Pricing a Forward on a Zero-Coupon Bond by Replication

# Price a forward contract on a zero-coupon bond


# Price a forward contract on a zero-coupon bond












I'm trying to calculate the price of a forward contract on a zero-coupon bond (ZCB). The forward contract matures at $t_1$ and the ZCB matures at $t_2$. So is the price of the forwards contract just the ratio of (price ZCB that matures at $t_2$) / (price ZCB that matures at $t_1$) ?

## Answer by LocalVolatility (score 6)

https://quant.stackexchange.com/a/32304

Another way to obtain this result is, as I mentioned in the comment, to think about how you would replicate the forward contract. It has the following cash-flow structure:

```
 type          | t                  | t1                      | t2
----------------------------------------------------------------------------
 forward       | 0                  | +P(t1, t2) - K          | 0
```

Here, I also use $P(t_1, t_2)$ to denote the time $t_1$ price of the zero coupon bond with maturity in $t_2$. $K$ is the fair delivery price of the forward contract.

You replicate this contract by taking a long position in the zero-coupon with maturity $t_2$ and financing the purchase by selling the zero coupon bond with maturity in $t_1$ for a notional that yields a current cash-inflow of $P(t, t_2)$. You get

```
 type          | t                  | t1                      | t2
-----------------------------------------------------------------------------
 long ZCB t2   | -P(t, t2)          | +P(t1, t2)              | 0
 short ZCB t1  | +P(t, t2)          | -P(t, t2) / P(t, t1)    | 
-----------------------------------------------------------------------------
 total         | 0                  | +P(t1, t2)              | 0
               |                    | -P(t, t2) / P(t, t1)    |
```

The portfolio has the same cash-flows as the forward in both $t$ and $t_2$. It has the same random cash-flow in $t_1$ ($+P(t_1, t_2)$) and thus the non-random cash-flows at this time also have to agree, i.e. $K = P(t, t_2) / P(t, t_1)$.

## Answer by Gordon (score 3)

https://quant.stackexchange.com/a/32288

Let $E^{t_1}$ be the expectation operator under the $t_1$-forwad probability measure $Q^{t_1}$, which takes the bond price process $\{P(t, t_1), \, 0\le t \le t_1\}$ as the numeraire. Then, the price of the forward contract, at time $t$, where $0\le t \le t_1$, is given by \begin{align*} E^{t_1}\big(P(t_1, t_2)\mid \mathcal{F}_t\big) &= E^{t_1}\left(\frac{P(t_1, t_2)}{P(t_1, t_1)}\mid \mathcal{F}_t\right)\\ &=\frac{P(t, t_2)}{P(t, t_1)}, \end{align*} as $\left\{\frac{P(t, t_2)}{P(t, t_1)}, 0\le t \le t_1\right\}$ is a martingale under the $t_1$-forwad probability measure $Q^{t_1}$. Here, $\mathcal{F}_t$ is the information set at time $t$.

## Answer by Gabriele Pompa (score 1)

https://quant.stackexchange.com/a/63417

Elaborating on @Gordon's answer. Denoting with $P(t,T,S)$ the $T$-forward price of a zero-coupon bond maturing in $S$ ($S \geq T$), we have the relation:

$$ P(t,T) P(t,T,S) = P(t,S) $$

which, in terms of spot ($L$) and forward ($F$) simply-compounded yields, which are related by the relation:

$$ (1+\tau(t,T) L(t,T))(1+\tau(T,S)F(t,T,S))=(1+\tau(t,S)L(t,S)) $$

from which the standard forward yield, defined as the (see e.g., Brigo-Mercurio (2001) chap 1

$$ F(t,T,S) = \frac{1}{\tau(T,S)} \left(\frac{P(t,T)}{P(t,S)} - 1\right) $$

can be interpreted as the appropriate yield matching the forward price of a zcb

$$ P(t,T,S) = \frac{1}{1+ \tau(T,S) F(t,T,S)} $$

where $\tau(x,y)$ is the time measure between times $x$ and $y$.

## Answer by user name (score 0)

https://quant.stackexchange.com/a/80529

A simple way is to think about what this forward contract will allow you to do.

Note that the 'forward price' of the asset refers to the price you must pay to purchase the $t_2$-bond at $t_1$, and let that be K.

so you will pay K at time $t_1$, and collect 1 at time $t_2$.

Now discount the money back to time $t$, we have $- K * P(t,t_1) + 1 * P(t,t_2)$ since discounting 1 unit of money at time s we just multiply it by $P(t,s)$. To ensure no arbitrage, we simply let that equal to 0, and get the value of K.