Deriving Forward-Rate Dynamics from Zero-Coupon Bond Volatility
Summary
The document derives the stochastic dynamics of a forward rate from the risk-neutral dynamics of zero-coupon bond prices. It first defines the instantaneous forward rate as the negative maturity derivative of the log bond price, then uses the stochastic-exponential solution for the bond to express the forward rate in terms of the maturity dependence of bond volatility.
Differentiating that representation yields a drift related to the maturity derivative of half the squared volatility and a diffusion related to the maturity derivative of volatility. For a finite maturity interval, the corresponding terms are expressed as differences in volatility and squared volatility across the two maturities, divided by the interval length. The derivation assumes the relevant maturity derivatives and stochastic integral manipulations are valid; the source offers no numerical example or discussion of model-specific constraints.
Key ideas
- The instantaneous forward rate can be written as the negative maturity derivative of the log zero-coupon bond price.
- The bond's risk-neutral stochastic-exponential solution provides a route to derive forward-rate dynamics.
- Forward-rate drift and diffusion depend on how bond volatility varies with maturity.
- A finite maturity interval expresses those dependencies through differences across the two maturities.
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Full text
# How to derive the expression for the forward rate?
# How to derive the expression for the forward rate?
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?:
$$df(t,T) = \frac{(\sigma_Z(t,T))^2 - (\sigma_Z(t,T))^2}{2\delta}dt + \frac{\sigma_Z(t,T) - \sigma_Z(t,T)}{\delta}dX_t$$
## Answer by oliversm (score 1)
https://quant.stackexchange.com/a/47507
## 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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.