Modified duration calculation

[fullofquestions]

I got my hands on the 4th edition of the FRM handbook and have a question about example 1-6. Basically it asks to compute the modified duration of a 2 year bond that pays a coupon semiannually. From the reading, the formula is presented rather quickly and that is all. The formula given is Dmodified = D / (1 + y). There is no mention of accounting for the compounding frequency. So all I am asking is for confirmation of the formula because the answer to the example does calculate Dmodified as = D / (1 + y/2), i.e. it accounts for compounding frequency.

Please confirm that Dmodified = D / (1 + y / compounding frequency)

So for example, for a bond compounding monthly, regardless of whether it matures in 1,2, 5 or 10 years, the modified duration = D / (1 + y /12).

I think the reason why I ask this is because when compounding is continuous, Dmodified approaches D, which to me is counterintuitive. I would expect continuous compounding to pay the outstanding earlier than otherwise, hence, lowering the risk to the interest rate. As it stands, the formula produces a lower modified duration for less frequent compounding and a higher modified duration for more frequent compounding...know what I mean?

Sorry for belaboring the point but I just want to make sure. Many thanks.


Just in case, here is the question: (FRM 4th edition 1-6)
consider a 2 year, 6% semi-anual bond currently yielding 5.2% on a bond equivalent basis. If the Macaulay duration of the bond is 1.92 years, its modified duration is cosest to...

 

[David Harper CFA FRM]

hi FoQ,

Yes, correct is Modified Duration = Macaulay Duration / (1+y/k) where k= number of compound periods per year; e.g., if monthly compounding, then Mod Duration = Mac Duration / (1+y/12)...
...as continuous compounding has k approaching infinitiy, continuous compounding is the only case where Mac duation (years or avg time to cash flow) = Mod duration (sensitivity or semi-elasticity).

and we can see that:
as k --> infinite, 1 + y/k tends toward 1 + 0, such that Mod duration = Mac Duration / 1 under continuous compounding

But for any DISCRETE compounding, Mod duration < Mac duration...
...the intuition is hard to come by, i think it's much easier to work with the first derivative to see why this is the case.

Thanks, David