# Calculating instantaneous forward rate from zero-coupon yield curve

I have a big dataset containing zero-coupon bond yields with different relative maturities. I fix a time horizon on my dataset and I want to calculate instantaneous forward rate. I'm going to write how I calculated:

The yield curve is given by: $Y(t,T)=-\frac{\log(P(t,T))}{T-t}$ formula.

So by inverting it we get bondprice:

$P(t,T)=\exp(-Y(t,T)(T-t))$

We get instantaneous forward rate from partial derivate of $\log(P(t,T))$ by $T$ so the formula I use is:

$f(t,T_k)=-\frac{\log(P(t,T_k))-\log(P(t,T_{k-1}))}{T_k-T_{k-1}}$.

where $T_0=0$.

My goal is to set up an observation matrix of instant. forward rates for volatility estimation in a model and I want to be sure if my pre-calculations are fine. Thanks for help in advanced.

Your overall approach is correct. However to my knowledge it is formally more appealing to work with a parameterized and smoothed yield curve.

Basically one assumes that the yield curve can be described by a smooth function $r(t,\alpha, \beta,\gamma)$ (mostly of three parameters)

Given a set of market data $Y(t,T_1)\dots Y(t, T_n)$ one looks for parameters $\alpha,\beta,\gamma$ so that the distance $\sum_{i=1}^n (r(T_i,\alpha,\beta,\gamma)-Y(t,T_i))^2$ is minimized (depending on the choice of $r$ one might have to use a numerical optimization routine) After $\alpha, \beta,\gamma$ have been found they are seen as fixed inputs.

This method has two significant advantages:

1. Due to the continuity of $r(t,\alpha, \beta,\gamma)$ one can calculate yields for maturities not quoted by the market via $r(T,\alpha, \beta,\gamma)$
2. $r(t,\alpha, \beta,\gamma)$ is smooth. Thus $P(t,T)=exp(-r(T-t,\alpha,\beta,\gamma)(T-t))$ is a smooth function and one can easily calculate $f(t,T)=-\frac{\partial P(t,T)}{\partial T}$

For more on yield curve construction I refer you to the Nelson–Siegel–Svensson model

• Thanks for the answer. So overall, my procedure is fine in these discrete points, but I could get a better solution with a smooth function that I can derivate by T? – user7778 Apr 21 '14 at 13:59
• exactly - using the smooth function entails optimization but it is generally the cleaner way to go. This is primarily done to get the quotes for all maturities. Also a parametric form smoothes away "erratic" and unplausible behaviour. – Probilitator Apr 21 '14 at 15:54