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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 $t_0$?

I.e.:

$$\frac{Z(t_0,t_2)}{Z(t_0,t_1)}=Z(t_0,t_1,t_2)$$

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:

$$Z(t_0,t_2)=Z(t_0,t_1)Z(t_0,t_1,t_2)$$

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?

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$Z(t_0,t_1,t_2)$ 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.
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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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