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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:

$dW_{i}(t)dW_{j}(t)=\rho_{ij}dt$.

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

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

Question:

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:

$\text{Black}_{\text{caplet},i,j}(P(T_{j};0),K,L(T_{i},T_{j};0),\sigma_{i})$

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

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1 Answer 1

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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.

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