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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

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    $\begingroup$ Have a look of this question. $\endgroup$ – Gordon Feb 12 at 14:00
  • $\begingroup$ @Gordon thanks for the link, but that doesn't answer my question $\endgroup$ – JohnLord Feb 13 at 10:11
  • $\begingroup$ You have the bond price formula, then you can derive the SDE. $\endgroup$ – Gordon Feb 13 at 13:58
  • $\begingroup$ I see your point. but what about showing that the second $dt$ term in my post is equal to zero? $\endgroup$ – JohnLord Feb 13 at 14:45
  • $\begingroup$ 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. $\endgroup$ – Gordon Feb 13 at 15:26
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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.

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  • $\begingroup$ "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? $\endgroup$ – Sanjay Feb 13 at 18:06
  • $\begingroup$ 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. $\endgroup$ – Gordon Feb 13 at 18:45
  • $\begingroup$ 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 $\endgroup$ – JohnLord Feb 14 at 8:30
  • $\begingroup$ If you write down the partial derivative by the definition, you will know where you were wrong. $\endgroup$ – Gordon Feb 14 at 14:01

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