6. Delta Ladder

[2]:

from fixedincome2025 import table

Overview

Recall that, given $D^{\$}$ and $C^{\$}$, we can estimate the change of the value of a bond portfolio for parallel yield curve shift
As the yield curve doesn’t always move parallelly, we need more granular control of the portfolio value for interest rate risk management
The Delta ladder, also known as the bucketed Delta, or index Delta, is a risk managing tool for this purpose

Having learned yield curve interpolation, we can obtain yield at any terms, and with that we can compute the value of arbitrary bonds/portfolios
Today’s curve is the base scenario, and the portfolio value computed with today’s curve is the base PV (present value)
Bumped curves:
- There are 14 buckets on the below curve
- Adding 1 bp to each bucket, we obtain 14 bumped curves
  - The first one has yields $\underline{4.24\%}, 4.18\%, 4.11\%, \ldots$
  - The second one has yields $4.23\%, \underline{4.19\%}, 4.11\%, \ldots$
Revaluate the same portfolio with the 14 bumped curves to obtain 14 PVs
These 14 PVs minus the base PV give you 14 PV differences, which form the Delta ladder

[9]:

table('yc_10022025').T

[9]:

	1 Mo	1.5 Mo	2 Mo	3 Mo	4 Mo	6 Mo	1 Yr	2 Yr	3 Yr	5 Yr	7 Yr	10 Yr	20 Yr	30 Yr
10/2/2025	4.23%	4.18%	4.11%	4.02%	3.96%	3.81%	3.62%	3.55%	3.56%	3.67%	3.86%	4.10%	4.66%	4.69%

Bump and revaluation is a numerical procedure to compute the Delta ladder for general products (even nonlinear!)
- If you have a pricer (say, for Bermudan swaption) that can give you a PV, you can compute the Delta ladder with it
But the Delta ladder of a ZCB can be computed directly if the term happens to be one of the curve point
- That is, ZCB is a curve instrument
- That is, given the representation of the Treasury yield curve we’ve been using. This is not the case for the SOFR curve

Assuming 10y yield is $4.02\%$, what’s the Delta ladder of a 10y ZCB paying $\$1$ at maturity?
First look at the Delta at the 10y bucket. What happens to the ZCB price if 10y yield is up 1bp?

ZCB price today is $e^{-TR} = e^{-10\times 0.0402} = 0.66898$
If the 10y yield is up $\Delta = 1$bp, the bond price change is $e^{-T(R+\Delta)} - e^{-TR}$
But for small yield change we have the approximation

\[\frac{e^{-T(R+\Delta)} - e^{-TR}}{\Delta} \approx -Te^{-TR}\]
Thus we have the bond price change (which is the Delta at the 10y bucket) \begin{align*} e^{-T(R+\Delta)} - e^{-TR} &\approx -Te^{-TR}\Delta \\ &= -10\times 0.66898\times 0.0001\\ &= -0.00066898 \end{align*}

Yield change at other points won’t affect the ZCB price a bit, so the Delta ladder is 0 at all buckets except 10y where the Delta is $-0.00066898$

Recall that
- A bond’s price change is approximately $-D^{\$}\Delta$ for small yield change $\Delta$
- A ZCB’s dollar duration $T\times\text{(current bond price)} = T e^{-TR}$
- Thus ZCB price change is $-Te^{-TR}\Delta$
We’ve just derived the same conclusion as Chapter 4 (and if you check Chapter 4 carefully you will find it’s the same derivation)
If the term happens to be one of the curve point, the Delta ladder of a ZCB is 0 at all buckets except at the term where the Delta is $-Te^{-TR}\Delta$

Assuming 2y yield is $3.46\%$, what’s the Delta ladder of a 2y ZCB paying $\$1$ at maturity?

[16]:

table('delta_ladder').T

[16]:

	1m	1.5m	2m	3m	4m	6m	1y	2y	3y	5y	7y	10y	20y	30y	Sum
Delta	-185	-231	-303	-275	-690	778	-3739	-2320	314	-65	23	-6	1	0	-6698

[16]:

table('delta_ladder').T

[16]:

	1m	1.5m	2m	3m	4m	6m	1y	2y	3y	5y	7y	10y	20y	30y	Sum
Delta	-185	-231	-303	-275	-690	778	-3739	-2320	314	-65	23	-6	1	0	-6698

PnL computation:
- Scenario 4: The 1m point of the curve is up 2 bps, while all other points unchanged
  - We will lose $185\times 2 = \$370$ on the portfolio
- Scenario 5: The 6m points is both up 1 bp and 1m point is up 2 bps, while all other points unchanged
  - Portfolio PnL will be $778-370 = \$408$

[16]:

table('delta_ladder').T

[16]:

	1m	1.5m	2m	3m	4m	6m	1y	2y	3y	5y	7y	10y	20y	30y	Sum
Delta	-185	-231	-303	-275	-690	778	-3739	-2320	314	-65	23	-6	1	0	-6698

PnL computation:
- Scenario 6: Curve parallel shifts up by 1 bp
  - Portfolio PnL will be $-185-231-303 \cdots -6 + 1 + 0 = -6698$, sum of bucketed delta
Recall that for small parallel curve shift $\Delta$, we have

\[-D^{\$} \Delta \approx P(\Delta) - P(0)\]
Thus, a bond portfolio’s negative dollar duration $-D^{\$}$ should be always close to its sum of bucketed delta times $10,000 = 1/(1 \text{ bp}) = 1/\Delta$, a conclusion consistent with the previous example and exercise

[3]:

table('yc_10022025_10032025')

[3]:

	10/2/2025	10/3/2025	Rates Movement (bps)	Portfolio Delta Ladder
1 Mo	4.23%	4.24%	1	-185
1.5 Mo	4.18%	4.17%	-1	-231
2 Mo	4.11%	4.11%	0	-303
3 Mo	4.02%	4.03%	1	-275
4 Mo	3.96%	3.96%	0	-690
6 Mo	3.81%	3.82%	1	778
1 Yr	3.62%	3.64%	2	-3739
2 Yr	3.55%	3.58%	3	-2320
3 Yr	3.56%	3.59%	3	314
5 Yr	3.67%	3.72%	5	-65
7 Yr	3.86%	3.90%	4	23
10 Yr	4.10%	4.13%	3	-6
20 Yr	4.66%	4.69%	3	1
30 Yr	4.69%	4.71%	2	0

PnL is the rate movement times delta, bucket by bucket, and sum up the products

\[(1)\times(-185) + (-1)\times(-231) + (0)\times(-303) + \cdots + (3)\times(1) + 2\times0 = -\$13,195\]