Interpolating Discount Factors with a Constant Continuous Forward Rate
Summary
The document derives how to interpolate a discount factor between two curve nodes when the continuously compounded forward rate is constant across the interval. Starting from the relationship between the endpoint discount factors, it expresses the forward rate as the log ratio of those factors divided by the time between nodes. The intermediate discount factor then follows by applying that constant rate over the shorter elapsed time.
For equally spaced nodes, the intermediate value is the geometric midpoint of the endpoint discount factors; the text also gives a generalized expression for any time between two endpoints. This is a curve-construction relationship rather than an empirical result. It relies on the stated constant-forward-rate assumption over the interval and uses continuously compounded rates with consistent time units. If the curve’s forward rate varies within the interval, this interpolation rule need not represent that shape.
Key ideas
- A constant continuously compounded forward rate links discount factors exponentially across an interval.\nThe forward rate can be recovered from the logarithm of the endpoint discount-factor ratio.\nThe intermediate discount factor is obtained by applying that rate over the elapsed time.\nFor equally spaced nodes, the midpoint discount factor is the geometric mean of the endpoints.\nThe interpolation assumes a constant forward rate between the specified nodes.
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Full text
# Constant continuous forward rate interpolation
# 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 $D_{11}$?
## Answer by Magic is in the chain (score 3, accepted)
https://quant.stackexchange.com/a/50434
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 $s<t<T$, 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:
$D_t=D_{s}e^{\frac{t-s}{T-s}\ln \frac{D_T}{D_s}}=D_{ s}e^{ln \left(\frac{D_T}{D_s}\right)^\frac{t-s}{T-s}}$
Thus,
$D_t=D_{ s} \left(\frac{D_T}{D_s}\right)^\frac{t-s}{T-s}=D_{ s} D_s^{-\frac{t-s}{T-s}}D_T^{\frac{t-s}{T-s}}=D_s^{\frac{T-t}{T-s}}D_T^{\frac{t-s}{T-s}}$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.