Bootstrapping a 3M LIBOR Curve from 6M Swaps and Basis Swaps

# Deriving a 3M libor curve from 6M libor swaps and 3M-6M libor basis swaps


# Deriving a 3M libor curve from 6M libor swaps and 3M-6M libor basis swaps












If I had a set of 6M Libor instruments and another set of 3M-6M basis swap instruments, how would I derive the 3M Libor curve?

Just bootstrap the 6M curve and the basis curve and add up the zero rates at the matching tenors, to give me the 3M zeros? Does this make sense?

## Answer by Phun (score 2)

https://quant.stackexchange.com/a/23003

Let $\delta$ be 3 month and consider points of interest $\{T_i\}_i$ evenly spaced with $T_{i+1} -T_i = 3 month$. The Forward Rate $F_m^n(t)$ from period m to n at time $t$ is defined by $$(1 + \delta (n-m) F_m^n(t)) = \frac{B(t,T_m)}{B(t,T_n)},$$ where $B(t,T_i)$ is the time $t$ value of a zero coupon bond that matures in $T_i$.

A swap rate $S_m^n(t)$ a time $t$ of a swap starting in $T_m$ and ending in $T_n$ can be written as $$ S_m^n(t)=\sum_{i=m}^{n-1} \frac{\delta B(t,T_{i+1})}{\sum_{j=m}^{n-1}\delta B(t,T_{j+1})}F_i^{i+1}(t) $$

It holds $$F_0^1(0) = (S_0^2(0) - \frac{1}{2+\delta F_1^2(0)}F_1^2(0))(1+1\frac{1}{1+\delta F_1^2(0)}) $$

Derivation

Use definition for Swap Rate

$$ \frac{B(0,T_1)}{B(0,T_1)+B(0,T_2)}F_0^1(0) + \frac{B(0,T_2)}{B(0,T_1)+B(0,T_2)}F_1^2(0) = S_0^2(0)$$

Now use

$$ (1 + \delta F_0^1(0))(1+\delta F_1^2(0)) = \frac{1}{B(0,T_2)} $$

and

$$ (1 + \delta F_0^1(0)) = \frac{1}{B(0,T_1)}$$ which leads to

$$ \frac{\frac{1}{(1 + \delta F_0^1(0))}}{\frac{1}{(1 + \delta F_0^1(0))} + \frac{1}{(1 + \delta F_0^1(0))(1+\delta F_1^2(0))}}F_0^1(0) + \frac{\frac{1}{(1 + \delta F_0^1(0))(1+\delta F_1^2(0))}}{\frac{1}{(1 + \delta F_0^1(0))} + \frac{1}{(1 + \delta F_0^1(0))(1+\delta F_1^2(0))}}F_1^2(0) = S_0^2(0)\\ \Leftrightarrow \frac{1}{1+\delta F_1^2(0)} F_0^1(0) + \frac{1}{2+\delta F_1^2(0)}F_1^2(0) = S_0^2(0) $$ which is equivalent to my the expression above.