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I'd like to compare the swap rates using OIS discounting $S_{a,b}^{OIS, 6M}$ and a swap rate $S_{a,b}$ not using OIS discounting: $$S_{a,b}^{OIS, 6M}=\sum_{a+1}^b L_{6M}(0,T_i,T_{i+6M})\times df^{OIS}(O,T_i)/\sum_{a+1}^b df^{OIS}(O,T_i)$$ and: $$S_{a,b}=\sum_{a+1}^b L(0,T_i,T_{i+6M})\times df(O,T_i)/\sum_{a+1}^b df(O,T_i)$$

Let's suppose for simplification that the rates are positive. Is there a way to compare both in general? Thanks you in advance.

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Please note that there is a slight indexing error in your formulas: for standard upfront swaps the libor rate paid on $T_i$ covers the period $[T_{i-6M}, T_i]$ so the correct formula for the swap rate is $$ S_{a,b}^{OIS,6M} = \sum_{i=a+1}^b L_{6M}(0, T_{i-6M}, T_{i}) \times df^{OIS}(0, T_i)/\sum_{i=a+1}^b df^{OIS}(0, T_i) $$

As can be seen from this formula, the swap rate is a weighted average of forward libor rates $L_{6M}(0, T_{i-6M}, T_{i})$, the weights being the discount factors divided by the fixed leg PV01: $$ w_i^{OIS} = df^{OIS}(0, T_i)/\sum_{k=a+1}^b df^{OIS}(0, T_k)$$ $$ w_i = df(0, T_i)/\sum_{k=a+1}^b df(0, T_k) $$

Assuming there is no convexity adjustment between OIS discounting forward libor rates and non OIS discounting forward libor rates, the difference between the OIS discounting swap rate and the non OIS discounting swap rate will result from different relative weights.

For instance if your non OIS rates are above OIS rates, then libor rates for short maturities will have a smaller weight in the OIS case than in the non OIS case. In a market configuration where the libor curve is increasing with maturity then the OIS discounting swap rate will be above the non OIS discounting swap rate.

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  • $\begingroup$ Yeah, sorry for the mistake in the formula. $\endgroup$ – ababoua May 29 '17 at 10:38
  • $\begingroup$ Just one more question: how di you imply that: $\omega_i^{OIS}>\omega_i$ from having $df^{OIS}>df$ and a upward sloping curve? $\endgroup$ – ababoua May 29 '17 at 14:21
  • $\begingroup$ Actually in the example $w_i^{OIS} < w_i$ for $i$ small and $w_i^{OIS} > w_i$ for $i$ large (these are weights that sum up to 1), but that property is not necessarily true even when non OIS rates are above OIS rates. But if the spread between non OIS rates and OIS rates is about constant then I think that property is true because then the ratio $df^{OIS} / df$ increases with maturity (bit of algebra required here). $\endgroup$ – Antoine Conze May 29 '17 at 14:41
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As you can see the swap rate is a weighted average of the forward Libors. Since Libor>OIS, we typically have df(OIS)>df, so the OIS discounted swap rate is more heavily weighted towards the back end of the swap. In an upward sloping yield curve which we currently have, this means that the OIS discounted swap rate is slightly higher than the non OIS.

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  • $\begingroup$ How do you infer that $\omega_i^{OIS}>\omega_i$ even if $df(OIS)>df$ $\endgroup$ – ababoua May 29 '17 at 10:47
  • $\begingroup$ Because the weights are the discount factors $\endgroup$ – dm63 May 30 '17 at 6:21
  • $\begingroup$ Well, the weights are not actually discoutn factors and this strictly superior relation cannot be verified since: $\sum_{i} \omega_i=1$ and $\sum_{i} \omega^{OIS}_i=1$ $\endgroup$ – ababoua May 30 '17 at 6:56
  • $\begingroup$ I agree with @conze comment above. Intuition is coming from the idea that term structure of libor-OIS spread is relatively well behaved. If this is constant the statement should be provable. $\endgroup$ – dm63 May 30 '17 at 21:25
  • $\begingroup$ Exactly.Could prove it for special cases but in general it's a bit trickier /unfeasable...Thank you!! $\endgroup$ – ababoua May 31 '17 at 6:55

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