I am currentyl reading Baxter&Rennie and I have a difficulty with understanding a derivation of formula for one function, $g(x,t,T)$ (this can be found on page 152 in the book). I know that there is a solution here Ho and lee derivation for short rates model. However, the authors say that the formula can be obtained by Ito, though I cannot see how I could use it in this case.
Thank you for your help
Here we provide another answer using Ito's calculus. It appears involved, but it also has its own interest.
Given the short rate dynamics \begin{align*} dr_t = \nu(r_t, t) dt + \rho(r_t, t) dW_t, \end{align*} we define the function \begin{align*} g(x, t, T) = -\ln E\left(e^{-\int_t^T r_s ds} \,\big|\, r_t = x\right). \end{align*} The forward rate $f(t, T)$ is then defined by \begin{align*} f(t, T) = \frac{\partial g}{\partial T}(r_t, t, T). \end{align*} Using Ito's lemma, \begin{align*} df(t, T) &= \frac{\partial^2 g}{\partial t \partial T} dt + \frac{\partial^2 g}{\partial r_t \partial T}dr_t + \frac{1}{2}\frac{\partial^3 g}{\partial^2 r_t \partial T}d\langle r, r\rangle_t\\ &=\left(\frac{\partial^2 g}{\partial t \partial T} + \frac{\partial^2 g}{\partial r_t \partial T} \nu(r_t, t) + \frac{1}{2}\rho(r_t, t)^2\frac{\partial^3 g}{\partial^2 r_t \partial T} \right)dt + \rho(r_t, t)\frac{\partial^2 g}{\partial r_t \partial T}dW_t\\ &= \sigma(t, T)\Sigma(t, T)dt + \sigma(t, T) dW_t, \end{align*} where \begin{align*} \sigma(t, T) &= \rho(r_t, t)\frac{\partial^2 g}{\partial r_t\partial T}(r_t, t, T), \\ \Sigma(t, T) &= \int_t^T\sigma(t, s) ds = \rho(r_t, t)\frac{\partial g}{\partial r_t}(r_t, t, T),\\ \sigma(t, T)\Sigma(t, T) &= \frac{\partial^2 g}{\partial t \partial T} + \frac{\partial^2 g}{\partial r_t \partial T} \nu(r_t, t) + \frac{1}{2}\rho(r_t, t)^2\frac{\partial^3 g}{\partial^2 r_t \partial T}. \end{align*} Then \begin{align*} \rho(r_t, t)^2\frac{\partial g}{\partial r_t}(r_t, t, T)\frac{\partial^2 g}{\partial r_t\partial T}(r_t, t, T) = \frac{\partial^2 g}{\partial t \partial T} + \frac{\partial^2 g}{\partial r_t \partial T} \nu(r_t, t) + \frac{1}{2}\rho(r_t, t)^2\frac{\partial^3 g}{\partial^2 r_t \partial T}, \end{align*} that is, \begin{align*} \frac{1}{2}\rho(r_t, t)^2\frac{\partial }{\partial T}\left[\left(\frac{\partial g}{\partial r_t}\right)^2 \right] = \frac{\partial^2 g}{\partial t \partial T} + \frac{\partial^2 g}{\partial r_t \partial T} \nu(r_t, t) + \frac{1}{2}\rho(r_t, t)^2\frac{\partial^3 g}{\partial^2 r_t \partial T}. \end{align*} Note that \begin{align*} \int_t^T \frac{\partial }{\partial u}\left[\left(\frac{\partial g}{\partial r_t}(r_t, t, u)\right)^2 \right]du &= \left(\frac{\partial g}{\partial r_t}(r_t, t, T)\right)^2,\\ \int_t^T \frac{\partial^2 g}{\partial t \partial u}(r_t, t, u) du &= \lim_{s\rightarrow t+} \frac{\partial }{\partial t}\int_s^T \frac{\partial g}{\partial u}(r_t, t, u) du\\ &= \lim_{s\rightarrow t+} \frac{\partial }{\partial t} \big(g(r_t, t, T) -g(r_t,t, s)\big)\\ &=\frac{\partial g}{\partial t} +r_t. \end{align*} See Addendum below for more details. Then, \begin{align*} \frac{1}{2}\rho(r_t, t)^2\left(\frac{\partial g}{\partial r_t}\right)^2 = \frac{\partial g}{\partial t} +r_t + \frac{\partial g}{\partial r_t} \nu(r_t, t) + \frac{1}{2}\rho(r_t, t)^2\frac{\partial^2 g}{\partial^2 r_t}. \tag{1} \end{align*} Note that, we can also obtain Equation $(1)$ using the PDE for the bond price $ P$ (see PDE for Pricing Interest Rate Derivatives) and then make the substitution $P=e^{-g}$.
Moreover, note that \begin{align*} r_t &= f(t, t)\\ &=f(0, t) - \int_0^t \sigma(s, t)\Sigma(s, t)ds + \int_0^t \sigma(s, t) dW_s. \end{align*} In Ho-Lee model, $\rho(r_t, t) = \sigma$ and $\nu(r_t, t)=\theta_t$. Then for any $t>0$, \begin{align*} Var(r_t - f(t, t)) = \int_0^t E\left(\sigma(s, t)-\sigma \right)^2ds = 0 \end{align*} That is, \begin{align*} \sigma(t, T) &= \sigma,\\ \frac{\partial^2 g}{\partial r_t\partial T}(r_t, t, T) &= 1, \\ \frac{\partial g}{\partial r_t}(r_t, t, T) &= T-t. \end{align*} From $(1)$, \begin{align*} \frac{1}{2} \sigma^2 (T-t)^2 = \frac{\partial g}{\partial t} +r_t + (T-t) \theta_t, \end{align*} Therefore, \begin{align*} \frac{\sigma^2}{6} (T-t)^3 = g(r_t, T, T) - g(r_t, t, T) +r_t(T-t) + \int_t^T (T-s) \theta_s ds. \end{align*} That is, \begin{align*} g(r_t, t, T) &= r_t(T-t) - \frac{\sigma^2}{6} (T-t)^3 + \int_t^T (T-s) \theta_s ds. \end{align*}
Addendum
We note that \begin{align*} \lim_{s\rightarrow t+} \frac{\partial }{\partial t}g(r_t, t, s) &=\lim_{s\rightarrow t+}\lim_{\delta \rightarrow 0+}\frac{-\ln E\left(e^{-\int_{t+\delta}^s r_udu}\mid \mathcal{F}_{t+\delta} \right) + \ln E\left(e^{-\int_t^s r_udu}\mid \mathcal{F}_t \right)}{\delta}\\ &=\lim_{s\rightarrow t+}\lim_{\delta \rightarrow 0+}\frac{-\ln E\left(e^{\int_t^{t+\delta} r_udu}e^{-\int_t^s r_udu}\mid \mathcal{F}_{t+\delta} \right) + \ln E\left(e^{-\int_t^s r_udu}\mid \mathcal{F}_t \right)}{\delta}\\ &=\lim_{s\rightarrow t+}\lim_{\delta \rightarrow 0+}\frac{-\int_t^{t+\delta} r_udu -\ln E\left(e^{-\int_t^s r_udu}\mid \mathcal{F}_{t+\delta} \right) + \ln E\left(e^{-\int_t^s r_udu}\mid \mathcal{F}_t \right)}{\delta}\\ &=\lim_{\delta \rightarrow 0+}\lim_{s\rightarrow t+}\frac{-\int_t^{t+\delta} r_udu -\ln E\left(e^{-\int_t^s r_udu}\mid \mathcal{F}_{t+\delta} \right) + \ln E\left(e^{-\int_t^s r_udu}\mid \mathcal{F}_t \right)}{\delta}\\ &=-r_t. \end{align*}
OK, so I think I have figured it out. I assumed that we need to use Ito's lemma here, however, it seems the authors mean to use Ito's isometry, which must be used to prove below equality $$\mathbb{E}_Q\Big(e^{-\sigma\int_t^T (T-u)dW_u} \mid r_t\Big)=e^{\frac{\sigma^2}{2}\int_t^T(T-u)^2 du}$$
We know that for normal-distibuted random variable $X$ (with mean $\mu$ and variance $\sigma ^2$) and $\lambda \in \mathbb{R}$ below formula holds: $$\mathbb{E}(e^{\lambda X} )= \exp \left(-\lambda \mu+ \frac{1}{2} \lambda^2 \sigma^2 \right)$$ from moment-generating fuction theorem.
The expected value of $-\sigma\int_t^T (T-u)dW_u$ is $0$.
The variance can be obtained from formula $$\mathbb{D}^2X=\mathbb{E}X^2-(\mathbb{E}X)^2$$
From Ito's isometry (this is the step with Ito I was looking for):
$$\mathbb{E}\left( -\sigma\int_t^T (T-u)dW_u \right)^2=\mathbb{E}\left( \sigma^2\int_t^T (T-u)^2du \right)=\sigma^2\int_t^T (T-u)^2du$$ since $\sigma^2\int_t^T (T-u)^2du$ is constant. Having $\mu$ and $\sigma^2$ we can apply formula for expected value from MGF and get the first equality