G] = aE [X1 | G] + bE [X2 | G].
3. Suppose X1 , X2 , . . . , Xn are independent and for each i we have P(Xi = 1)
n
1 3
= P(Xi = в€’1) = 2 . Let Sn = i=1 Xi . Show that Mn = Sn в€’ 3nSn is a martingale.
4. Let Xi and Sn be as in Problem 3. Let П†(x) = 1 (ex +eв€’x ). Show that Mn = eaSn П†(a)в€’n
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