Deriving Zero-Coupon Bond Prices from Instantaneous Forward Rates

# zero-coupon bond and forward rate


# zero-coupon bond and forward rate












My understanding, in a discrete-time setting, the relationship between a zero-coupon bond price and forward rates is:

$$p(t,T)=\frac{1}{\Pi_{j=1}^{T-1}f(t,j)}.$$

where $p(t,T)$ represents the price of the zero-coupon bond at time $t$ paying a sure dollar at $T$, and $f(t,S)$ is the forward rate between $t$ and $S$.

In the continuous time setting, my understanding is:

$$p(t,T)=e^{-\int_t^Tf(t,s)ds}.$$

How do we get to this?

Is it:

$$\lim_{k\rightarrow T-1}\frac{1}{\Pi_{j=1}^k f(t,j)}=p(t,T)=e^{-\int_t^Tf(t,s)ds}$$

My Question:

1) Is this correct?

2) If so, how do you prove the third equation?

3) If not correct, how do you get from the first to the second equation?

## Answer by Magic is in the chain (score 2, accepted)

https://quant.stackexchange.com/a/49884

I will try a simplified approach:

Let $P(t,T)$ represent the price at time t of a zero coupon that pays 1 at time T. If you divide the period between t and T into n sub-intervals, assume $F \left( t; t_{i-1}, t_{i}\right)$ represent the simple forward rate at time t for the interval between $i-1$ and $i$, where we assume the length of each interval is equal to $\Delta t$. Then you can write the price as follows:

$P(t,T)=\prod_{i=1}^{n}{\frac{1}{1+F \left( t; t_{i-1}, t_{i}\right) \Delta t }}$

Re-arrange to:

$P(t,T)\prod_{i=1}^{n}{\left(1+F \left( t; t_{i-1}, t_{i}\right) \Delta t \right)}=1$

Now let's say we increase the number of compounding in each interval (these are the sub-intervals of length $\Delta t$, and instead of simple compounding within each of these intervals, we are increasing the compounding frequency. m=1 reproduces the original):

$P(t,T)\prod_{i=1}^{n}{\left(1+\frac{F \left( t; t_{i-1}, t_{i}\right)}m \Delta t \right)^m}=1$

Now let m tend to infinity (continuous compounding within each sub-interval):

$\lim_{m \to \infty} \; P(t,T)\prod_{i=1}^{n}{\left(1+\frac{F \left( t; t_{i-1}, t_{i}\right)}m \Delta t \right)^m}=1$

And recall the basic identity: $e^x=\lim_{m \to \infty} \left(1+\frac{x}{m}\right)^m$

$P(t,T)\prod_{i=1}^{n}{e^{F \left( t; t_{i-1}, t_{i}\right) \Delta t}}=1$

Which then becomes, noting product of exponential is exponential of sum of exponents:

$P(t,T)e^ {\sum_{i=1}^{n}{F \left( t; t_{i-1}, t_{i}\right)\Delta t}}=1$

Now if you let n goes to infinity, then the sum will become integral, and you then just need to shift the exponential term to the right hand side, and you will get the desired formula after applying continuous time interpretation of the instantaneous forward rate:

$P(t,T)=e^{-\int_t^T{f(t,s)ds}}$