## Ho Lee model

We derive the Ho Lee term structure model formulae.

### Identifying the Model

The short rate under the continuous time version of Ho Lee model has the following dynamics:

$$d r_{t}= \theta_{t} dt + \sigma d w_{t}$$

Where $$\sigma$$ is the volatility of the short rate, and the time dependent drift, $$\theta_{t}$$, is to be determined from the current term structure or the zero coupon bond prices. In the remainder of this sub-section, we use the current zero coupon bond prices to derive formula for $$\theta_{t}$$. The current zero coupon bond price can be represented as:

$$P \left( 0, T\right)=E \left[ e^{-\int_{0}^{T}{r_{t} dt}}\right]$$

Thus we need to solve the equation representing the dynamics of the short rate for $$r_{t}$$, then integrate $$r_{t}$$ over the maturity of the zero coupon bond, and then plug the resulting equation in the above formula. Integrating the short rate equation from 0 to t, we get

$$\int_{0}^{t}{d r_{u}}= \int_{0}^{t}{\theta_{u} du} + \int_{0}^{t}{\sigma d w_{u}}$$ $$r_{t}=r_{0} + \int_{0}^{t}{\theta_{u} du} + \sigma\int_{0}^{t}{d w_{u}}$$

Now integrating the short rate from 0 to T (maturity of the bond), we get

$$\int_{0}^{T}{r_{t} dt}=\int_{0}^{T}{r_{0} dt} + \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}+ \sigma\int_{0}^{T}{\int_{0}^{t}{d w_{u}} dt}$$ $$=r_{0} T + \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}+ \sigma\int_{0}^{T}{\int_{u}^{T}{dt d w_{u}} }$$ $$=r_{0} T + \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}+ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}}$$

Now, to facilitate the derivation of the Bond price formula, we derive the expected value and variance of $$\int_{s}^{T}{r_{t} dt}$$:

$$E \left[ \int_{0}^{T}{r_{t} dt} \right]=E \left[ r_{0} T + \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}+ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}} \right]$$ $$=r_{0} T + \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}$$ $$V \left[ \int_{0}^{T}{r_{t} dt} \right]=V \left[ r_{0} T + \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}+ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}} \right]$$ $$=V \left[ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}} \right]$$ $$={\sigma}^{2} \int_{0}^{T}{ {\left( T-u \right)}^{2} du}$$ $$= \frac{{\sigma}^{2}{T}^{3}}{3}$$

Now, we use the current bond prices/term structure to derive formula for $$\theta_{t}$$ in terms of current observables:

$$P \left( 0, T\right)=E \left[ e^{-\int_{0}^{T}{r_{t} dt}}\right]= e^{-E \left[ \int_{0}^{T}{r_{t} dt} \right]+\frac{1}{2}V \left[ \int_{0}^{T}{r_{t} dt} \right] }$$ $$=e^{-r_{0} T - \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}+{\sigma}^{2} \frac{{T}^{3}}{6}}$$

Taking log of both sides enable us to derive formula for $$\int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}$$ :

$$ln P \left( 0, T\right)=-r_{0} T - \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}+{\sigma}^{2} \frac{{T}^{3}}{6}$$ $$\int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}=-ln P \left( 0, T\right)-r_{0} T +\frac{{\sigma}^{2}{T}^{3}}{6}$$

We can now derive explicit expressions for the other terms involving $$\theta$$ :

$$\int_{0}^{T}{\theta_{u} du}=\frac {\partial} {\partial T} \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}$$ $$= \frac {\partial} {\partial T} \left(-ln P \left( 0, T\right)-r_{0} T +\frac{{\sigma}^{2}{T}^{3}}{6} \right)$$ $$= \frac {\partial} {\partial T} \left(-ln P \left( 0, T \right) \right) -r_{0} + \frac{{\sigma}^{2}{T}^{2}}{2}$$ $$= \frac {\partial} {\partial T} \left( \int_{0}^{T}{f \left( 0,u \right) du} \right)-r_{0} + \frac{{\sigma}^{2}{T}^{2}}{2}$$ $$= f \left( 0,T \right)-r_{0} + \frac{{\sigma}^{2}{T}^{2}}{2}$$ $$\theta_{T} =\frac {\partial} {\partial T} \int_{0}^{T}{\theta_{u} du}$$ $$=\frac {\partial} {\partial T} \left( f \left( 0,T \right)-r_{0} + \frac{{\sigma}^{2}{T}^{2}}{2} \right)$$ $$=\frac {\partial} {\partial T} f \left( 0,T \right) + {\sigma}^{2} T$$

Hence, the Ho Lee model can be represented as:

$$d r_{t}= \theta_{t} dt + \sigma d w_{t}=\left( \frac {\partial} {\partial t} f \left( 0,t \right) + {\sigma}^{2} t\right)dt + \sigma d w_{t}$$

The solution of which, given the current value $$r_{0}$$, is,

$$r_{t}=r_{0} + \int_{0}^{t}{\theta_{u} du} + \sigma\int_{0}^{t}{d w_{u}}$$ $$r_{t}=r_{0} + \left( f \left( 0,t \right)-r_{0} + \frac{{\sigma}^{2}{t}^{2}}{2} \right) + \sigma\int_{0}^{t}{d w_{u}}$$ $$r_{t}= f \left( 0,t \right)+ \frac{{\sigma}^{2}{t}^{2}}{2} + \sigma\int_{0}^{t}{d w_{u}}$$

Which is Gaussian with mean and variance given by,

$$E \left[ r_{t} \right]=E \left[ f \left( 0,t \right)+ \frac{{\sigma}^{2}{t}^{2}}{2} + \sigma\int_{0}^{t}{d w_{u}} \right]= f \left( 0,t \right)+ \frac{{\sigma}^{2}{t}^{2}}{2}$$ $$V \left[ r_{t} \right]= V \left[ f \left( 0,t \right)+ \frac{{\sigma}^{2}{t}^{2}}{2} + \sigma\int_{0}^{t}{d w_{u}} \right] = {\sigma}^{2}{t}$$

For completeness, we also show below the solution of the term structure

$$\int_{0}^{T}{r_{t} dt}=r_{0} T + \int_{0}^{T}{\int_{0}^{t}{\theta_{u} du} dt}+ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}}$$ $$=r_{0} T -ln P \left( 0, T\right)-r_{0} T +\frac{{\sigma}^{2}{T}^{3}}{6}+ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}}$$ $$=-ln P \left( 0, T\right)+\frac{{\sigma}^{2}{T}^{3}}{6}+ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}}$$

Which again is Gaussian with mean and variance given by,

$$E \left[ \int_{0}^{T}{r_{t} dt} \right]=E \left[ -ln P \left( 0, T\right)+\frac{{\sigma}^{2}{T}^{3}}{6}+ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}}\right]= -ln P \left( 0, T\right)+\frac{{\sigma}^{2}{T}^{3}}{6}$$ $$V \left[ \int_{0}^{T}{r_{t} dt} \right]= V \left[ -ln P \left( 0, T\right)+\frac{{\sigma}^{2}{T}^{3}}{6}+ \sigma\int_{0}^{T}{\left( T-u \right)d w_{u}} \right] = {\sigma}^{2}\int_{0}^{T}{{\left( T-u \right)}^{2}du} =\frac{{\sigma}^{2}{T}^{3}}{3}$$

It is easily verified that the current bond prices are recovered:

$$P \left( 0, T\right)=E \left[ e^{-\int_{0}^{T}{r_{t} dt}}\right]= e^{-E \left[ \int_{0}^{T}{r_{t} dt} \right]+\frac{1}{2}V \left[ \int_{0}^{T}{r_{t} dt} \right] }$$ $$=e^{ln P \left( 0, T\right)-\frac{{\sigma}^{2}{T}^{3}}{6} +\frac{{\sigma}^{2}{T}^{3}}{6}}=P \left( 0, T\right)$$