Merton model

We derive the Merton model term structure model formulae.

Alternative Derivation of Bond price formula

Now we present an alternative way of deriving the bond price formula. We assume that the price of T-maturity zero coupon bond at time t given current rate of \( r_{t} \) can be represented as:

$$ P \left( t, T \right)= A \left( t, T \right) e^{ -r_{t} B \left(t, T \right) }$$ $$ dP \left( t, T\right)= \frac {\partial P}{\partial t}dt + \frac {\partial P}{\partial r}dr + \frac{1}{2} \frac{\partial^2 P}{\partial r^2}{dr^2} $$

Now:

$$ \frac {\partial P}{\partial t}= \frac {\partial }{\partial t} \left( A e^{ -r B}\right)=e^{ -r B}\frac {\partial A}{\partial t} + A e^{ -r B}\frac {\partial}{\partial t} \left( -r B \right)$$ $$ = e^{ -r B}\frac {\partial A}{\partial t} -r A e^{ -r B}\frac {\partial B}{\partial t}=\frac{P}{A} \frac {\partial A}{\partial t} -r P\frac {\partial B}{\partial t} $$ $$ \frac {\partial P}{\partial r}= \frac {\partial }{\partial r} \left( A e^{ -r B}\right)= A e^{ -r B}\frac {\partial}{\partial r} \left( -r B \right)=-P B$$ $$ \frac {\partial^2 P}{\partial r^2}= \frac {\partial }{\partial r} \left( -P B\right)= -B\frac {\partial}{\partial r} \left( P \right)=P B^2$$

Thus:

$$ dP \left( t, T\right)= \frac {\partial P}{\partial t}dt + \frac {\partial P}{\partial r}dr + \frac{1}{2} \frac{\partial^2 P}{\partial r^2}{dr^2} $$ $$ dP \left( t, T\right)= \left( \frac{P}{A} \frac {\partial A}{\partial t} -r P\frac {\partial B}{\partial t} \right) dt + \left( -P B \right) dr + \frac{1}{2} \left( P B^2 \right) {dr^2} $$ $$ \frac{dP}{P}= \left( \frac{1}{A} \frac {\partial A}{\partial t} -r \frac {\partial B}{\partial t} \right) dt - B dr + \frac{1}{2} B^2 {dr^2} $$ $$ \frac{dP}{P}= \left(\frac{1}{A} \frac {\partial A}{\partial t} -r \frac {\partial B}{\partial t} \right) dt - B \left(\theta dt + \sigma d w_{t} \right) + \frac{1}{2} B^2 {\sigma}^2 dt $$ $$ \frac{dP}{P}= \left(\frac{1}{A} \frac {\partial A}{\partial t} -r \frac {\partial B}{\partial t} - \theta B + \frac{1}{2} B^2 {\sigma}^2\right) dt - \sigma B d w_{t} $$

The expected return of this under the risk neutral measure must be equal to the risk free rate:

$$ \frac{1}{A} \frac {\partial A}{\partial t} -r \frac {\partial B}{\partial t} - \theta B + \frac{1}{2} B^2 {\sigma}^2=r $$ $$ \frac{1}{A} \frac {\partial A}{\partial t} - \theta B + \frac{1}{2} B^2 {\sigma}^2=r \left( 1+ \frac {\partial B}{\partial t} \right) $$

As this equality holds for all values of r, which does not appear on the left hand side, it implies that

$$ \frac{1}{A} \frac {\partial A}{\partial t} - \theta B + \frac{1}{2} B^2 {\sigma}^2=0 $$ $$ 1+ \frac {\partial B}{\partial t} =0 $$

And given that the price of a zero coupon bond at maturity P(T,T)=1, the function \( P=A e^{-r B}\) implies B(T,T)=0 and A(T,T)=1. Now, rearranging and integrating the second equation from t to T, we get

$$ 1+ \frac {\partial B}{\partial t} =0 $$ $$ d B\left( t,T \right)=-dt $$ $$ \int_{t}^{T}{d B \left( u,T \right) } =-\int_{t}^{T}{du} $$ $$ B \left( T,T \right)-B \left( t,T \right) =-\left( T-t \right) $$ $$ B \left( t,T \right) = T-t $$

Substituting into the second equation and integrating, we get:

$$ \frac{1}{A} \frac {\partial A}{\partial t} - \theta \left( T-t \right) + \frac{1}{2} \left( T-t \right)^2 {\sigma}^2=0 $$ $$ \frac{1}{A} \frac {\partial A}{\partial t}= \theta \left( T-t \right) - \frac{1}{2} \left( T-t \right)^2 {\sigma}^2 $$ $$ \int_{t}^{T}{ \frac {d A \left( u,T \right)}{A \left( u,T \right)} } = \theta \int_{t}^{T}{\left( T-u \right) du} - \frac{{\sigma}^2}{2}\int_{t}^{T}{\left( T-u \right)^2 du} $$ $$ ln {A \left( T,T \right) } - ln {A \left( t,T \right) } = - \theta \frac{\left( T-t \right)^2}{2} + {{\sigma}^2}\frac{\left( T-t \right)^3}{6} $$ $$ A \left( t,T \right) =e^{\theta \frac{\left( T-t \right)^2}{2} - {{\sigma}^2} \frac{\left( T-t \right)^3}{6} } $$

Thus:

$$ P \left( t, T \right)= A \left( t, T \right) e^{ -r_{t} B \left(t, T \right) }$$ $$ P \left( t, T \right)= e^{\theta \frac{\left( T-t \right)^2}{2} - {{\sigma}^2} \frac{\left( T-u \right)^3}{6}} e^{ -r_{t} \left( T-t \right) } $$ $$ P \left( t, T \right)= e^{-r_{t} \left(T-t\right)+\theta \frac{\left( T-t \right)^2}{2} - {{\sigma}^2} \frac{\left( T-t \right)^3}{6}} $$