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Here is the expression of a basis floating versus floating swap where the first term is a forward CMS Swap leg and the second one is a forward BOR leg where X is the margin that would make equal both leg?

Suppose X and all other parameters are know except the CA term? How can I express the CA term related to the others?

  • $\begingroup$ Any reply please? $\endgroup$
    – Bond007
    Commented Jan 17, 2017 at 16:04

1 Answer 1


As far as I can tell, not being an expert in basis swap pricing, this is just algebra -

$$ X_{n,c} = \frac{\sum_{i=1}^n \left( S_{i,c}'(0) + {\rm\bf CA}(S_{i,c}'; \delta)\right) P(0,T_i')}{\sum_{i=1}^n P(0,T_i')} - \frac{1 - P(0,T_n')}{\delta \sum_{i=1}^n P(0,T_i')} $$

which rearranges to

$$ \sum_{i=1}^n {\rm\bf CA}(S_{i,c}'; \delta) P(0,T_i') = - \sum_{i=1}^n S_{i,c}'(0)P(0,T_i') + \frac{1 - P(0,T_n')}{\delta} + X_{n,c} \sum_{i=1}^n P(0,T_i') $$

You can't reduce it any further, since there are multiple CA terms (one for each $i$) not just one.

As pointed out in the comments, you can bootstrap the curve if you have $X_{n,c}$ for multiple $n$. For example, for $n = 1$ you can derive

$$ {\rm\bf CA}(S_{1,c}'; \delta) = - S_{1,c}'(0) + \frac{1 - P(0,T_1')}{\delta} + X_{1,c} $$

For any other $n$, assuming that you have already computed ${\rm\bf CA}(S_{k,c}';\delta)$ for $k=1,\dots,n-1$ then then you have

$$ {\rm\bf CA}(S_{n,c}';\delta) = \frac{ - \sum_{i=1}^n S_{i,c}'(0)P(0,T_i') + \frac{1 - P(0,T_n')}{\delta} + X_{n,c} \sum_{i=1}^n P(0,T_i') - \sum_{i=1}^{n-1} {\rm\bf CA}(S_{i,c}'; \delta) P(0,T_i') }{P(0,T_n') } $$

which allows you to compute all values of ${\rm\bf CA}(S_{n,c}';\delta)$ iteratively.

  • $\begingroup$ pretty sure this is the point of bootstrapping though - you set n=1, that makes the algebra tractable for what you have, solve that, and the sub it into the equation for n=2, and then repeat... $\endgroup$
    – will
    Commented Jan 18, 2017 at 23:38

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