Pricing In-Arrears Caplets by Changing Forward Measures

# Caplet "in arrears" pricing formula


# 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?

## Answer by Gordon (score 11, accepted)

https://quant.stackexchange.com/a/60323

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 $Q_{t_2}$.

## Answer by B_B (score 4)

https://quant.stackexchange.com/a/60321

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.