# Determining bond price based on diffusion process for the short rate model

Suppose the diffusion process for the short rate $$r_t$$ under the risk-neutral measure $$Q$$ is given by: $$dr_t = \theta(t)dt+\sigma dZ_t$$

where $$Z_t$$ is a Brownian motion.

I am trying to show that the parameter $$\theta(t)$$ is related to the slope of forward rate curve: $$\theta(t) = \left. \frac{\partial F}{\partial T}(0,T) \right|_{T=t} + \sigma^2 t$$ where: $$\partial F(0,T) = -\frac{\partial \log P(0, T)}{T}$$

I am lost as to where to start. Any help would be appreciated

• Pls look up Ho Lee drift. Jun 1, 2021 at 11:06

(1) First, integrate twice the short rate SDE to get $$\int_0^t r_u du$$, you will find out that it's gaussian with this distribution: \begin{aligned} -\int_0^t r_udu &\sim \mathcal{N}(m_t = -r(0)t - \int_0^t \int_0^u \theta(v)dv, v_t = \frac{\sigma^2t^3}{6}) \end{aligned} This might help you get there: Integral of Brownian motion w.r.t. time
(2) compute the zero-coupon bond price given by the model: $$P(0,t) = \mathbb{E} \left[e^{-\int_0^t r_udu}\right]=e^{m_t + \frac{v_t}{2}}$$
(2) Then, take the logarithm and differentiate twice to get the desired expression: $$\theta(t) = \left. \frac{\partial F(0,T) }{\partial T} \right|_{T=t} + \sigma^2 t$$
• You can redo the same steps but with maturity $T$ and conditionally on $0 < t < T$ (as now you know $\theta(t)$ expression). Jun 2, 2021 at 9:52