Assume that the continuously compounded forward rate is constant between two node points. What is the interpolated discount factor between these two points?

So you have the two discount factors $D_{10}$ and $D_{12}$. What is $D_{11}$?


1 Answer 1


Assume the (annualised, continuously compounded) forward rate between two nodes, say $t_{10}$ and $t_{12}$, is constant, say $ f_{10,12}$, then the discount factors of the two consecutive knots will be linked as follows:

$D_{12}=D_{10}e^{-f_{10,12} \left(t_{12}-t_{10}\right)}=D_{10}e^{-2f_{10,12}}$

From which is then easy to infer the formula for $t_{11}$,

$D_{11}=D_{10}e^{-f_{10,12} \left(t_{11}-t_{10}\right)}=D_{10}e^{-f_{10,12}}$

or alternatively, you can use


Re-first comment, we can rearrange the first equation to get f in terms of D's:

$D_{12}= D_{10}e^{-f_{10,12} \left(t_{12}-t_{10}\right)}$

$\frac{D_{12}}{ D_{10}}=e^{-f_{10,12} \left(t_{12}-t_{10}\right)}$

$\ln \frac{D_{12}}{ D_{10}}=-f_{10,12} \left(t_{12}-t_{10}\right)$

$f_{10,12} =-\frac{1}{t_{12}-t_{10}}\ln \frac{D_{12}}{ D_{10}}=\frac{1}{t_{12}-t_{10}}\ln \frac{D_{10}}{ D_{12}}$

Re-second comment, assume $s<t<T$, I assume your $\hat t=s$ in this sense. So, as per above, the $D_t$ and $D_s$ will be linked as follows:

$D_t=D_{ s}e^{-f(t-s)}$

Now I think you are assuming that f is constant across tenors:

$f=-\frac{1}{T-s}\ln \frac{D_T}{D_s}$

Substitute this f into the previous equation, and then rearrange the term in the exponent so that we can cancel e and ln:

$D_t=D_{s}e^{\frac{t-s}{T-s}\ln \frac{D_T}{D_s}}=D_{ s}e^{ln \left(\frac{D_T}{D_s}\right)^\frac{t-s}{T-s}}$


$D_t=D_{ s} \left(\frac{D_T}{D_s}\right)^\frac{t-s}{T-s}=D_{ s} D_s^{-\frac{t-s}{T-s}}D_T^{\frac{t-s}{T-s}}=D_s^{\frac{T-t}{T-s}}D_T^{\frac{t-s}{T-s}}$

  • $\begingroup$ Can you add the formula for $f_{10,12}$? I.e. rearrange your first equation. $\endgroup$
    – emcor
    Dec 25, 2019 at 15:46
  • $\begingroup$ Thanks! have added further details in the answer $\endgroup$ Dec 25, 2019 at 16:02
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    $\begingroup$ When I plug $f$ into $D11$, I get: $$ D_t^*=D_{\hat t}e^{-f(t-\hat t)}=D_{\hat t}e^{-\frac{\ln(D_{\hat t}/D_{T})}{T-\hat t}(t-\hat t)}=D_T\frac{t-\hat t}{T-\hat t} $$ Is that correct? $\endgroup$
    – emcor
    Dec 25, 2019 at 16:08
  • $\begingroup$ Thanks I like this comment! Have added further details! $\endgroup$ Dec 25, 2019 at 16:48
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    $\begingroup$ By the second last line, you mean $D_t=D_{s}e^{\frac{t-s}{T-s}\ln \frac{D_T}{D_s}}$? As s<t, $D_t =D_se^{-f(t-s)}$ is fine, i think, and the two minus signs cancel (f also has a minus sign). $\endgroup$ Dec 25, 2019 at 17:10

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