## 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:

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