We trivially have that:

$$\frac{Z(t_0,t_1)}{Z(t_0,t_2)}=1+\tau L(t_0,t_1,t_2)$$

Where $L(t_0,t_1,t_2)$ is the forward Libor between $t_1$ and $t_2$, as of $t_0$.

Simply inverting this relationship then yields:

$$\frac{Z(t_0,t_2)}{Z(t_0,t_1)}=\frac{1}{1+\tau L(t_0,t_1,t_2)}$$

Could one interpret $\frac{1}{1+\tau L(t_0,t_1,t_2)}$ as a forward starting zero-coupon bond between $t_1$ and $t_2$, as of $t_0$?



If the above is true, then suppose we want to value a Caplet "set in arrears" (i.e. pay-off described in my last question).

This caplet pays $(L(t_1,t_1,t_2)-K)^{+}$ at time $t_1$. Valuing this caplet at $t_0$, choosing $Z(t_0,t_2)$ as Numeraire, we have:

$$C(t_0, T=t_1)=Z(t_0,t_2)\mathbb{E}^{t_2}_{t_0}\left[\frac{(L(t_1,t_1,t_2)-K)^{+}}{Z(t_1,t_2)}\right]$$

Using the identity:


I get:

$$C(t_0, T=t_1)=Z(t_0,t_1)Z(t_0,t_1,t_2)\mathbb{E}^{t_2}_{t_0}\left[\frac{(L(t_1,t_1,t_2)-K)^{+}}{Z(t_1,t_2)}\right]=\\=Z(t_0,t_1)\mathbb{E}^{t_2}_{t_0}\left[(L(t_1,t_1,t_2)-K)^{+}\right]$$

And the problem at hand now seems trivial, since $L(t_1,t_1,t_2)$ is a martingale under $Z(t_0,t_2)$.

The above cannot be correct, since the answer is different to what @Gordon derived in my previous question linked above. So where have I gone wrong here?


$Z(t_0,t_1,t_2)$ is the $t_1$-forward price of the ZC bond with maturity $t_2$, as of $t_0$. We have: $$ Z(t_0,t_1,t_2) = E_{t_0}^{t_1}[Z(t_1,t_2)]\not= Z(t_1,t_2).$$

With a not-trivially stochastic index, there is no way to take out $Z(t_1,t_2)^{-1}$ from under your conditional expectation operator until the running $t_0$ hits $t_1$. It is not $t_0$-measurable.

Note: To clarify @Novice555 comment below, we have ($L(t_1,t_2)=L(t_1,t_1,t_2)$):

$$ E^{t_1}_{0}[L(t_1,t_2)] \stackrel{(1)}{=} E^{t_2}_{0}\left[\frac{dQ^{t_1}}{dQ^{t_2}}\big\vert_{t_1} L(t_1,t_2) \right] $$

where $$ \frac{dQ^{t_1}}{dQ^{t_2}}\big\vert_{s} = \frac{Z(s,t_1)/Z(0,t_1)}{Z(s,t_2)/Z(0,t_2)},$$

hence $$ E^{t_1}_{0}[L(t_1,t_2)] = Z(0,t_1,t_2)E^{t_2}_{0}\left[Z(t_1,t_2)^{-1} L(t_1,t_2) \right]$$

$$ = Z(0,t_1,t_2) E^{t_2}_{0}\left[ L(t_1,t_2) \right] + Z(0,t_1,t_2) E^{t_2}_{0}\left[ L(t_1,t_2)^2\right] $$

$$ = Z(0,t_1,t_2) L(0,t_1,t_2) + Z(0,t_1,t_2) (t_2-t_1) E^{t_2}_{0}\left[ L(t_1,t_2)^2\right] $$

Note that all this can also be obtained from @Gordon's answer on caplets where the strike $K$ is set to $0$.

Also note that for ZC bonds, the same approach (replace $L$ by $Z$ in (1)) gives:

$$ E^{t_1}_{0}[Z(t_1,t_2)] = Z(0,t_1,t_2) E^{t_2}_{0}\left[Z(t_1,t_2)^{-1} Z(t_1,t_2) \right] = Z(0,t_1,t_2) $$

One way to sum up this subject is:

  • the forward price of a ZC bond, $Z(\cdot, t_1,t_2)$, is a martingale under the $t_1$-forward measure, while
  • the forward interest rate, $L(\cdot, t_1,t_2)$, is a martingale under the $t_2$-forward measure.
  • $\begingroup$ For my clarity, are we saying that: $$Z(t_0,t_1,t_2)=\frac{1}{1+\tau L(t_0,t_1,t_2)}=\frac{Z(t_0,t_2)}{Z(t_0,t_1)}=\mathbb{E}^{t_1}_{t_0}[Z(t_1,t_2)]=\mathbb{E}^{t_1}_{t_0}\left[\frac{Z(t_1,t_2)}{Z(t_1,t_1)}\right]$$ So from the above we have: $$\frac{Z(t_0,t_2)}{Z(t_0,t_1)}=\mathbb{E}^{t_1}_{t_0}\left[\frac{Z(t_1,t_2)}{Z(t_1,t_1)}\right]$$ So that basically means that $Z(t\leq t_1,t_1,t_2)$ is a martingale under numeraire $Z(t,t_1)$. Furthermore the above implies that $$\frac{Z(t\leq t_1,t_2)}{Z(t \leq t_1,t_1)}$$ is also a martingale under the $t_1-forward$ numeraire? $\endgroup$ – Novice555 Mar 14 at 18:10
  • $\begingroup$ Just answered to your first question (now deleted :) ). Does the note help with your second question above? $\endgroup$ – ir7 Mar 14 at 18:13
  • $\begingroup$ Confusingly, the above would also imply that: $$\frac{1}{1+\tau L(t,t_1,t_2)}$$ is a martingale under the $t_1$-forward measure. Whilst we have that $L(t,t_1,t_2)$ is a martingale under the $t_2$-forward measure. Seems rather too pretty to be true, no? :) $\endgroup$ – Novice555 Mar 14 at 18:13
  • 1
    $\begingroup$ Yes, you confirmed what I ask above. Seems really quite an impressive result... thank you very much indeed. $\endgroup$ – Novice555 Mar 14 at 18:15
  • 2
    $\begingroup$ It's certainly very important, if not impressive :). We can expect this by looking at the denominator of the ratio that determines each of them, each ratio being a martingale under suitable forward measures (by the definition of these 'forward measures'). Forward Z is $P(.,t_2)/P(.,t_1)$, while forward L is a linear function of $P(.,t_1)/P(.,t_2)$. $\endgroup$ – ir7 Mar 14 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.