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For example, suppose we have an interest rate curve bootstrapped from multiple instruments, at the short end, we used eurodollar future (up to 2Y), at the longer end, we used interest rate swap (3Y to 30Y).

Usually, if we want to know the delta of a trade for a specific tenor (e.g 2Y) of the curve, we can shock the 2Y eurodollar market quote(price) by 1 basis point and reprice the trade, then calculate the difference.

Suppose we are now interested in how the trade is sensitive to a 2Y IR Swap (we didn't use it to bootstrap now, we used the 2Y EuroDollar future), how can we calculate the delta without replacing the instrument at the 2Y tenor?

Is it risk resampling? Any advice is highly appreciated!

Thank you!

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    $\begingroup$ Does this answer your question? Interest Rate Swap Delta ladder, under OIS Discounting $\endgroup$ Sep 3 at 12:26
  • $\begingroup$ Please see also quant.stackexchange.com/questions/60059 and quant.stackexchange.com/questions/59507 $\endgroup$ Sep 3 at 12:54
  • $\begingroup$ @DimitriVulis thanks! For the second link in the second comment, wondering whose Jacobian we are calculating and how we calculate the Swap sensitivity to ED Future? Suppose they are both 2Y, does the ED Future quotes changes 1 bps mean that only the change of the last cash flow exchange of swap? And by fixing the other tenor rates fixed (0.5Y, 1Y, 1.5Y), we can get a new swap rate? so that it's the swap sensitivity to a ED Future? Thanks again! $\endgroup$
    – Parting
    Sep 3 at 16:18
  • $\begingroup$ It's actually just what @Kermittfrog wrote in his excellent answer. May I suggest this book: Marc Henrard. Interest Rate Modelling in the Multi-Curve Framework: Foundations, Evolution and Implementation (2014), chapter 5 explains it very well I think $\endgroup$ Sep 3 at 16:58
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Let's assume that the relevant pillar $t$ of your curve is currently (exclusively) calibrated using the reference instrument $f_0$ at market quote $q_0$. The instrument could be a swap, a forward rate agreement, tenor basis swap...

In what follows, I simplify somewhat in using scalar expressions; in practice you may see gradients / vector valued functions popping up.

Outline

The sensitivity of an instrument $F$ to a very small change in the quote $q_0$, say $\mathrm{d}q_0$, when measured through the effect of the quote on the zero rate $r_t$, is approximately:

$$ \begin{align} \mathrm{d}F&=\frac{\partial F}{\partial r}\mathrm{d}r\\ &=\left.\frac{\partial F}{\partial r}\left(-\left.\frac{\partial f_0}{\partial q}\right/\frac{\partial f_0}{\partial r}\right)\mathrm{d}q_0\right|_{r=r_t,q=q_0} \end{align} $$

where the term in the bracket stems from a requirement on our calibration instrument, evaluated at the perfectly calibrated zero rate level $r_t$

$$ \begin{align} \mathrm{d}f_0(q_0,r_t)\stackrel{!}{=}0&\stackrel{!}{=}\left.\frac{\partial f_0}{\partial q}\mathrm{d}q+\frac{\partial f_0}{\partial r}\mathrm{d}r\right|_{r=r_t,q=q_0}\\ \Rightarrow\quad\quad\quad\mathrm{d}r&=\left.-\left.\frac{\partial f_0}{\partial q}\right/\frac{\partial f_0}{\partial r}\mathrm{d}q\right|_{r=r_t,q=q_0} \end{align} $$

Let us intrudce another calibration instrument $f_1$ with corresponding quote $q_1$ that is supposed to replace $f_0$. Assume further, that both calibration instruments will yield the same zero rate $r_t$. Then, we should be able to simply replace $\mathrm{d}r$ from $f_0$ with that from $f_1$:

$$ \begin{align} \mathrm{d}F&=\left.\frac{\partial F}{\partial r}\left(-\left.\frac{\partial f_0}{\partial q}\right/\frac{\partial f_1}{\partial r}\right)\mathrm{d}q_1\right|_{r=r_t,q=q_1} \end{align} $$

where we have used the already existing valuation curve, and we of course assume that the 'other' instrument is perfectly priced at par as well. Thus, we can use the 'other' instrument's sensitivities, as valued versus the already bootstrapped curve. I.e., you can get your sensis from your front office system or such.

Example

Assume a single curve world. Our reference instruments are are vanilla swaps with annual fix/float payments and annual forward rate agreements. The rates are in agreement, i.e. we can use both for bootstrapping and arrive at the same discount factors and continuously compounded zero rates:

tenor  swap rate%  forward rate%  Discount factor  zero %
1      1.0000      1.00000        0.99009901       0.995033
2      2.0000      3.03030        0.96097845       1.990165
3      3.0000      5.13455        0.91404629       2.995802

Let's focus on the third tenor. The simple swap bootrapper yields (with quote $c=3.00\%$)

$$ S(c,r_1,r_2,r_3)=c(e^{-r_1}+e^{-2r_2}+e^{-3r_3})-(1-e^{-3r_3})=0 $$

with sensitivities

$$ \begin{align} \frac{\partial S}{\partial c} &= (e^{-r_1}+e^{-2r_2}+e^{-3r_3})=2.86512374824004\\ \frac{\partial S}{\partial r_3} &=-3D_3(1+c)=-2.82440302853815 \end{align} $$

If we had used a simple forward rate agreement with quote $f=5.13455\%$

$$ F(f,r_2,r_3)=D_3\left(Fwd(2\to 3)-f\right)=D_2-D_3(1+f)=e^{-2r_2}-e^{-3r_3}(1+f)=0 $$

with sensitivities

$$ \begin{align} \frac{\partial F}{\partial f} &= -D_3=-0.914046287552799\\ \frac{\partial F}{\partial r_3} &=3D_3(1+f)=2.88293535235877 \end{align} $$

and thus

$$ \begin{align} \mathrm{d}r&=1.01442\mathrm{d}c \quad\quad \mathrm{vs\ \ swap}\\ \mathrm{d}r&=0.31705\mathrm{d}f \quad\quad \mathrm{vs\ \ forward} \end{align} $$

A 1bp shift in the forward rate will shift the zero by 0.317 bp, whereas a 1bp shift in the swap rate will shift the zero by approx. 1bp as well.

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