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The following RN dynamics of a ZCB maturing at time is given:

$$\frac{dZ(t,T)}{Z(t,T)} = r_tdt + \sigma_Z(t,T)dX_t$$

and the forward rate is given:

$$f(t,T,T+\delta) = \frac{ln(Z(t,T)) - ln(Z(t,T,T+\delta))}{\delta}$$


How to use Ito lemma to get the SDE for forward rate as follows?:

$$df(t,T) = \frac{(\sigma_Z(t,T))^2 - (\sigma_Z(t,T))^2}{2\delta}dt + \frac{\sigma_Z(t,T) - \sigma_Z(t,T)}{\delta}dX_t$$

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  • $\begingroup$ The final SDE you have in the question as written is wrong (it is trivially zero), which can be seen as the left hand side has no dependence on $\delta$ whereas the right hand side does. Furthermore, in the right hand side both terms are zero. $\endgroup$ – oliversm Sep 5 at 15:08
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A definition

We note that $f(t;T)$ is defined as $$ f(t;T) = \lim_{\delta \to 0} f(t;T,t+\delta) \equiv -\frac{1}{Z(t;T)} \frac{\partial}{\partial T}Z(t;T). $$

We know the solution for $Z$

We know that the solution for the ZCB is given by the stochastic/Dolean exponential $$ Z(t;T) = Z(t_0;T)\exp\left(\int_{t_0}^t \left(r(s) - \frac{\sigma^2(s;T)}{2}\right)\mathrm{d}s + \int_{t_0}^t \sigma(s;T) \mathrm{d}X(s) \right) $$ for $t \geq t_0$, where for brevity we have dropped the $Z$ in $\sigma_Z$.

Combining the results

Putting the solution for $Z$ into the equation for $f$ gives \begin{align} f(t;T) & = -\frac{\partial}{\partial T}\left(\int_{t_0}^t \left(r(s) - \frac{\sigma^2(s;T)}{2}\right)\mathrm{d}s + \int_{t_0}^t \sigma(s;T) \mathrm{d}X(s) \right) \\ & = -\int_{t_0}^t \frac{\partial}{\partial T}\left(\frac{\sigma^2(s;T)}{2}\right)\mathrm{d}s - \int_{t_0}^t \frac{\partial}{\partial T}\sigma(s;T) \mathrm{d}X(s) \end{align} from which it we can read off the infinitesimal change $$ \mathrm{d}f(t;T) = -\frac{\partial}{\partial T}\left(\frac{\sigma^2(t;T)}{2}\right)\mathrm{d}t - \frac{\partial}{\partial T}\sigma(t;T) \mathrm{d}X(t). $$

So if the above is what you meant by the expression $f(t;T)$ then this is the desired result.

Returning to a finite time perturbation of size $\delta$

If you wish to reinsert some small $\delta$ term as a perturbation from $T$, then we use the reverse our definition of the partial derivative where $$ -\frac{\partial}{\partial T} g(t;T) \equiv \lim_{\delta \to 0} \frac{g(t;T) - g(t;T+\delta)}{\delta}. $$ Doing this for our expression for $f$ gives $$ \mathrm{d}f(t;T,T+\delta) = \left(\frac{\sigma^2(t;T) - \sigma^2(t;T+\delta)}{2\delta}\right)\mathrm{d}t + \left(\frac{\sigma(t;T) - \sigma(t;T+\delta)}{\delta}\right) \mathrm{d}X(t), $$ which seems what you were after.

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