Tuesday 11 October 2011

Anatomy of a convert - yield translators


In my last post, I mentioned that I'd examine the ways in which the various levels of rate compounding could be inter-translated.  But first, why is there a need to do this at all?  The answer is to facilitate comparison between various bonds (and swaps, for that matter).

Usually, if two financial instruments can be valued to a present value cash amount, then this explicitly facilitates the comparison.  Good enough for many instruments.  With fixed income instruments like convertibles (actually, any bonds), there's a second dimension people like to compare across disparate instrument types - the yield.  At its most general, you'd like to know the internal rate of return so you can more directly compare one bond to another - including also to a so-called risk free government security.  If you had the internal rate of return of a convertible, then you can directly read off how much greater this bond's internal rate of return is compared with, for example, British gilts by simply subtracting the gilt rate from the bond internal rate of return.  This then gives you an idea of how much additional return you can expect from that bond.

Always remember in what follows, these are just in essence quoting conventions you'll be translating.  And you will need to do this because there are many bond markets out there each with its own historically determined quoting convention which traders in that market obey when quoting rates.  So inter-market comparisons require yield translation.  For any rate quoted with an expectation of the simple interest formula being applied to the principal, there will always be a corresponding other rate which will give you exactly the same final amount, but with $n$ discrete compoundings per year, or with continuous compounding.  Think about this a moment.  I've heard people say that continuous compounding is somehow an approximation when used in academic finance, since real instruments are simple or discretely compounded.  You should now realise, if you didn't already, that for each and every discrete fixed interest instrument on the planet, there's a continuously compounded rate which will present value the cash flows to precisely the same value as the discrete formula.

Remember too, your basic atomic operation on a single bond payment period is that of simple interest.  You're likely to see simple interest in instruments which last a short enough time to have only a single payment period.  This time round, I'll introduce the subscript $s$ to indicate all the variables of this formula relate to simple interest but I'll still assume the principal is £1 and that the rate is annualised over a term running from now and lasting for $t_s$ years.  You will get back $(1+r_st_s)$.  Continuous compounding returns you $e^{r_ct_c}$.  Finally, I'll give two versions of discrete compounding, with different compounding frequencies $n_1$ and $n_2$ and times $t_1$ and $t_2$.  This will allow me to show you how to translate from one discrete convention into a separate one.  Since I'm using 1 and 2 as the subscript, I'll continue for discrete compounding, using $r_1$ and $r_2$ for the discrete-to-discrete case.  If I'm just referring to a single discrete compounding, I'll drop the subscript.  In which case £1 discretely compounfed $n$ times per year for $t$ years will return you $(1 + \frac{r}{n})^{nt}$ in the end.



  1. Simple to continuous.   $(1+r_st_s) = e^{r_ct_c}$ which means $ r_ct_c = \ln( 1+r_st_s)$ and as a result  $ r_c = \frac{\ln( 1+r_st_s)}{t_c}$
  2. Continuous to simple.  Here you start from the same equation as 1 above but quickly move to $r_s =  \frac{e^{r_ct_c}-1}{t_s}$
  3. Simple to simple.  Why not, eh, for completeness?  This is easy,   $(1+r_1t_1) = (1+r_2t_2)$ and the 1s drop off leading you to notice that you're just time scaling one rate into another $r_1=r_2 \frac{t_2}{t_1}$
  4. Continuous to continuous.  Start with $  e^{r_1t_1} =   e^{r_2t_2}$, take logs and you're back to the situation of simple to simple, where you're just scaling one rate into another in proportion to the time ratio  $r_1=r_2 \frac{t_2}{t_1}$
  5. Simple to discrete.   First as usual the equation, $(1+r_st_s)  = (1+\frac{r}{n})^{nt}$  Then take logs, $\ln(1+r_st_s) = nt \ln (1+\frac{r}{n})$ and divide through by $nt$ so that you get $\ln (1+\frac{r}{n}) = \frac{ \ln(1+r_st_s) }{nt}$.  Raise to the power of $e$ to get $ (1+\frac{r}{n}) = e^{\frac{ \ln(1+r_st_s) }{nt}}$.  After that, take the 1 across, then multiply by $n$ so that $r= n(e^{\frac{ \ln(1+r_st_s) }{nt}}-1)$
  6. Discrete to simple, starts at the same place as 5, but is much easier since $ r_s= \frac{(1+\frac{r}{n})^{nt}-1}{t_s}$ right away.  
  7. Continuous to discrete.  Equate   $   e^{r_ct_c}  = (1+\frac{r}{n})^{nt}$.  Taking logs you see that $r_ct_c =  nt \ln (1+\frac{r}{n})$  and so $ \ln (1+\frac{r}{n}) = \frac{r_ct_c}{nt}$.  Finally $r=n(e^{ \frac{r_ct_c}{nt}}-1)$
  8. Discrete to continuous.  This time the equation in 7 becomes $r_ct_c = \ln{ (1+\frac{r}{n})^{nt} }$ and so straight away $r_c = \frac{ \ln{ (1+\frac{r}{n})^{nt}} }{t_c}$ and if you want to do the conversion as quickly as possible you'll eliminate the power so that  $r_c = \frac{ nt \ln{ (1+\frac{r}{n})} }{t_c}$
  9. Discrete to discrete. Start with  $(1+\frac{r_1}{n_1})^{n_1t_1} =  (1+\frac{r_2}{n_2})^{n_2t_2}$.  Take logs.  $  n_1t_1 \ln(1+\frac{r_1}{n_1}) =   n_2t_2 \ln(1+\frac{r_2}{n_2})$.  So  $   \ln(1+\frac{r_1}{n_1}) =  \frac{ n_2t_2}{ n_1t_1 } \ln(1+\frac{r_2}{n_2})$ and when you raise to $e$ again you get  $(1+\frac{r_1}{n_1}) =  e^{\frac{ n_2t_2}{ n_1t_1 } \ln(1+\frac{r_2}{n_2})}$.  In that case   $r_1 =  n_1(e^{\frac{ n_2t_2}{ n_1t_1 } \ln(1+\frac{r_2}{n_2})}-1)$.  As you can see, this is bound to be the most computationally demanding converter.

1 comment:

  1. If you're finding the maths renders too small then CTRL '+' to increase the font size to your comfort level.

    Math symbol rendering works well on Chrome and Firefox for me, but doesn't work at all on IE. Sorry, I don't care enough to look into it.

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