# Constant continuous forward rate interpolation

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 $$D_{11}$$?

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

$$D_{11}=D_{12}e^{f_{10,12}}$$

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 $$s, I assume your $$\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:

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

Thus,

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

• Can you add the formula for $f_{10,12}$? I.e. rearrange your first equation. – emcor Dec 25 '19 at 15:46
• Thanks! have added further details in the answer – Magic is in the chain Dec 25 '19 at 16:02
• 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? – emcor Dec 25 '19 at 16:08
• Thanks I like this comment! Have added further details! – Magic is in the chain Dec 25 '19 at 16:48
• 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). – Magic is in the chain Dec 25 '19 at 17:10