Aquí ofrecemos otra respuesta utilizando el cálculo de Ito. Parece complicado, pero también tiene su interés.
Dada la dinámica de los tipos cortos \begin{align*} dr_t = \nu(r_t, t) dt + \rho(r_t, t) dW_t, \end{align*} definimos la función \begin{align*} g(x, t, T) = -\ln E\left(e^{-\int_t^T r_s ds} \,\big|\, r_t = x\right). \end{align*} El tipo de interés a plazo $f(t, T)$ se define entonces por \begin{align*} f(t, T) = \frac{\partial g}{\partial T}(r_t, t, T). \end{align*} Usando el lema de Ito, \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*} donde \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*} Entonces \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*} es decir, \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*} Tenga en cuenta que \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*} Para más detalles, véase el anexo. Entonces, \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*} Obsérvese que, también podemos obtener la ecuación $(1)$ utilizando la EDP para el precio del bono $ P$ (ver PDE para la determinación del precio de los derivados de tipos de interés ) y luego hacer la sustitución $P=e^{-g}$ .
Además, hay que tener en cuenta que \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*} En el modelo Ho-Lee, $\rho(r_t, t) = \sigma$ y $\nu(r_t, t)=\theta_t$ . Entonces, para cualquier $t>0$ , \begin{align*} Var(r_t - f(t, t)) = \int_0^t E\left(\sigma(s, t)-\sigma \right)^2ds = 0 \end{align*} Eso es, \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*} Desde $(1)$ , \begin{align*} \frac{1}{2} \sigma^2 (T-t)^2 = \frac{\partial g}{\partial t} +r_t + (T-t) \theta_t, \end{align*} Por lo tanto, \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*} Eso es, \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*}
Anexo
Observamos que \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*}