Ho-Lee Model

Every fixed income model is determined by the continuously compounded stochastic forward rate f_t. This corresponds to the repurchase agreement rate at time t. One unit invested at time t pays 1 + f_t\,dt = \exp(f_t\,dt) at time t + dt. Rolling over one unit invested at time 0 has realized return R_t = \exp(\int_0^t f_s\,ds) at time t. The stochastic discount is D_t = 1/R_t.

The price at time t of a zero coupon bond maturing at u is D_t(u) = E_t[D_u]/D_t = E_t[\exp(-\int_t^u f_s\,ds)] where E_t is the conditional expectation given information at time t. The forward curve f_t(u) at time t is defined by {D_t(u) = \exp(-\int_t^u f_t(s)\,ds)}. We write D(t) for D_0(t) and f(t) for f_0(t), today’s discount and forward curves.

Hint. Use D(t) = E[\exp(-\int_0^t f_s\,ds)] = \exp(-\int_0^t f(s)\,ds) and compute the derivative with respect to t

This shows f(t) is the par coupon of a forward contract paying f_t - f(t) at t.

Hint: Use D_t(u) = E_t[\exp(-\int_t^u f_s\,ds)] = \exp(-\int_t^u f_t(s)\,ds) and compute the derivative with respect to u.

This shows f_t(u) is the par coupon at time t of a forward contract paying f_u - f_t(u) at u.

The futures quote at time t of a contract settling at u to f_u is φ_t(u) = E_t[f_u], t\le u. Futures quotes are naturally occurring martingales. We write φ(t) for φ_0(t).

From above we have {f(t)E[D_t] = E[f_t D_t] = E[f_t]E[D_t] + \operatorname{Cov}(f_t, D_t)} so {φ(t) - f(t) = -\operatorname{Cov}(f_t, D_t)/D(t)}. This is positive since the forward and discount are negatively correlated. For most models it is approximately proportional to the the square of time to expiration.

The stochastic forward rate determines all quantities relevant to the dynamics of fixed income instruments. See Yield Curve Model for details.

Model

The Ho-Lee model assumes the stochastic forward rate is f_t = φ(t) + σ(t)\cdot B_t where φ(t) is the futures quote at time t, σ(t) is the vector-valued volatility, and B_t is vector-valued standard Brownian motion.

The stochastic discount is D_t = \exp(-\int_0^t f_s\,ds) = \exp(-\int_0^t φ(s) + σ(s)\cdot B_s\,ds).

Exercise: Show \int_0^t σ(s)\cdot B_s\,ds = \int_0^t (Σ(t) - Σ(s))\cdot dB_s where {Σ(t) = \int_0^t σ(s)\,ds}.

Hint Use d(Σ(t)\cdot B_t) = Σ(t)\cdot dB_t + Σ'(t)\cdot B_t\,dt.

Solution

Integrating we have {Σ(t)\cdot B_t - Σ(0)\cdot B_0 = \int_0^t Σ(s)\cdot dB_s + \int_0^t Σ'(s)\cdot B_s\,ds}. The result follows from {B_t = \int_0^t dB_s} and σ(t) = Σ'(t).

This shows D_t = \exp\bigl(-\int_0^t φ(s)\,ds - \int_0^t (Σ(t) - Σ(s)\cdot dB_s\bigr).

Exercise. Show \operatorname{Var}(\int_0^t (Σ(t) - Σ(s))\cdot dB_s) = \int_0^t \|Σ(t) - Σ(s)\|^2\,ds.

Hint: Use the Ito isometry E[(\int_0^t X_s\cdot dB_s)^2] = E[\int_0^t \|X_s\|^2\,ds].

Since the exponent is normally distributed and E[\exp(N)] = \exp(E[N] + \operatorname{Var}(N)/2) if N is normal we have D(t) = E[D_t] = \exp\bigl(-\int_0^t φ(s)\,ds + \int_0^t \|Σ(t) - Σ(s)\|^2/2\,ds\bigr)

Exercise. Show if σ is constant then D(t) = \exp(-\int_0^t φ(s)\,ds + \|σ\|^2 t^3/6).

Hint: \int_0^t (t - s)^2\,ds = t^3/3.

Now we determine the forward curve.

Exercise. Show (d/dt) \int_0^t \|Σ(t) - Σ(s)\|^2\,ds = 2σ(t)\cdot \int_0^t Σ(t) - Σ(s)\,ds.

Hint. Use the Leibniz integral rule (d/dt)\int_0^t F(t,s)\,ds = F(t, t) + \int_0^t (\partial/\partial t) F(t, s)\,ds.

Solution

Let F(t,s) = \|Σ(t) - Σ(s)\|^2 so (\partial/\partial t) F(t,s) = 2(Σ(t) - Σ(s))\cdot σ(t) and {(d/dt) \int_0^t \|Σ(t) - Σ(s)\|^2\,ds = 0 + \int_0^t 2(Σ(t) - Σ(s))\cdot σ(t)\,ds}.

Since D(t) = \exp(-\int_0^t f(s)\,ds) the forward curve is f(t) = φ(t) - σ(t)\cdot \int_0^t Σ(t) - Σ(s)\,ds.

Exercise. If σ is constant then {f(t) = φ(t) - \|σ\|^2 t^2/2}.

Hint: Use Σ(t) = σt.

There is an explicit formula for convexity in the Ho-Lee model.

Exercise. Show \operatorname{Cov}(f_t, D_t)/D(t) = -σ(t)\cdot \int_0^t sσ(s)\,ds.

Hint: Use E[f(N) \exp(M)] = E[f(N + \operatorname{Cov}(N,M))] E[\exp(M)] if N and M are jointly normal to show \operatorname{Cov}(N, \exp(M)) = \operatorname{Cov}(N,M) E[\exp(M)]. Recall \operatorname{Cov}(B_t, B_s) = sI for s\le t.

Solution

We have \operatorname{Cov}(f_t, D_t) = \operatorname{Cov}(σ(t)\cdot B_t,-\int_0^t σ(s)\cdot dB_s) E[D_t] = -σ(t) \cdot \int_0^t σ(s)s\,ds\,D(t).

Exercise. If σ is constant show \operatorname{Cov}(f_t, D_t)/D(t) = -\|σ\|^2 t^2/2.

Dynamics

For the Ho-Lee model {D_t(u) = E_t[\exp(-\int_t^u φ(s) + σ(s)\cdot B_s\,ds)]}.

Exercise. Show \int_t^u σ(s)\cdot B_s\,ds = \int_t^u Σ(u) - Σ(s)\cdot dB_s + (Σ(u) - Σ(t))\cdot B_t where {Σ(t) = \int_0^t σ(s)\,ds}.

Hint Use d(Σ(t)\cdot B_t) = Σ(t)\cdot dB_t + Σ'(t)\cdot B_t\,dt.

Solution

We have \begin{aligned} \int_t^u σ(s)\cdot B_s\,ds &= -\int_t^u Σ(t)\cdot dB_s + Σ(u)\cdot B_u - Σ(t)\cdot B_t \\ &= -\int_t^u Σ(s)\cdot dB_s + Σ(u)\cdot B_u - Σ(u)\cdot B_t + Σ(u)\cdot B_t - Σ(t)\cdot B_t \\ &= -\int_t^u Σ(s)\cdot dB_s + Σ(u)\cdot \int_t^u dB_s + Σ(u)\cdot B_t - Σ(t)\cdot B_t \\ &= \int_t^u Σ(u) - Σ(s)\cdot dB_s + (Σ(u) - Σ(t))\cdot B_t \\ \end{aligned}

Exercise. Show \operatorname{Var}(\int_t^u Σ(u) - Σ(s)\,dB_s) = \int_t^u (Σ(u) - Σ(s))^2\,ds.

Exercise. Show E_t[\exp(\int_t^u Λ(s)\,dB_s)] = \exp(\int_t^u Λ(s)^2/2\,ds).

Hint: Use {X_t = \exp(\int_0^t Λ(s)\,dB_s - \int_0^t Λ(s)^2/2\,ds)} is a martingale.

Solution

1 = E_t[X_u/X_t] = E_t[\exp(\int_t^u Λ(s)\,dB_s - \int_t^u Λ(s)^2/2\,ds)]

Note the right hand side is not random and {E_t[\exp(-\int_t^u Λ(s)\,dB_s)] = \exp(\int_t^u Λ(s)^2/2\,ds)} by replacing Λ with -Λ. We use this below.

Using the previous two exercises, the price at t of a zero coupon bond maturing at u in the Ho-Lee model is \begin{aligned} D_t(u) &= E_t[D_u/D_t] \\ &= E_t[\exp(-\int_t^u φ(s) + σ(s) B_s\,ds)] \\ &= E_t[\exp\bigl(-\int_t^u φ(s)\,ds - \int_t^u Σ(u) - Σ(s)\,dB_s - (Σ(u) - Σ(t)) B_t\bigr)] \\ &= \exp(\int_t^u -φ(s)\,ds + \int_t^u (Σ(u) - Σ(s))^2/2\,ds - (Σ(u) - Σ(t)) B_t) \\ \end{aligned}

Exercise. If σ is constant show D_t(u) = \exp(\int_t^u -φ(s)\,ds + σ^2(u - t)^3/6 - σ(u - t) B_t).

Note D_t(u) is lognormal with log-variance \operatorname{Var}(\log D_t) = (Σ(u) - Σ(t))^2 t and expected value E[D_t(u)] = \exp(\int_t^u -φ(s)\,ds + \int_t^u (Σ(u) - Σ(s))^2/2\,ds + (Σ(u) - Σ(t))^2t/2)

Exercise. If σ is constant show E[D_t(u)] = \exp\bigl(\int_t^u -φ(s)\,ds + σ^2(u - t)^3/6 + σ^2(u - t)^2 t/2\bigr).

Note \begin{aligned} \frac{D(u)}{D(t)} &= \exp\bigl(\int_t^u -φ(s)\,ds + \int_0^u (Σ(u) - Σ(s))^2/2\,ds - \int_0^t (Σ(t) - Σ(s))^2/2\,ds\bigr) \\ &= \exp\bigl(\int_t^u -φ(s)\,ds + \int_t^u (Σ(u) - Σ(s))^2/2\,ds + \int_0^t U(s)/2\,ds\bigr) \\ &= E[D_t(u)] \exp\bigl(-(Σ(u) - Σ(t))^2t/2 + \int_0^t U(s)/2\,ds\bigr) \\ \end{aligned} where {U(s) = (Σ(u) - Σ(s))^2 - (Σ(t) - Σ(s))^2 = Σ(u)^2 - Σ(t)^2 - 2(Σ(u) - Σ(t))Σ(s))}. We have {\int_0^t U(s)/2\,ds = (Σ(u)^2 - Σ(t)^2)t/2 - (Σ(u) - Σ(t)) \int_0^t Σ(s)\,ds} so \begin{aligned} E[D_t(u)] &= \frac{D(u)}{D(t)} \exp\bigl(-(Σ(u) - Σ(t))^2t/2 + (Σ(u)^2 - Σ(t)^2)t/2 + (Σ(u) - Σ(t))\int_0^t Σ(s)\,ds\bigr) \\ &= \frac{D(u)}{D(t)} \exp\bigl((Σ(u) - Σ(t))Σ(t) + (Σ(u) - Σ(t))\int_0^t Σ(s)\,ds\bigr) \\ \end{aligned}

Exercise. If σ is contant show D_t(u) = D(u)/D(t) \exp(-σ^2 ut(u - t)/2 - σ(u - t) B_t)

Now we determine the forward curve.

Exercise Show (\partial/\partial u)\int_t^u (Σ(u) - Σ(s))^2\,ds = 2σ(u)\int_t^u (Σ(u) - Σ(s))\,ds.

Hint. Use (\partial/\partial u)\int_t^u F(u,s)\,ds = F(u, u) + \int_t^u (\partial/\partial u) F(u, s)\,ds.

Solution

Let F(u,s) = (Σ(u) - Σ(s))^2 so (\partial/\partial u) F(u,s) = 2(Σ(u) - Σ(s)) σ(u) and {(d/du) \int_t^u (Σ(u) - Σ(s))^2\,ds = 0 + \int_t^u 2(Σ(u) - Σ(s)) σ(u)\,ds}.

Since D_t(u) = \exp(-\int_t^u f_t(s)\,ds) we have f_t(u) = φ(u) - σ(u)\int_t^u Σ(u) - Σ(s)\,ds + σ(u) B_t.

Exercise. If σ is constant show f_t(u) = φ(u) - σ^2(u - t)^2/2 + σ B_t.

The Ho-Lee model has a closed form solution for the dynamics of zero-coupon bond prices.

Options

A caplet with strike k and expiration t pays {\max\{f_t - k, 0\} = (f_t - k)^+} and a floorlet pays {(k - f_t)^+} at t. The risk-neutral value of a floorlet is p = E[(k - f_t)^+D_t].

Exercise Show p = E[(k + σ^2 t^2/2 - f_t)^+]D(t).

Hint. Recall E[f(M) e^N] = E[f(M + \operatorname{Cov}(M, N))]E[e^N] if M and N are jointly normal.

Solution

We have \operatorname{Cov}(f_t, \log D_t) = \operatorname{Cov}(σ B_t, -\int_0^t σ B_s\,ds) = -σ^2 \int_0^t s\,ds = -σ^2 t^2/2.

Note the floorlet value can be calculated using the Bachelier model. If F = f + sZ where Z is standard normal then E[(k - F)^+] = (k - F)\Phi(z) + sφ(z) where z = (k - f)/s, \Phi is the standard normal cumulative distribution, and φ = \Phi' is the standard normal density function.

Exercise. Find a closed form solution for the floorlet value E[(k - f_t)^+ D_t].

A forward contract is specified by an interval [t,u], a forward rate f, and a day count basis \delta. It has cash flows -1 at t and 1 + f\delta(t,u) at u where the day count fraction \delta(t,u) is approximately equal to the time in years from t to u.

Exercise. The price of the forward contract is zero at time s \le t if and only if f = (D_s(t)/D_s(u) - 1)/\delta(t,u)

Hint: 0 = E_s[-D_t + (1 + f\delta)D_u].

We call F_s^\delta(t,u) = (D_s(t)/D_s(u) - 1)/\delta(t,u) the par forward at time s over [t,u] for day count basis \delta.

Exercise Show E_s[F_t(t,u))\delta(t,u)D_u]] = E_s[D_t - D_u].

A forward contract paying in arrears is also specified by an interval [t,u], a forward rate f, and a day count basis \delta. It has a single cash flow (f - F_t^\delta(t,u))\delta(t,u) at u. Note F_t(t,u) is the forward rate at t over the interval [t, u]. The effective date of the contract is t and the termination date is u.

Exercise. Show E_s[-D_t + (1 + f\delta)D_u] = E_s[(f - F_t(t,u))\delta(t,u)D_u].

Hint: Use the previous exercise.

Suppose a fixed income instrument pays c_j at times u_j. Its value at time t is {P_t = \sum_{u_j > t} c_j D_t(u_j)}. If a European option pays g(P_t) at time t its value is E[g(P_t) D_t] = E[h(\dots, D_t(u_j),\dots) D_t]

We can approximate this with a lognormal having expected value {E[P_t] = \sum_{u_j > t} c_j E[D_t(u_j)]} and variance {\operatorname{Var}(P_t) = \sum_{u_j, u_k > t} c_j c_k \operatorname{Cov}(D_t(u_j), D_t(u_k))}. A European option paying g(P_t) at time t has value E[g(P_t)D_t]. Note g(P_T) = h(B_t) since D(t,u) is a function of B_t. The option value is E[h(B_t)D_t] = E[h(B_t + \int_0^t s σ(s)\,ds)] D(t).

Exercise. Show \int_0^t s σ(s)\,ds = t Σ(t) - \int_0^t Σ(s)\,ds.

Ho, Thomas S. Y., and Sang-Bin Lee. “Term Structure Movements and Pricing Interest Rate Contingent Claims.” The Journal of Finance 41, no. 5 (1986): 1011–29. https://doi.org/10.2307/2328161.

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Ho-Lee Model¹

Model

Dynamics

Options

Reference