# Getting $df(t,T)$ when given $d\ln P(t,T)$ and $f(t,T)=-\frac{\partial}{\partial T} \ln P(t,T)$

Let the HJM dynamics of $$\ln P(t,T)$$ (log of bond prices) given by (In the risk neutral measure ) :

$$d \ln P(t,T) = \mathcal{O}( dt) - \sigma_P (t,T) dW(t)$$

Knowing that $$f(t,T)=-\frac{\partial}{\partial T} \ln P(t,T)$$ I want to compute $$df(t,T)$$. (The dynamics of the instantaneous forward rate)

For that I tried applying Itô, but I'm stuck at defining the variables driving $$f(t,T)$$. I usually define a $$\phi$$ depending on time and the random variable and then apply Itô. But here I'm confused as I could choose time being $$t$$ or $$T$$.

So my first question is : what is the rule of thumb for defining $$\phi$$ to which I apply Itô?

My second question : is applying Itô the right way to get $$df(t,T)$$ ?

Thanks

• You need to treat $t$ and $T$ as independent. That is, for differential with respect $t$, you can treat $T$ as a constant. Commented Feb 27, 2019 at 17:06
• @Gordon could you show how to get $df(t,T)$ by computation? Commented Feb 28, 2019 at 13:47

We assume that \begin{align*} d \ln P(t,T) = \mu(t, T) dt - \sigma (t,T) dW(t). \end{align*} Then, \begin{align*} \ln P(t,T) = \ln P(0,T) + \int_0^t \mu(s, T) ds - \int_0^t \sigma (s,T) dW(s). \end{align*} Moreover, \begin{align*} f(t, T) &= -\frac{\partial\ln P(t,T)}{\partial T} \\ &= -\frac{\partial\ln P(0,T)}{\partial T} - \int_0^t \frac{\partial\mu(s, T)}{\partial T} ds + \int_0^t \frac{\partial\sigma (s,T)}{\partial T} dW(s), \end{align*} and \begin{align*} d f(t,T) = \frac{\partial\mu(t, T)}{\partial T} dt + \frac{\partial\sigma (t,T)}{\partial T} dW(t). \end{align*}
Note that, under the risk-neutral probability measure, \begin{align*} \mu(t, T) = r_t - \frac{1}{2}\sigma^2 (t,T). \end{align*} Then, \begin{align*} d f(t,T) = \sigma(t, T)\frac{\partial\sigma(t, T)}{\partial T} dt + \frac{\partial\sigma (t,T)}{\partial T} dW(t). \end{align*}