Deriving the Ho–Lee Bond Price Function with Itô Calculus

# Ho Lee model in Baxter&Rennie


# Ho Lee model in Baxter&Rennie












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

## Answer by Gordon (score 5)

https://quant.stackexchange.com/a/41461

> 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*}

## Answer by siwy9 (score 1)

https://quant.stackexchange.com/a/41349

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