Let $P(t,T)$ be the the value of a contract at time $t$. This contract guarantees its holder the payment of $1$ at time $T$.

consider $t<T<S$, when the interest rate is non-deterministic, do we have $$\frac{P(t,S)}{P(T,S)}=P(t,T)$$ ?

I think the answer is no, but Brigo gives it a proof when he calculate the forward rates(Page 11, paragraph after equation (1.18), the following is an image of page 11)

enter image description here

the main idea is:

consider $A:=1/P(T,S)$ as an amount of currency held at $S$, on the one hand, its value at $t$ is $P(t,S)/P(T,S)$; on the other hand, its value at $T$ is $1$, then discount it back to $t$, we get its value at $t$ is $P(t,T)$, hence $$\frac{P(t,S)}{P(T,S)}=P(t,T)$$

Would you mind telling me what's wrong with this proof?

  • $\begingroup$ Why does the amount $A:=1/P(T,S)$ held at $S$ has the value $P(t,S)/P(T,S)$ at $t$? $\endgroup$ – Gordon Sep 26 '18 at 17:52
  • $\begingroup$ @Gordon by the def of $P(t,S)$: $1$ at $S$ equals $P(t,S)$ at $t$, hence $A$ at $S$ equals $P(t,S)A$ at $t$. $\endgroup$ – Lookout Sep 26 '18 at 18:02
  • $\begingroup$ If $A$ is known at $t$, your argument is fine. Bu$1/P(T, S)$ is unknown at time $t$, for $t < T$. $\endgroup$ – Gordon Sep 26 '18 at 18:17

Just adding the valuation date/time to the bonds identifier will make it clearer. See below:

enter image description here

  • $\begingroup$ I don't understand "It is easy to see from the above:...". I think you simply replicate an FRA, right? so it should be $P(t;t,S)/P(t;t,T)=1+(S-T)P(t;T,S)$? $\endgroup$ – Lookout Sep 27 '18 at 3:32
  • $\begingroup$ It is indeed replication, but it is replicating the forward bond. By definition, agreeing to buy the forward bond at P(t;T,S) means you pay this price, -P(t;T, S) at time T to get +1 at S, and the above strategy just replicates these flows. The final step is to use the usual arbitrage argument, and that is what ‘it is easy to see’ is alluding to, $\endgroup$ – Magic is in the chain Sep 27 '18 at 6:44

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