Caplet “in arrears” pricing formula

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?

• 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”. – Gabriele Pompa Jan 2 at 15:48
• @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. – Novice555 Jan 2 at 18:27
• 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’). – dm63 Jan 3 at 4:11
• @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? – Novice555 Jan 3 at 8:08
• 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 – 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 $$0 and, for $$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 $$Q_{t_2}$$.

• I like your solution. Very elegant and therefore better than mine. :) – B_B Jan 4 at 21:58
• Thanks @B_B. I aimed to make it as simple as possible. – Gordon Jan 5 at 0:35
• Gordon, this community would not be the same without you. Pls stay active. For ever :) – Jan Stuller Jan 5 at 11:13
• $$=P(0,t_2)(1+(t_1 - t_2)K)E_{t_2}((L(t_1;t_1,t_2)-K)^+ +2P(0, t_2)(t_2-t_1)\int_K^\infty(L(t_1;t_1,t_2)-J)^+\,dJ$$ – dm63 Jan 5 at 11:23
• @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? – 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:

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

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:

$$H(L(T_{p},T_{p},T_{E}))=(L(T_{p},T_{p},T_{E})-K)_{+}.$$

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 $$T_{f}=T_{p}=T_{S}$$):

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

EDIT:

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

Look at the Gordon's solution.