# 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?

• What do you mean by bootstrap the 6M curve AND basis curve? You start by bootstrapping the 6M curve by repricing par swaps to 0 (assuming the same 6M curve is used for discounting) Once you have the 6M curve the process of bootstrapping is identical in principle but in lieu of par swaps, basis swap spreads are repriced. You end up with both 6M and 3M curves at the end so no addition of zero rates is needed. Clarify in case i have misinterpreted your question. Also, what level of detail are you looking for? – compilation-error Feb 25 '16 at 3:35

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.

• Can you please elaborate? I don't see how a basis swap would equal a forward rate. – learner101 Feb 24 '16 at 3:09
• Fixed my answer. Missread the question the first time (mixed 6M Swap Rate and 6M Libor Rate up) – Phun Feb 24 '16 at 21:18
• How is the 3m-6m basis swap involved? – Gordon Feb 24 '16 at 21:51