Deriving Forward-Rate Dynamics from Bond-Price Dynamics

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


# 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

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/44332

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*}