There are some questions and answers on this site which touch upon this topic, but none actually show step-by-step on how to bootstrap a coupon OIS Swap curve to construct a zero-curve for discounting.

Bootstrapping a bond curve is easy: say we have three bonds with annual coupons and maturities 1 year, 2 years and 3 years. These bonds trade at prices $PV_1$, $PV_2$ and $PV_3$, with face-values $N$ and annual percentage coupons $C_1$, $C_2$ & $C_3$.

The 1y tenor zero-rate "$x$" simply solves $PV_1=\frac{N+C_1}{1+x}$.

The 2y tenor zero-rate "$y$" then solves $PV_2=\frac{C_2}{1+x}+\frac{N+C_2}{(1+y)^2}$.

The 3y tenor zero-rate "$z$" then solves $PV_3=\frac{C_3}{1+x}+\frac{C_3}{(1+y)^2}+\frac{N+C_3}{(1+z)^3}$.

My question is this: if we have three OIS swap with maturities 1y, 2y and 3y, and their (annual) fixed rates are $r_1$, $r_2$ and $r_3$ respectively, how can we bootsrap these swaps? What would be the equivalent $PV_1$, $PV_2$ and $PV_3$ on these swaps?


From Pricing and Hedging Swaps by Paul Miron and Philip Swannell:

Here I will take the input rates: $r_{1y}$, $r_{2y}$, $r_{3y}$ and create the DF values for each tenor $df_{1y}$, $df_{2y}$, $df_{3y}$, and thus create the zero coupon swap curve rates $z_{1y}$, $z_{2y}$, $z_{3y}$.

The book demonstrates how this formula represents both the fixed and floating cashflow of the swap (assuming fixed principle):

$PV(\text{swap_1y}) = -Pdf_0 + Pr_{1y}\alpha_{0,1y}df_{1y} + Pdf_{1y}$

$P = \text{principle}$

$df_x = \text{discount factor at some tenor } x$

$\alpha_{a, b} = \text{year fraction (using the day count basis of the fixed leg of the swap) between tenors } a \text{ and } b$

$r_x = \text{quote for the fixed leg of an annual swap for some tenor } x$

So since we know that for a swap $PV(\text{swap_1y}) = 0$ we can then see that:

$df_{1y} = \frac{df_0}{1+r_{1y}\alpha{0,1y}}$

Therefore we can extend this to the case of 2Y and 3Y:

$PV(\text{swap_2y}) = -Pdf_0 + Pr_{2y}\alpha_{0,1y}df_{1y} + Pr_{2y}\alpha_{1y,2y}df_{2y} + Pdf_{2y}$

$PV(\text{swap_3y}) = -Pdf_0 + Pr_{3y}\alpha_{0,1y}df_{1y} + Pr_{3y}\alpha_{1y,2y}df_{2y} + Pr_{3y}\alpha_{2y,3y}df_{3y} + Pdf_{3y}$

Again setting $PV(\text{swap_2y}) = 0$ and $PV(\text{swap_3y}) = 0$ we have:

$df_{2y} = \frac{df_0-r_{2y}\alpha_{0,1y}df_{1y}}{1+r_{2y}\alpha_{1y,2y}}$

$df_{3y} = \frac{df_0-r_{3y}(\alpha_{0,1y}df_{1y} + \alpha_{1y,2y}df_{2y})}{1+r_{3y}\alpha_{2y,3y}}$

At this point we have bootstrapped the curve to 3Y. In order to then create the zero curve values I can perform for any tenor $x$:

$z_{x} = \frac{1}{df_x}^\frac{1}{t_{0,x}} - 1$

$t_{a, b} = \text{Year fraction of your choice, suppose ACT/ACT, from } a \text{ to } b$

  • $\begingroup$ Thank you. Just to confirm: I suppose that $df_0=1$ by definition? $\endgroup$
    – Novice555
    Nov 11 '20 at 7:20
  • 1
    $\begingroup$ Yep that's right. I like to leave it written like that since when you want to do forward swap calculations you will replace it with the df at the tenor that the swap begins $\endgroup$
    – Dillon
    Nov 11 '20 at 21:05

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