Let $n \geq 2$, and consider a tenor discretization: $0 = T_{0} < T_{1} < ... < T_{n}$ and associated forward rates evaluated at time $t$, as $L_{i}(t):=L(T_{i},T_{i+1};t)$ for any $i = 0,...,n-1$.

Furthermore we assume lognormal dynamics under the measure $\mathbb P$,

$$ dL_{i}(t)=L_{i}(t)(\mu_{i}^{\mathbb P}(t)dt +\sigma_{i}(t)dW_{i}(t)),$$

Where $W_{i}$ is a Brownian motion, and furthermore, we have the following correlation structure:


We define the following money market account $M$ that takes on two arguments $T_{i} < T_{j}$:

$M(T_{i}, T_{j})=\frac{1}{T_{j}-T_{i}}\left(\prod\limits_{k=i}^{j-1}\left(1+L_{k}(T_{k})(T_{k+1}-T_{k})\right)-1\right)\; (*)$


Given strike $K>0$, how can I value to the following caplet-type product:

It pays $\max(M(T_{i},T_{j})-K,0)$ at time $T_{j}$?

My thoughts:

If we were in the Black model and only evaluating the caplet that pays $\max(L(T_{i},T_{j};T_{i})-K,0)$ at time $T_{j}$, it would simply be a case of taking the "terminal" measure $\mathbb Q^{P(T_{j})}$, and using the black formula:


I am unsure how to do the same for $(*)$. Any ideas?


1 Answer 1


We assume that, under the $T_i$-forward measure, \begin{align*} dL_i = L_i(t)\sigma_i(t)g_i(t)dW_t^i, \end{align*} where $g_i(t)=\pmb{1}_{t \le T_i}$. Then, for $k=i, \ldots, j$, under the $T_j$-forward measure, \begin{align*} dL_k = L_k(t)\sigma_k(t)g_k(t)\bigg(dW_t^j - \sum_{l=k}^{j-1}\frac{\rho_{k,l}\Delta_l\sigma_l(t)g_l(t)L_l(t)}{1+\Delta_l L_l(t)}dt\bigg), \end{align*} where $\Delta_l = T_{l+1}-T_l$. Let \begin{align*} M_t=\frac{1}{T_j-T_i}\left(\prod_{k=i}^{j-1}\big(1+L_{k}(t)\Delta_k\big)-1\right), \end{align*} and \begin{align*} \Sigma_t = \sum_{k=i}^{j-1}\ln\big(1+L_{k}(t)\Delta_k\big). \end{align*} Then $M_{T_j} = M(T_i, T_j)$. Moreover, \begin{align*} d\Sigma_t &= \sum_{k=i}^{j-1}\bigg(\frac{\Delta_kdL_{k}(t)}{1+ L_k(t)\Delta_k} -\frac{1}{2} \frac{\Delta_k^2d\langle L_k, L_k\rangle_t}{\big(1+ L_k(t)\Delta_k\big)^2}\bigg)\\ &=\sum_{k=i}^{j-1}\frac{\Delta_k \sigma_k(t)g_k(t)L_{k}(t)}{1+ L_k(t)\Delta_k}\bigg(dW_t^j \\ &\qquad-\sum_{l=k}^{j-1}\frac{\rho_{k,l}\Delta_l\sigma_l(t)g_l(t)L_l(t)}{1+\Delta_l L_l(t)}dt -\frac{1}{2} \frac{\Delta_k\sigma_k(t)g_k(t)L_{k}(t)}{1+ L_k(t)\Delta_k}dt\bigg), \end{align*} and \begin{align*} dM_t &= \frac{1}{T_j-T_i}d\left(e^{\Sigma_t}-1\right)\\ &= \frac{1}{T_j-T_i} e^{\Sigma_t}\left(d\Sigma_t + \frac{1}{2} d\langle\Sigma, \, \Sigma\rangle_t\right)\\ &=M_t \frac{1+(T_j-T_i)M_t}{(T_j-T_i)M_t}\sum_{k=i}^{j-1}\frac{\Delta_k \sigma_k(t)g_k(t)L_{k}(t)}{1+ L_k(t)\Delta_k}\Bigg(dW_t^j \\ &\qquad\qquad-\bigg(\sum_{l=k}^{j-1}\frac{\rho_{k,l}\Delta_l\sigma_l(t)g_l(t)L_l(t)}{1+\Delta_l L_l(t)} +\frac{1}{2} \frac{\Delta_k\sigma_k(t)g_k(t)L_{k}(t)}{1+ L_k(t)\Delta_k}\\ &\qquad\qquad\qquad\qquad- \frac{1}{2} \sum_{k=i}^{j-1}\frac{\Delta_k \sigma_k(t)g_k(t)L_{k}(t)}{1+ L_k(t)\Delta_k}\bigg)dt\Bigg). \end{align*} Approximating all coefficients with their values at time 0, you can have an analytical approximation for $M$ and then a Black style caplet valuation formula.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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