B(t, T ) = β(t)E t

β(T )

Let™s derive the SDE satis¬ed by B(t, T ). Let Nt = E [1/β(T ) | Ft ]. This is a

martingale. By the martingale representation theorem,

t

Nt = E [1/β(T )] + Hs dWs

0

for some adapted integrand Hs . So B(t, T ) = β(t)Nt . Here T is ¬xed. By Ito™s product

formula,

dB(t, T ) = β(t)dNt + Nt dβ(t)

= β(t)Ht dWt + Nt r(t)β(t)dt

= β(t)Ht dWt + B(t, T )r(t)dt,

98

and we thus have

dB(t, T ) = β(t)Ht dWt + B(t, T )r(t)dt. (25.1)

Forward rates. We now discuss forward rates. If one holds T ¬xed and graphs B(t, T ) as

a function of t, the graph will not clearly show the behavior of r. One sometimes speci¬es

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 + µ.

At the present time we are at time t ¤ 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 + µ. Our outlay of money at time t is

B(t, T ) ’ N B(t, T + µ) = 0.

If we set

N = B(t, T )/B(t, T + µ),

our outlay at time t is 0. At time T we receive $1. At time T +µ we pay B(t, T )/B(t, T +µ).

The e¬ective rate of interest R over the time period T to T + µ is

B(t, T )

eµR = .

B(t, T + µ)

Solving for R, we have

log B(t, T ) ’ log B(t, T + µ)

R= .

µ

We now let µ ’ 0. We de¬ne the forward rate by

‚

f (t, T ) = ’ log B(t, T ). (25.2)

‚T

Sometimes interest rates are speci¬ed 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

‚

log B(t, u)du = ’ log B(t, u) |u=T

f (t, u)du = ’ u=t

‚u

t t

= ’ 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

’ f (t,u)du

B(t, T ) = e . (25.3)

t

99

Recovering r from f . Next, let us show how to recover r(t) from the forward rates. We

have T

’ r(u)du

| Ft .

B(t, T ) = E e t

Di¬erentiating,

‚ T

’ r(u)du

B(t, T ) = E ’ r(T )e | Ft .

t

‚T

Evaluating this when T = t, we obtain

E [’r(t) | Ft ] = ’r(t). (25.4)

On the other hand, from (25.3) we have

‚ T

’ f (t,u)du

B(t, T ) = ’f (t, T )e .

t

‚T

Setting T = t we obtain ’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) speci¬es the forward

rates:

df (t, T ) = σ(t, T )dWt + ±(t, T )dt. (26.1)

Let us derive the SDE that B(t, T ) satis¬es. Let

T T

— —

± (t, T ) = ±(t, u)du, σ (t, T ) = σ(t, u)du.

t t

T

Since B(t, T ) = exp(’ t f (t, u)du), we derive the SDE for B by using Ito™s formula with

T

the function ex and Xt = ’ t f (t, u)du. We have

T

dXt = f (t, t)dt ’ df (t, u)du

t

T

= r(t)dt ’ [±(t, u)dt + σ(t, u)dWt ] du

t

T T

= r(t)dt ’ ±(t, u)du dt ’ σ(t, u)du dWt

t t

— —

= r(t)dt ’ ± (t, T )dt ’ σ (t, T )dWt .

Therefore, using Ito™s formula,

dB(t, T ) = B(t, T )dXt + 2 B(t, T )(σ — (t, T ))2 dt

1

= B(t, T ) r(t) ’ ±— + 2 (σ — )2 dt ’ σ — 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 ’ σ — B(t, T )dWt .

Comparing with (26.1), we see that if P is the risk-neutral measure, we have ±— = 2 (σ — )2 .

1

See Note 1 for more on this.

Hull and White model

In this model, the interest rate r is speci¬ed as the solution to the SDE

dr(t) = σ(t)dWt + (a(t) ’ b(t)r(t))dt. (26.2)

Here σ, a, b are deterministic functions. The stochastic integral term introduces random-

ness, while the a ’ br term causes a drift toward a(t)/b(t). (Note that if σ(t) = σ, a(t) =

a, b(t) = b are constants and σ = 0, then the solution to (26.2) becomes r(t) = a/b.)

101

t

(26.2) is one of those SDE™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) ’ b(t)r(t) dt + eK(t) [σ(t)dWt ]

= eK(t) a(t)dt + eK(t) [σ(t)dWt ].

Integrating both sides,

t t

K(t) K(u)

eK(u) σ(u)dWu .

e r(t) = r(0) + e a(u)du +

0 0

Multiplying both sides by e’K(t) , we have the explicit solution

t t

’K(t) K(u)

eK(u) σ(u)dWu .

r(t) = e r(0) + e a(u)du +

0 0

If F (u) is deterministic, then

t

F (ui )(Wui+1 ’ Wui ).

F (u)dWu = lim

0

From undergraduate probability, linear combinations of Gaussian r.v.™s (Gaussian = nor-

mal) are Gaussian, and also limits of Gaussian r.v.™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

’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

’2K(t)

e2K(u) σ(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

’ 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 doesn™t make sense. The Cox-Ingersoll-

Ross model avoids this by modeling r by the SDE

dr(t) = (a ’ br(t))dt + σ r(t)dWt .

102

The di¬erence 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 ¬‚uctuations in

1

r(t) are larger than they are in the Hull and White model. Provided a ≥ 2 σ 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 θ(t) be a

t 1t

function of t, let Mt = exp(’ 0 θ(u)dWu ’ 2 0 θ(u)2 du) and de¬ne P(A) = E [MT ; A] for

A ∈ FT . By the Girsanov theorem,

dB(t, T ) = B(t, T ) r(t) ’ ±— + 2 (σ — )2 + σ — θ]dt ’ σ — B(t, T )dWt ,

1

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

±— = 1 (σ — )2 + σ — θ.

2

Di¬erentiating with respect to T , we obtain

±(t, T ) = σ(t, T )σ — (t, T ) + σ(t, T )θ(t).

If we try to solve this equation for θ, there is no reason o¬-hand that θ depends only on t and

not T . However, if θ 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 = ’1) = 2 . Let Sn = i=1 Xi . Show that Mn = Sn ’ 3nSn is a martingale.

4. Let Xi and Sn be as in Problem 3. Let φ(x) = 1 (ex +e’x ). Show that Mn = eaSn φ(a)’n