I am trying to understand the proof of the zero bond price $Z(t)$ of the Ho-Lee model which is the unique solution of the following SDE:
$$ dZ(t) = -Z(t) [ \sigma(T-t)dW(t) + [ \int_t^T \alpha(t,u)du - \frac{1}{2} \sigma^2(T-t)^2]dt] \tag{1} $$
Now using the definition of $$Z(t) := \frac{P(t,T)}{B(t)} \tag{2}$$ where $$P(t,T)= e^{-\int_t^Tf(0,s)ds - \int_0^t \int_t^T \alpha(s,u)duds - \sigma W(t)(T-t)} \tag{3} $$ and $$ B(t) = e^{\int_0^t f(0,s)ds +\int_0^t \int_s^t \alpha(s,u)duds + \sigma \int_0^tW(s)ds} \tag{4} $$
I calculate $Z(t)$ as: $$ Z(t) = P(0,T) e^{\int_0^ta(s)dW(s) -\frac{1}{2}\int_0^ta(s)^2ds + \int_0^tb(s)ds} \tag{5} $$ where $$ a(s) = -\sigma (T-s) \tag{6} $$ $$b(s) = \frac{1}{2}a(s)^2 +\int_s^T\alpha(s,u)du \tag{7} $$
Now onto my question: I am trying to calculate $dZ(t)$ as follows: \begin{align} \frac{dZ(t)}{dt} &= \frac{d}{dt} [P(0,T). e^{Y(t)}] \\ &= Z(t). dY(t) \tag{8} \end{align}
where $Y(t)$ is the term in the exponent of (5)
However, my $\frac{dY(t)}{dt}$ is \begin{align} \frac{dY(t)}{dt} &= (a(t)-a(0))(W(t)-W(0)) - \frac{1}{2}(a(t)^2 - a(0)^2) +b(t)-b(0) \\ &= \sigma(t)W(t) + \int_t^T\alpha(t,u)du - \int_0^T\alpha(0,u)du \tag{9} \end{align}
which is clearly not consistent with what I want in (1).
Can anyone help me out in determining the mistake? How should I rectify my $dY(t)/dt$ in (9)?
A bit too complicated and confusing nomenclature/notation. The short rate in the Ho-Lee model is $$\tag{1} r_t=r_0+\Theta(t)+\sigma\,W_t\,,\quad\text{ or equivalently, }\quad dr_t=\theta(t)\,dt+\sigma\,dW_t $$ where $\Theta(t)=\int_0^t\theta(s)\,ds\,.$
Obviously $r_0$ can be absorbed into the function $\Theta(t)$ which I will do from now on. Then \begin{align} \int_0^t r_s\,ds&=\int_0^t\Theta(s)\,ds+\sigma\int_0^tW_s\,ds= \int_0^t\Theta(s)\,ds+\sigma\int_0^t(t-s)\,dW_s\\ &=\int_0^t\Theta(s)\,ds+\sigma\,t\,W_t-\sigma\int_0^ts\,dW_s\,.\tag{2} \end{align} Likewise \begin{align} \int_t^T r_s\,ds&= \int_t^T\Theta(s)\,ds+\sigma\,T\,W_T-\sigma\,t\,W_t-\sigma \int_t^T s\,dW_s\\ &=\int_t^T\Theta(s)\,ds+\sigma\,(T-t)W_t+\sigma \int_t^T(T-s)\,dW_s\,.\tag{3} \end{align} The zero bond price is \begin{align} P(t,T)&=\mathbb E\Big[\exp\Big(-\int_t^T r_s\,ds\Big)\Big|{\cal F}_t\Big]\\ &=\exp\Big(-\int_t^T \Theta(s)\,ds-\sigma\,(T-t)W_t\Big)\\ &\quad\times\mathbb E\Big[\exp\Big(-\sigma\int_t^T (T-s)\,dW_s\Big)\Big|{\cal F}_t\Big]\\ &=\exp\Big(-\int_t^T \Theta(s)\,ds-\sigma\,(T-t)W_t+\sigma\int_0^t (T-s)\,dW_s\Big)\\ &\quad\times\mathbb E\Big[\exp\Big(-\sigma\int_0^T (T-s)\,dW_s\Big)\Big|{\cal F}_t\Big]\,.\tag{4} \end{align} Let's concentrate on the conditional expectation in the last line. From \begin{align}\tag{5} \Big\langle\int_0^. (T-s)\,dW_s \Big\rangle_t=\int_0^t(T-s)^2\,ds=\frac{(t-T)^3+T^3}{3} \end{align} it follows that \begin{align}\tag{6} \exp\Big(-\sigma\int_0^t(T-s)\,dW_s-\sigma^2\frac{(t-T)^3+T^3}{6}\Big) \end{align} is a martingale in $t\in[0,T]\,.$ Therefore, \begin{align} &\mathbb E\Big[\exp\Big(-\sigma\int_0^T(T-s)\,dW_s\Big)\Big|{\cal F}_t\Big]\\ &= \exp\Big(-\sigma\int_0^t(T-s)\,dW_s-\sigma^2\frac{(t-T)^3+T^3}{6}\Big)\exp\Big(\sigma^2\frac{T^3}{6}\Big)\\ &=\exp\Big(-\sigma\int_0^t(T-s)\,dW_s-\sigma^2\frac{(t-T)^3}{6}\Big)\,.\tag{7} \end{align} In total $$\tag{8} \boxed{\quad P(t,T)=\exp\Big(-\int_t^T \Theta(s)\,ds-\sigma\,(T-t)W_t-\sigma^2\frac{(t-T)^3}{6}\Big)\,. \quad} $$ To get the SDE that this satisfies we use Ito's formula: \begin{align}\require{cancel} dP(t,T)&=P(t,T)\Big\{\Theta(t)+\sigma\,W_t-\cancel{\sigma^2\frac{(t-T)^2}{2}}\Big\}\,dt\\ &\quad-P(t,T)\,\sigma\,(T-t)\,dW_t+\cancel{P(t,T)\sigma^2\frac{(T-t)^2}{2}\,dt}\tag{9} \end{align} so that $$\tag{10} \boxed{\quad \frac{dP(t,T)}{P(t,T)}=\Big\{\Theta(t)+\sigma\,W_t\Big\}\,dt-\sigma\,(T-t)\,dW_t\,. \quad} $$ Note that this is equal to $$\tag{11} \boxed{\quad \frac{dP(t,T)}{P(t,T)}=r_t\,dt-\sigma\,(T-t)\,dW_t \quad} $$ as it should.
To find the HJM forward rates $f(t,T)$ that are related with the zero bond by $$\tag{12} P(t,T)=\exp\Big(-\int_t^Tf(t,s)\,ds\Big)\,, $$ we note that from (8) $$\tag{13} f(t,T)=-\frac{d}{dT}\log P(t,T)=\Theta(T)+\sigma\,W_t-\sigma^2\frac{(T-t)^2}{2}\,. $$ As expected, $f(t,t)=r_t$ and $$\tag{14} f(0,T)=\Theta(T)-\sigma^2\frac{T^2}{2}\,. $$ Since \begin{align}\tag{15} f(t,T)&=f(0,T)+\sigma^2\frac{T^2}{2}-\sigma^2\frac{(T-t)^2}{2}+\sigma\,W_t \end{align} the equations (8) and (12) can also be written as $$\tag{16} P(t,T) =\exp\Big(-\int_t^Tf(0,s)\,ds-\sigma\,(T-t)\,W_t-\sigma^2\frac{T^3-t^3-(T-t)^3}{6}\Big)\,. $$ A final remark on the drift of the forward rate $f(t,T)\,$: from HJM we know that this must be of the form $$\tag{17} \alpha(t,T)=\sigma(t,T)\int_t^T\sigma(t,s)\,ds $$ where $\sigma(t,T)$ is $f$'s volatility function. In the Ho-Lee model this is constant, $\sigma(t,T)=\sigma\,,$ so that $$\tag{18} \alpha(t,T)=\sigma^2\,(T-t)\,. $$ Then, from $\int_0^t\alpha(u,T)\,du=\sigma^2\frac{T^2-(T-t)^2}{2}\,$, we see that (15) can be written in the familiar HJM form $$\tag{19} f(t,T)=f(0,T)+\int_0^t\alpha(s,T)\,ds+\int_0^t\sigma(s,T)\,dW_s\,. $$ With the Ho-Lee $\alpha(t,T)$ from (18) we can also write (16) as $$\tag{20} \boxed{\, P(t,T) =\exp\Big(-\int_t^Tf(0,s)\,ds-\sigma\,(T-t)\,W_t-\int_t^T \int_0^t\alpha(u,s)\,du\,ds\Big)\,. \,} $$