title: Ho-Lee Model[^1]
author: Keith A. Lewis
institute: KALX, LLC
classoption: fleqn
fleqn: true
abstract: Normal short rate.
...
\newcommand\mb[1]{\mathbf{#1}}
\newcommand{\Var}{\operatorname{Var}}
\newcommand{\Cov}{\operatorname{Cov}}
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.
Exercise. Show $0 = E[(f_t - f(t))D_t]$.
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 $t$
Solution
We have $(d/dt)E[\exp(-\int_0^ t f_s\,ds)] = E[-D_t f_t]$ and
$(d/dt)\exp(-\int_0^t f(s)\,ds) = -D(t) f(t)$.
The result follows from $D(t) = E[D_t]$.
This shows $f(t)$ is the par coupon of a forward contract paying $f_t - f(t)$ at $t$.
Exercise. Show $0 = E_t[(f_u - f_t(u))D_u]$, $t\le u$.
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 ${f(t)E[D_t] = E[f_t D_t] = E[f_t]E[D_t] + \Cov(f_t, D_t)}$ so
${φ(t) - f(t) = -\Cov(f_t, D_t)/D(t)}$. 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.
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 $\Var(\int_0^t (Σ(t) - Σ(s))\cdot dB_s) = \int_0^t |Σ(t) - Σ(s)|^2,ds$.
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
$E[\exp(N)] = \exp(E[N] + \Var(N)/2)$ 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 $\Cov(f_t, D_t)/D(t) = -σ(t)\cdot \int_0^t sσ(s),ds$.
Hint: Use $E[f(N) \exp(M)] = E[f(N + \Cov(N,M))] E[\exp(M)]$
if $N$ and $M$ are jointly normal to show
$\Cov(N, \exp(M)) = \Cov(N,M) E[\exp(M)]$. Recall $\Cov(B_t, B_s) = sI$ for $s\le t$.
Solution
We have $\Cov(f_t, D_t) = \Cov(σ(t)\cdot B_t,-\int_0^t σ(s)\cdot dB_s) E[D_t]
= -σ(t) \cdot \int_0^t σ(s)s\,ds\,D(t)$.
Exercise. If $σ$ is constant show $\Cov(f_t, D_t)/D(t) = -|σ|^2 t^2/2$.
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 $\Var(\int_t^u Σ(u) - Σ(s),dB_s) = \int_t^u (Σ(u) - Σ(s))^2,ds$.
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 $\Var(\log D_t) = (Σ(u) - Σ(t))^2 t$
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.
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 $E[f(M) e^N] = E[f(M + \Cov(M, N))]E[e^N]$ if $M$ and $N$ are jointly normal.
Solution
We have $\Cov(f_t, \log D_t) = \Cov(σ B_t, -\int_0^t σ B_s\,ds)
= -σ^2 \int_0^t s\,ds = -σ^2 t^2/2$.
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 ${\Var(P_t) = \sum_{u_j, u_k > t} c_j c_k \Cov(D_t(u_j), D_t(u_k))}$.
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 $\int_0^t s σ(s),ds = t Σ(t) - \int_0^t Σ(s),ds$.
[^1]:
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.