In this post here it is shown that the forward Libor rate $L(t,t_1,t_2)$, with $0 \leq t \leq t_1$, must be a martingale under the T-forward measure associated with the zero coupon bond $P(t,t_2)$ that matures at time $t_2$.

Pricing a caplet that "matures" at $t_2$ then becomes trivial (i.e. a caplet where the Libor sets at $t_1$ but the payment occurs at $t_2$):

$$C(t_0, T=t_2)=P(t_0,t_2)\mathbb{E}^{P_{t_2}}\left[\frac{(L(t_1,t_1,t_2)-K)^{+}}{P(t_2,t_2)}\right]=P(t_0,t_2)\mathbb{E}^{P_{t_2}}\left[(L(t_1,t_1,t_2)-K)^{+}\right]=P(t_0,t_2)Black76(K,L(t_0,t_1,t_2))$$

However, suppose the caplet matures at $t_1$ (which actually seems more natural, because that is when the Libor $L(t,t_1,t_2)$ sets), then we have:

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

It's not immediately obvious how to evaluate this expectation (we could potentially chose $P(t,t_1)$ as Numeraire instead, but then we'd need to come up with a more complicated process for the Libor $L(t,t_1, t_2)$ under this Numeraire than just an exponential driftless martingale process that we use under the $P(t,t_2)$ as numeraire, so that doesn't really solve the issue).

How do we price such caplet? Libor market model?

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    $\begingroup$ This is a Libor in-arrear caplet. Situations like this call for a convexity adjustment. You could have a look at Brigo-Mercurio, “Interest Rate Models - Theory and Practice” (2006), sec. 13.8.2 “The Convexity Adjustment Technique”. $\endgroup$ – Gabriele Pompa Jan 2 at 15:48
  • $\begingroup$ @GabrielePompa: thank you. Any chance you could post the crux of it as an answer pls? I don't have access to that book at the moment. $\endgroup$ – Novice555 Jan 2 at 18:27
  • $\begingroup$ Hi it seems to me that you have it reversed. The caplet expiring $t_1$ And paying out on $t_2$ is easier to price because the relevant forward rate is a martingale, as you point out. It is the caplet expiring $t_2$ and also paying $t_2$ that is more difficult (the ‘arrears caplet’). $\endgroup$ – dm63 Jan 3 at 4:11
  • $\begingroup$ @dm63: the relevant forward rate $L(t,t_1,t_2)$ that sets at $t_1$ is indeed a martingale, but under $P(t,t_2)$, i.e. a bond that matures at $t_2$. So as of $t_1$, when the caplet matures and the Libor $L(t_1,t_1,t_2)$ had set, $P(t_1,t_2)$ had not yet matured: so it is not equal to 1 yet, and as a random variable, can't be taken out of teh expectation, no? $\endgroup$ – Novice555 Jan 3 at 8:08
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    $\begingroup$ Hi in a standard caplet, expiration is at $t_1$ a d payment is at $t_2$. This is why there is confusion on nomenclature. The case where both setting and payment occurs at t_1 is the arrears caplet requiring the convexity adjustment. I will try to complete your proof $\endgroup$ – dm63 Jan 3 at 16:32

Let $P(t, T)$ be the price at time $t$ of a zero-coupon bond with maturity $T$ and unit face value. Consider the pricing of the caplet with payoff $(L(t_1; t_1, t_2)-K)^+$ at time $t_1$, where $0<t_1 < t_2$ and, for $0\le s \le t_1$, \begin{align*} L(s; t_1, t_2) = \frac{1}{t_2-t_1}\left(\frac{P(s, t_1)}{P(s, t_2)}-1\right) \end{align*} is the forward rate set at time $s$ for the calculation period $(t_1, t_2]$. Let $Q_{t_1}$ and $Q_{t_2}$ be the respective $t_1$- and $t_2$-forward probability measures, and $E_{t_1}$ and $E_{t_2}$ be the corresponding expectation operators.

Note that, for $0\le s \le t_1$, \begin{align*} \frac{dQ_{t_1}}{dQ_{t_2}}\big|_{s} &= \frac{P(0, t_2)}{P(0, t_1)}\frac{P(s, t_1)}{P(s, t_2)}\\ &=\frac{P(0, t_2)}{P(0, t_1)}\Big(1+ (t_2-t_1)L(s; t_1, t_2) \Big). \end{align*} We assume that, under the $t_2$-forward probability measure $Q_{t_2}$, \begin{align*} dL(t; t_1, t_2) = \sigma L(t; t_1, t_2) dW_t, \end{align*} for $0\le t \le t_1$, where $\sigma$ is the volatility and $\{W_t,\, t \ge 0\}$ is a standard Brownian motion. Then, the value of the caplet is given by \begin{align*} &\ P(0, t_1) E_{t_1}\big((L(t_1; t_1, t_2)-K)^+\big) \\ =&\ P(0, t_1) E_{t_2}\left(\frac{dQ_{t_1}}{dQ_{t_2}}\big|_{t_1}(L(t_1; t_1, t_2)-K)^+\right)\\ =&\ P(0, t_2) E_{t_2}\left(\Big(1+ (t_2-t_1)L(t_1; t_1, t_2) \Big)(L(t_1; t_1, t_2)-K)^+ \right)\\ =&\ P(0, t_2) E_{t_2}\big((L(t_1; t_1, t_2)-K)^+\big) + P(0, t_2)(t_2-t_1) E_{t_2}\big(L(t_1; t_1, t_2)(L(t_1; t_1, t_2)-K)^+\big). \end{align*} The remaining computations are straightforward, given the dynamics of $L(t_1; t_1, t_2)$ under $Q_{t_2}$.

  • $\begingroup$ I like your solution. Very elegant and therefore better than mine. :) $\endgroup$ – B_B Jan 4 at 21:58
  • $\begingroup$ Thanks @B_B. I aimed to make it as simple as possible. $\endgroup$ – Gordon Jan 5 at 0:35
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    $\begingroup$ Gordon, this community would not be the same without you. Pls stay active. For ever :) $\endgroup$ – Jan Stuller Jan 5 at 11:13
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    $\begingroup$ $$=P(0,t_2)(1+(t_1 - t_2)K)E_{t_2}((L(t_1;t_1,t_2)-K)^+ +2P(0, t_2)(t_2-t_1)\int_K^\infty(L(t_1;t_1,t_2)-J)^+\,dJ$$ $\endgroup$ – dm63 Jan 5 at 11:23
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    $\begingroup$ @dm63 I think your formula is incorrect. It should be rather $\int_{K}^{+\infty}(z-K)^{2}f(z)dz$ where $f(z)$ is a pdf of $L(t_{1};t_{1},t_{2})$. Note that, your formula is a price at time $0$ and $L(t_{1};t_{1},t_{2})$ is a random variable at time $0$, so it does not make any sense, does it? $\endgroup$ – B_B Jan 5 at 14:26

Case I

Let us consider a derivative with a payoff $H(L(T_{f},T_{S},T_{E}))$ which is paid at time $T_{p}$.

Note that:

  • $T_{f}$ - LIBOR fixing date;
  • $T_{S}$ - LIBOR start date;
  • $T_{E}$ - LIBOR maturity date;
  • $T_{p}$ - derivative payment date.

Also, $T_{f}=T_{S}=t_{1}$ and $T_{E}=T_{p}=t_{2}$ in the question.

In your first case $H(L(T_{f},T_{S},T_{E}))=(L(T_{f},T_{S},T_{E})-K)_{+}$ and the no-arbitrage price is:


where $\mathbb{E}^{T_{p}}[\ \cdot\ ]$ is the expectation under $T_{p}$-forward measure (with $P(t,T_{p})$ as numeraire) and $T_{p}=T_{E}$.

Case II

In your second case the derivative payment date $T_{p}$ is the same as the fixing date $T_{f}$ and the LIBOR has maturity $T_{E}=T_{p}+3M$, so the underling is simply LIBOR rate in-arrear.

To be more specific, in your case: $T_{f}=T_{p}=T_{S}=t_{1}$ and $T_{E}=t_{2}$.

The payoff of the second derivative (fixed at time $T_{p}$ and paid at time $T_{p}$) looks as follows:


It means, in order to price the derivative we need to change $T_{p}$-forward measure to $T_{E}$-forward measure (keeping in mind that $T_{f}=T_{p}=T_{S}$):

$$C(t)=P(t,T_{p})\cdot\mathbb{E}^{T_{p}}\Big[H(L(T_{f},T_{S},T_{E})|\mathcal{F}_{t}\Big]$$ $$=P(t,T_{p})\cdot\mathbb{E}^{T_{p}}\Big[H(L(T_{p},T_{p},T_{E}))|\mathcal{F}_{t}\Big]$$ $$=P(t,T_{E})\cdot\mathbb{E}^{T_{E}}\left[H(L(T_{p},T_{p},T_{E}))\cdot\frac{P(T_{p},T_{p})}{P(T_{p},T_{E})}{}|\mathcal{F}_{t}\right]$$ $$=P(t,T_{E})\cdot\mathbb{E}^{T_{E}}\left[H(L(T_{p},T_{p},T_{E}))\cdot\frac{1}{P(T_{p},T_{E})}|\mathcal{F}_{t}\right]$$ $$=P(t,t_{2})\cdot\mathbb{E}^{t_{2}}\left[H(L(t_{1},t_{1},t_{2}))\cdot\frac{1}{P(t_{1},t_{2})}|\mathcal{F}_{t}\right]$$


$$=P(t,t_{2})\cdot\mathbb{E}^{t_{2}}\Big[(L(t_{1},t_{1},t_{2})-K)^{+}\cdot(1+(t_{2}-t_{1})L(t_{1},t_{1},t_{2}))|\mathcal{F}_{t}\Big]$$ $$=P(t,t_{2})\cdot\Big(\mathbb{E}^{t_{2}}\Big[(L(t_{1},t_{1},t_{2})-K)^{+}|\mathcal{F}_{t}\Big]+(t_{2}-t_{1})\cdot\mathbb{E}^{t_{2}}\Big[L(t_{1},t_{1},t_{2})\cdot(L(t_{1},t_{1},t_{2})-K)^{+}|\mathcal{F}_{t}\Big]\Big).$$

Look at the Gordon's solution.


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