Bond dynamics in Ho Lee model

The short rate in the Ho-Lee model is given by :

$$dr_t=\left( \frac{df(0,t)}{dt} +\sigma^2t\right)dt + \sigma dW_t$$

I'm trying to find the bond dynamics given by :

$$dP(t,T)/P(t,T)=r_tdt-\sigma(T-t)dW_t$$

I started from :

$$P(t,T)=E_t[e^{-\int_t^T r_sds}]$$

and I applied Itô to the function $$P(t,T)=\phi(t,r)$$:

$$d\phi(t,r) = \frac{\partial \phi(t,r)}{\partial t}dt+\frac{\partial \phi(t,r)}{\partial r} dr_t+ \frac{1}{2} \frac{\partial^2\phi(t,r)}{\partial r^2}(dr_t)^2$$

I computed the derivatives :

$$\frac{\partial \phi(t,r)}{\partial t}=r_tP(t,T)$$

$$\frac{\partial \phi(t,r)}{\partial r} = -(T-t)P(t,T)$$

$$\frac{1}{2} \frac{\partial^2\phi(t,r)}{\partial r^2} = (T-t)^2P(t,T)$$

Assembling everything I get :

$$dP(t,T)/P(t,T) = r_tdt-(T-t)\sigma dW_t +\left[ \frac{1}{2}(T-t)^2\sigma^2-(T-t)\left( \frac{df(0,t)}{dt}+\sigma^2t \right) \right] dt$$

I don't know how to get rid of the last $$dt$$ term. Any Help? Or did I get the derivatives wrong? I checked them several times but I don't see where the probem comes from. Thank you

• Have a look of this question. – Gordon Feb 12 '19 at 14:00
• @Gordon thanks for the link, but that doesn't answer my question – JohnLord Feb 13 '19 at 10:11
• You have the bond price formula, then you can derive the SDE. – Gordon Feb 13 '19 at 13:58
• I see your point. but what about showing that the second $dt$ term in my post is equal to zero? – JohnLord Feb 13 '19 at 14:45
• Your derivative $\frac{\partial \phi(t,r)}{\partial t}$ does not appear correct. When you take the derivative, you need to be mindful for the conditional expectation. – Gordon Feb 13 '19 at 15:26

1 Answer

When taking the partial derivative $$\frac{\partial}{\partial t}$$ in a conditional expectation, not only the parameter $$t$$ within the expectation needs to be considered, the information set $$\mathscr{F}_t$$ should also be considered.

For this particular question, based on an answer to this question, \begin{align*} P(t, T) = e^{-(T-t)r_t - \int_t^T (T-u)\theta_u du + \frac{\sigma^2}{6}(T-t)^3}, \end{align*} where \begin{align*} \theta_t &= \frac{df(0,t)}{dt} +\sigma^2t. \end{align*} Then, \begin{align*} \frac{\partial P(t, T)}{\partial t} &= P(t, T)\Big(r_t + (T-t) \theta_t -\frac{\sigma^2}{2}(T-t)^2\Big)\\ &= P(t, T)\Big(r_t + (T-t) \Big(\frac{df(0,t)}{dt} +\sigma^2t\Big) -\frac{\sigma^2}{2}(T-t)^2\Big). \end{align*} The remaining is now straightforward.

• "When taking the partial derivative $\frac{\partial}{\partial t}$ in a conditional expectation, not only the parameter $t$ within the expectation needs to be considered, the information set $\mathscr{F}_t$ should also be considered." Can you suggest good reading material on this topic. Namely how to take find derivatives of expectations and conditional expectation? – Sanjay Feb 13 '19 at 18:06
• Have a look of this question. However, I would not suggest this in practice, instead, you can compute the conditional expectation first, and then take the derivative. – Gordon Feb 13 '19 at 18:45
• Thank you for the answer. Is it possible to show how to take the derivative of the conditional expectation directly as I tried to do? That way you will correct me as clearly I'm doing something really wrong – JohnLord Feb 14 '19 at 8:30
• If you write down the partial derivative by the definition, you will know where you were wrong. – Gordon Feb 14 '19 at 14:01