New answers tagged

1

No. $s_1$ is dependent on $X$ in the sense that the value of the swap at inception must equal zero (or close to it). This is what your equation is actually showing. $$ \frac{s_1}{1+r_1} + \frac{s_1}{(1+r_1)^2} = 1$$ The $1$ is also the NPV of the floating leg assuming no spread. Banks calibrate $s_1$ so that the NPVs of the floating and fixed legs are equal ...


0

Yes I guess in theory they should be the same value, but only because in practise there is no arbitrage between the 2 approaches presuming you end up with the same instrument. I guess you mean that. getting a EUR net present value from USD curves adjusted for EURUSD basis is perhaps 1 way of calculating interest rate differentials. Presumably it gives a ...


0

I have traded those convexity adjustments for many years. Any decent model of these adjustments allows the user to vary the correlation as they please, rather than assuming something. If it is of interest, the implied correlations usually trade significantly under 1, especially in periods when the curve is volatile. ie when forward rates might be going in ...


0

I assume that this decomposition is possible in the case of a variance swap as variance is decomposable in the sense that $$ V = VAR(R_1+ \cdots + R_N) \approx \frac1N \sum_{i=1}^N R_i^2 . $$ This for any $n \le N$ we can write $$ V \approx \frac1N \sum_{i=1}^n R_i^2 + \frac1N \sum_{i=n+1}^N R_i^2 = \frac n N \frac1n \sum_{i=1}^n R_i^2 + \frac{N-n}{N} ...



Top 50 recent answers are included