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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:

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