Why Swap Values Can Match While Caplet Values Differ

# Why would valuation for a swap be the same on the backward and forward rate but not a caplet


# Why would valuation for a swap be the same on the backward and forward rate but not a caplet












Consider for time discretization $0 = T_{0} < T_{1} <... < S < T < T_{n}$, and the corresponding forward rates and backward rate:

$\text{Forward rate: }L(S,T;t)$

$\text{Backward Rate: }I(S,T):=\frac{1}{T-S}(\frac{R(T)}{R(S)}-1)$

where $$R(\{T_{0} ,...,T_{n}\};t):=P(T_{m(t)+1};t)\prod\limits_{i=0}^{m(t)}(1+L_{i}(T_{i},T_{i+1};T_{i})(T_{i+1}-T_{i})),\; \; \;\\ m(t):=\max\{{i\in \{0,...,n}\}:T_{i}\leq t\}$$

Now why would the Monte Carlo valuation of the both the

$\textbf{Caplet on Forward Rate}$ and the $\textbf{Caplet on Backward Rate}$ not be equal

BUT the Monte Carlo valuation of both the

$\textbf{Swap on Forward Rate}$ and the $\textbf{Swap on Backward Rate}$

are (nearly) equal.

Why is this the case?

## Answer by Kurt G. (score 1, accepted)

https://quant.stackexchange.com/a/69532

When the distance between the $T_i$ is small then \begin{align}\tag{1} \frac{R(T)}{R(S)}=\prod_{i=m(S)}^{m(T)}1+L(T_i,T_{i+1};T_i)\Delta_i\approx\exp\left(\int_S^Tr(u)\,du\right)\,. \end{align} On the other hand, $$\tag{2} 1+L(S,T,S)\Delta=\frac{1}{P(S,T)} $$ where $P(S,T)$ is the conditional zero bond price $$ P(S,T)=\mathbb E\Big[\exp\Big(-\int_S^Tr(u)\,du\Big) \Big|{\cal F}_S\Big]\,. $$ This shows $$\tag{3} \mathbb E\Big[\exp\Big(-\int_0^Tr(u)\,du\Big)\frac{R(T)}{R(S)}\Big]=\mathbb E\Big[\exp\Big(-\int_0^Sr(u)\,du\Big)\Big]=P(0,S)\,. $$ Using the fact that $P(S,T)$ is ${\cal F}_S$-measurable (i.e. known at time $S$) the tower property of conditional expectations implies \begin{align} &\textstyle\mathbb E\Big[\exp\Big(-\int_0^Tr(u)\,du\Big)\frac{1}{P(S,T)}\Big]\\[3mm] &=\textstyle\mathbb E\Big[\mathbb E\Big[\exp\Big(-\int_0^Tr(u)\,du\Big)\frac{1}{P(S,T)} \Big|{\cal F}_S\Big]\Big]\\[3mm] &\textstyle=\mathbb E\Big[ \frac{\exp(-\int_0^Sr(u)\,du)}{P(S,T)}\underbrace{\mathbb E\Big[\exp\Big(-\int_S^Tr(u)\,du\Big)\Big|{\cal F}_S\Big]}_{P(S,T)}\Big]\\[3mm] &\textstyle=\mathbb E\Big[\exp\Big(-\int_0^Sr(u)\,du\Big)\Big]=P(0,S)\,.\tag{4} \end{align} $$ \boxed{\text{The fact that (3) and (4) are equal shows that the swap PVs must agree exactly.}} $$ If instead you have options on $L(S,T,S)$ and $I(S,T)$ you can see from (1) that $I(S,T)$ has a larger volatility since it takes time until $T$ until this overnight rate is known. In contrast, the libor $L(S,T,S)$ is known at time $S<T$ and therefore has a lower volatility.