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Using zero coupon Treasury curve, I discounted a 10% coupon bond. Using the same curve, I discounted a 5% coupon bond. Both these bonds have the same maturity. Since I discount both these bonds using the same curve, I should get two yields where the difference reflects the coupon effect.

I put in these two yields into Bloomberg and theoretically the OAS on both of them should match but it doesn't. Shouldn't the OAS match because the coupon effect is already accounted for. What would them be different?

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  • $\begingroup$ The OAS is option adjusted spread - so these bonds feature optionalities? $\endgroup$ – Kermittfrog Apr 25 '20 at 8:58
  • $\begingroup$ These are Treasury bonds with no optionality $\endgroup$ – VanillaCall Apr 25 '20 at 11:30
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    $\begingroup$ Are the curves used to do the initial PV and the OAS 100% consistent, including discount rate, day count convention, compounding convention, etc.? $\endgroup$ – Helin Apr 27 '20 at 0:04
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    $\begingroup$ There was an issue with my day count. I was using ACT/360 instead of ACT/ACT for bonds. This worked. In addition, I can also use par rate curve to discount my cash flows since the par curve captures the coupon effects. So any spread to the par curve is coupon adjusted as well $\endgroup$ – VanillaCall May 9 '20 at 19:29
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The pricing could be different for other reasons. For example, a 10 year bond issued this week will likely be more liquidity traded than a 30 year bond issued 20 years ago. The newly issued bond may have a liquidity premium.

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  • $\begingroup$ I'm discounting both bonds using the same zero Treasury coupon curve though so the only difference between them is really the coupon itself. $\endgroup$ – VanillaCall Apr 26 '20 at 15:37
  • $\begingroup$ When you discount them using that curve, are you able to match the market price? $\endgroup$ – Charles Fox Apr 26 '20 at 17:01
  • $\begingroup$ Yes, I'm able to match the market price.. $\endgroup$ – VanillaCall Apr 26 '20 at 20:28
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Treasuries properly discounted should have no z spread and oas.

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  • $\begingroup$ whoever down voted this is wrong $\endgroup$ – Edward Watson May 22 at 19:20
  • $\begingroup$ the reason being is that what you're really doing here if you have a z spread is assuming a smoothed spline or select treasuries to discount the cash flows, because if you didn't, the treasury curve and it's associated strips are not really arbitrageable (dealers do reconstitute). That's why you almost never see treasury oas or z spread in dealers treasury rv packages. $\endgroup$ – Edward Watson May 22 at 20:30
  • $\begingroup$ Hi @Edward Watson. I downvoted this because I feel it was inaccurate. I can think of only one scenario where z-spreads for Treasuries would be zero – an on-the-run spline with number of knot points / basis functions matching number of instruments. In this case, the on-the-run issues would have zero OAS, but still, off-the-run issues would have non-zero OAS. Typically though, splines are built with a large number of off-the-run issues and a small number of knots, so nearly all issues would have non-zero OAS, providing rich/cheap signals. $\endgroup$ – Helin May 23 at 3:11
  • $\begingroup$ [cont...] I also happen to have RV reports from JPM, MS, and Barclays in front of me – all of them report "spline OAS" or "fitted spreads." I personally was involved in building government splines and RV reports for dozens of countries years ago, so I'm a bit surprised by your answer. Anyways, it's possible I'm misinterpreting what you're saying. Happy to engage further. $\endgroup$ – Helin May 23 at 3:13
  • $\begingroup$ I would agree with what you're saying and I definitely like using oas to a spline or spread to a spline even more so that say oas to ois. But discount rates via strips are observable and treasuries prices equate to those so discount rates. It's a technical point. But of course that wouldn't be useful for treasury rv as nothing would appear cheap or rich. For instance, if we take that smoothed curve to discount structured products, we ignore the actual shape of the curve, which is misleading. $\endgroup$ – Edward Watson May 23 at 12:49

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