## Hull White model

We derive the one factor Hull White term structure model formulae.

### Identifying the model

The short rate under the One Factor Hull White model has the following dynamics:

$$d r_{t}= \kappa \left( \theta_{t}-r_{t} \right) dt + \sigma d w_{t}$$

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 zero coupon bond prices. The derivation of Hull White short rate specification is simplified when the short rate is decomposed into two components:

$$r_{t}= x_{t} + \alpha_{t}$$

Where $$d x_{t}= -\kappa x_{t} dt + \sigma d w_{t}$$ with $$x_{0}=0$$ , and $$\alpha_{t}$$ is deterministic function of t.

Thus:

$$d r_{t} = d x_{t}+d \alpha_{t}$$ $$d r_{t} = -\kappa x_{t} dt + \sigma d w_{t}+d \alpha_{t}$$ $$d r_{t} = -\kappa \left( r_{t}- \alpha_{t}\right) dt + \sigma d w_{t}+d \alpha_{t}$$ $$d r_{t} = \kappa \left( \alpha_{t} + \frac{1}{\kappa} \frac{\partial \alpha_{t}}{\partial t} - r_{t} \right) dt + \sigma d w_{t}$$

Comparing this to the equation we started with reveals that our yet to be determined $$\theta_{t}$$ is $$\theta_{t}=\alpha_{t} + \frac{1}{\kappa} \frac{\partial \alpha_{t}}{\partial t}$$

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:

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

To determine $$-\int_{0}^{T}{x_{t} dt}$$ we need to solve $$d x_{t}= -\kappa x_{t} dt + \sigma d w_{t}$$ and then integrate $$x_{t}$$ from 0 to t

$$d x_{t}= -\kappa x_{t} dt + \sigma d w_{t}$$ $$d x_{t} + \kappa x_{t} dt = \sigma d w_{t}$$ $$e^{\kappa t} d x_{t} + \kappa e^{\kappa t} x_{t} dt = e^{\kappa t} \sigma d w_{t}$$ $$d \left( e^{\kappa t} x_{t} \right) = e^{\kappa t} \sigma d w_{t}$$ $$\int_{0}^{t} {d \left( e^{k u} x_{u} \right)} = \sigma\int_{0}^{t} {e^{k u} d w_{u}}$$ $$e^{\kappa t} x_{t} -e^{0} x_{0} = \sigma\int_{0}^{t} {e^{k u} d w_{u}}$$ $$x_{t} = x_{0}e^{-\kappa t} +\sigma\int_{0}^{t} {e^{-k \left( t- u \right) } d w_{u}}$$ $$x_{t} = \sigma\int_{0}^{t} {e^{-k \left( t- u \right) } d w_{u}}$$

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

$$\int_{0}^{T}{x_{t} dt}= \sigma\int_{0}^{T}{\int_{0}^{t}{ e^{-k \left( t- u \right)} d w_{u}} dt}$$ $$=\sigma\int_{0}^{T}{\int_{u}^{T}{ e^{-k \left( t- u \right)} dt d w_{u}} }$$ $$= \frac{\sigma}{\kappa} \int_{0}^{T}{ \left( 1- e^{-k \left( T- u \right)} \right) d w_{u}}$$

Now, to facilitate the derivation of the Bond price formula, we derive the expected value and variance of $$\int_{0}^{T}{x_{t} dt}$$:

$$E \left[ \int_{0}^{T}{x_{t} dt} \right]=E \left[ \frac{\sigma}{\kappa} \int_{0}^{T}{ \left( 1- e^{-k \left( T- u \right)} \right) d w_{u}} \right]=0$$ $$V \left[ \int_{0}^{T}{x_{t} dt} \right]=V \left[ \frac{\sigma}{\kappa} \int_{0}^{T}{ \left( 1- e^{-k \left( T- u \right)} \right) d w_{u}} \right]$$ $$=\frac{{\sigma}^2}{{\kappa}^2} \int_{0}^{T}{ \left( 1- e^{-k \left( T- u \right)} \right)^{2} du}$$ $$=\frac{{\sigma}^2}{{\kappa}^2} \int_{0}^{T}{ \left( 1+e^{-2 k \left( T- u \right)}-2 e^{-k \left( T- u \right)} \right) du}$$ $$= \frac{{\sigma}^2}{{\kappa}^2} \int_{0}^{T}{du}+\frac{{\sigma}^2}{{\kappa}^2} \int_{0}^{T}{ e^{-2 k \left( T- u \right)} du}-\frac{2{\sigma}^2}{{\kappa}^2} \int_{0}^{T}{ e^{-k \left( T- u \right)} du}$$ $$= \frac{{\sigma}^2 T}{{\kappa}^2} + \frac{{\sigma}^2}{2{\kappa}^3} \left( 1-e^{-2 \kappa T} \right) -\frac{2{\sigma}^2}{{\kappa}^3} \left( 1-e^{-\kappa T}\right)$$ $$= \frac{{\sigma}^2}{2{\kappa}^3} \left( 2 \kappa T+ 1-e^{-2 \kappa T} -4 + 4 e^{-\kappa T} \right)$$ $$= \frac{{\sigma}^2}{2{\kappa}^3} \left( 2 \kappa T -3 -e^{-2 \kappa T} + 4 e^{-\kappa T} \right)$$

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

$$P \left( 0, T\right)=e^{-\int_{0}^{T}{\alpha_{t} dt}}E \left[ e^{-\int_{0}^{T}{x_{t} dt}}\right] = e^{-\int_{0}^{T}{\alpha_{t} dt}} e^{-E \left[ \int_{0}^{T}{x_{t} dt} \right]+\frac{1}{2}V \left[ \int_{0}^{T}{x_{t} dt} \right] }$$ $$= e^{-\int_{0}^{T}{\alpha_{t} dt}} e^{0 + \frac{{\sigma}^2}{4{\kappa}^3} \left( 2 \kappa T -3 -e^{-2 \kappa T} + 4 e^{-\kappa T} \right) }$$ $$= e^{-\int_{0}^{T}{\alpha_{t} dt}+ \frac{{\sigma}^2}{4{\kappa}^3} \left( 2 \kappa T -3 -e^{-2 \kappa T} + 4 e^{-\kappa T} \right) }$$

Taking log of both sides enables us to derive formula for $$\int_{0}^{T}{\alpha_{t} dt}$$ :

$$ln P \left( 0, T\right)=-\int_{0}^{T}{\alpha_{t} dt}+ \frac{{\sigma}^2}{4{\kappa}^3} \left( 2 \kappa T -3 -e^{-2 \kappa T} + 4 e^{-\kappa T} \right)$$ $$\int_{0}^{T}{\alpha_{t} dt}=-ln P \left( 0, T\right)+ \frac{{\sigma}^2}{4{\kappa}^3} \left( 2 \kappa T -3 -e^{-2 \kappa T} + 4 e^{-\kappa T} \right)$$

Differentiating twice with respect to T :

$$\frac {\partial} {\partial T} {\int_{0}^{T}{\alpha_{t} dt}}=\frac {\partial} {\partial T} \left( -ln P \left( 0, T\right)+ \frac{{\sigma}^2}{4{\kappa}^3} \left( 2 \kappa T -3 -e^{-2 \kappa T} + 4 e^{-\kappa T} \right) \right)$$ $$\alpha_{T}=\frac {\partial} {\partial T} \left( \int_{0}^{T}{f \left( 0,u \right) du} \right)+ \frac{{\sigma}^2}{4{\kappa}^3} \frac {\partial} {\partial T} \left( 2 \kappa T -3 -e^{-2 \kappa T} + 4 e^{-\kappa T} \right)$$ $$=f \left( 0,T \right)+ \frac{{\sigma}^2}{4{\kappa}^3} \left( 2 \kappa + 2 \kappa e^{-2 \kappa T} - 4 \kappa e^{-\kappa T} \right)$$ $$\frac {\partial \alpha_{T}}{\partial T} =\frac {\partial} {\partial T} f \left( 0,T \right)+\frac{{\sigma}^2}{4{\kappa}^3} \frac {\partial} {\partial T} \left( 2 \kappa + 2 \kappa e^{-2 \kappa T} - 4 \kappa e^{-\kappa T} \right)$$ $$=\frac {\partial f \left( 0,T \right)} {\partial T} + \frac{{\sigma}^2}{4{\kappa}^3} \left( - 4 {\kappa}^2 e^{-2 \kappa T} + 4 {\kappa}^2 e^{-\kappa T} \right)$$ $$=\frac {\partial f \left( 0,T \right)} {\partial T} + \frac{{\sigma}^2}{\kappa} \left( e^{-\kappa T} - e^{-2 \kappa T} \right)$$

Hence, the Hull White model can be represented as:

$$\alpha_{t} + \frac{1}{\kappa} \frac{\partial \alpha_{t}}{\partial t} =f \left( 0,t \right)+\frac{{\sigma}^2}{4{\kappa}^3} \left( 2 \kappa + 2 \kappa e^{-2 \kappa t} - 4 \kappa e^{-\kappa t} \right) +\frac{1}{\kappa} \frac{\partial f \left( 0,t \right)} {\partial t} + \frac{{\sigma}^2}{{\kappa}^2} \left( e^{-\kappa t} - e^{-2 \kappa t} \right)$$ $$=f \left( 0,t \right)+\frac{1}{\kappa} \frac{\partial f \left( 0,t \right)} {\partial t} + \frac{{\sigma}^2}{4{\kappa}^2} \left( 2 + 2 e^{-2 \kappa t}- 4 e^{-\kappa t} + 4 e^{-\kappa t} - 4 e^{-2 \kappa t} \right)$$ $$=f \left( 0,t \right)+\frac{1}{\kappa} \frac{\partial f \left( 0,t \right)} {\partial t} + \frac{{\sigma}^2}{2{\kappa}^2} \left( 1 - e^{-2 \kappa t} \right)$$ $$d r_{t} = \kappa \left( \alpha_{t} + \frac{1}{\kappa} \frac{\partial \alpha_{t}}{\partial t} - r_{t} \right) dt + \sigma d w_{t}$$ $$d r_{t} = \kappa \left( f \left( 0,t \right)+\frac{1}{\kappa} \frac{\partial f \left( 0,t \right)} {\partial t} + \frac{{\sigma}^2}{2{\kappa}^2} \left( 1 - e^{-2 \kappa t} \right)- r_{t} \right) dt + \sigma d w_{t}$$