# Ho and lee derivation for short rates model

A silly question that is bugging me. I am working my way through Baxter and Rennie (again) and I am getting my wires crossed on the short rate models in particular the straight forward Ho and Lee analysis.

So given the SDE (under the $\mathbb{Q}$ measure) $$dr_t = \sigma dW_t +\theta_t dt$$ where $\theta_t$ is both deterministic and bounded and $\sigma$ is constant. This becomes $$r_t = f(0,t) + \sigma W_t +\int_0^t \theta_s ds$$ (I hope).

How do I compute the integral $$\int_t^T r_sds$$

Basically it comes down to computing $$\int_t^T W_sds$$ and $$\int_t^T \int_0^s \theta_k dk.$$ From first integral is just book work (though it be nice to see a derivation here other than by parts?) It is the later which I am not sure about, as the result is apparently $$\int_t^T (T-s)\theta_s ds$$ Which I am puzzled by?

$\textbf{edit}$ Actually an important piece of information is that I am trying to compute $$-\log\mathbb{E}_{\mathbb{Q}}\left(\mathrm{e}^{-\int_t^T r_sds}\vert r_t = x\right)=x(T-t) -\frac{1}{6}\sigma^2 (T-t)^3 + \int_0^T (T-s)\theta_sds$$

• For integral $\int_t^T\int_0^s \theta_k dk ds$, you can using the switching order technique. Note that the integration domain is a triangle, you need to pay some attention on the integral limits. – Gordon Dec 22 '14 at 14:21
• @gordon cheers for your comment . I changed the limits of the integrand (and get the functional form) but for the integral wrt s I seem to get (0,T)? Is that what you meant by pay attention :)? I feel it is a silly step that I have long forgotten from my analysis days!! – Chinny84 Dec 22 '14 at 14:35

For any $s \geq t$, note that \begin{align*} r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du. \end{align*} Then, \begin{align*} \int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds + \int_t^T \int_t^s\theta_u du ds\\ &=(T-t)r_t + \sigma\int_t^T\int_u^T ds\, dW_u +\int_t^T\int_u^T\theta_u ds du\\ &=(T-t)r_t + \sigma\int_t^T (T-u)dW_u +\int_t^T (T-u) \theta_u du. \end{align*} Moreover, \begin{align*} E_Q\Big(e^{-\int_t^T r_s ds} \mid r_t \Big) &= e^{-(T-t)r_t - \int_t^T (T-u) \theta_u du}E_Q\Big(e^{-\sigma\int_t^T (T-u)dW_u} \mid r_t\Big)\\ &=e^{-(T-t)r_t - \int_t^T (T-u) \theta_u du}e^{\frac{\sigma^2}{2}\int_t^T(T-u)^2 du} \\ &=e^{-(T-t)r_t - \int_t^T (T-u)\theta_u du + \frac{\sigma^2}{6}(T-t)^3}. \end{align*} That is, \begin{align*} -\ln E_Q\Big(e^{-\int_t^T r_s ds} \mid r_t \Big) = (T-t)r_t + \int_t^T (T-u)\theta_u du - \frac{\sigma^2}{6}(T-t)^3. \end{align*} If you really need the integral $\int_t^T\int_0^s \theta_u du ds$, you can proceed as follows: \begin{align*} \int_t^T\int_0^s \theta_u du ds &= \int_t^T\int_0^t \theta_u du ds + \int_t^T\int_t^s \theta_u du ds \\ &=(T-t)\int_0^t \theta_u ds + \int_t^T\int_u^T \theta_u ds du\\ &=(T-t)\int_0^t \theta_u ds + \int_t^T (T-u)\theta_u du. \end{align*}