Deriving Compound Bond Prices from Hull-White Short-Rate Dynamics

# Pricing of the compound coupon bond with PDE


# Pricing of the compound coupon bond with PDE












I am now studying finance math using Steven E.Shereve's book. Using Interest Rate models, We can the price for zero-coupon with maturity price $1$ under Hull-White interest rate model[page 274] and Cox-Ingersoll-Ross positive constants model[page 275].

My question is that the formula for price of zero-coupon bond can be found in various literature, and I wonder why it is not covered in any book about a compound coupon bond pricing formula. Please propose the price of compound coupon in Hull-White interest rate model or let me know the relevant references. Compound coupon is a bond that does not pay interest before maturity and pays interest and principal at maturity, which is used in many financial sites, such as housing bonds.

My attempt to find price of compound coupon is as follows.

Statement Given probability space ($\Omega$, $F$, $\mathbb{P}$ ) and, under the Hull-White interest rate model for the short rate $r$, given by \begin{align*} dr_{t}=\left(b(t)-a r_{t} \right) dt + \sigma_{r} dW_{t}^{r} \end{align*} the no-arbitrage price at time $t$ of a compound bond expiring at time $T$ is as follows : $B(t,T,r) = \mathbb{E}^{*}(e^{\int_{t}^{T}{r_{s}ds}} \, \vert \, r_t = r)$ with the initial condition $B(T,T,r) = 1$. Then, $B(t,T,r)$ can be calculated as $B(t,T,r) = e^{-C(t,T)r-A(r,T)} $ where \begin{align*} C(t,T) &= \dfrac{1}{a} \left( {e^{a(t-T)-1}} \right) , \\ A(t,T) &= \dfrac{\sigma_{r}^{2}}{4 a^{3}} \left( - e^{-2a(T-t)} +4e^{-a(T-r)} +2a(T-t)-3 \right) + \int_{t}^{T}{\dfrac{b(s)}{a} \left( 1-e^{a(s-T)} \right) ds}. \end{align*}

Proof for Statement Under the Hull-White interest rate model given by \begin{align*} dr_{t}=\left(b(t)-a r_{t} \right) dt + \sigma_{r} dW_{t}^{r} \end{align*} the no-arbitrage price at time $t$ of a compound coupon bond with an expiry time $T$ is follows : \begin{align*} & B(t,T,r) = \mathbb{E}^{*}(e^{\int_{t}^{T}{r_{s}ds}} \, \vert \, r_t = r) \\ \end{align*} Then, by using Feynmann-Kac formula, the solution of $B(t,T,r)$ satisfies the following partial differential equation: \begin{align*} & \dfrac{\partial B}{\partial t} + \left( {b(t)-a r_t } \right) \dfrac{\partial B}{\partial r} + \dfrac{1}{2} \sigma_{r}^{2} \dfrac{\partial^{2}B}{\partial r^{2}} + r \, B = 0 \\ & B(T,T,r) = 1 \end{align*} and assuming the solution in the form of $B(t,T,r) = e^{-C(t,T)r-A(r,T)} $, We get two ordinary differential equations \begin{align*} & - \dfrac{d C(t,T)}{d t} + a C(t,T)+1 =0 \\ & - \dfrac{d A(t,T)}{d t}-b(t)C(t,T)+\dfrac{1}{2} \sigma_{r}^{2}{ \lbrace C(t,T) \rbrace } ^{2} = 0 \end{align*} $C(t,T)$ and $A(t,T)$ satisfying the above equation are as given in statement.

How about my try ? Thank you for reading it.