Mapping Par Swap DV01s to Zero-Rate Sensitivities

# Bounds for Par vs Zero DV01


# Bounds for Par vs Zero DV01












Let’s say I have a swap portfolio and a vector of Par sensitivities (DV01‘s) for N nodes of a curve. Let’s call the vector P = (P_1,…,P_N). To derive the sensitivities w.r.t zero rates, we could of course do this numerically via bump and reprice or some jacobian transformation. Let’s call this Z = (Z_1,…,Z_N). Without doing these calculations, is there any reliable (and ideally simple) estimate/formula on Z_i - P_i)?

## Answer by Attack68 (score 3, accepted)

https://quant.stackexchange.com/a/76952

This is a better answer than the last, although this works on exactly same principle, it considers multiple rates and differentiates the discount factors which is needed to get the answer right.

You can derive the Jacobian analytically

I will change the notation to make it easier to type than before. Suppose you have par rates, $S_j$, and zero coupon swap rates, $Z_j$ (these are not continuously compounded zero rates but I imagine its pretty close).

These are all dependent upon the discount factors, $v_j$. I will assume all annual periods so $d_i = 1.0$.

$$S_j = \frac{1-v_j}{\sum_{i=1}^j d_i v_i}, \quad 1+Z_j = \left ( \frac{1}{v_j} \right ) ^{1/j} $$

Equating the DF at maturity gives:

$$ 1- S_j \sum_1^j d_iv_i = (1+Z_j)^{-j} $$

Noting that, $$ \frac{\partial v_i}{\partial Z_k} = -i(1+Z_i)^{-i-1} \delta_k^i $$ One can derive the following formula with basic calculus:

$$ \frac{\partial S_j}{\partial Z_k} = \frac{1}{\sum_1^j d_i v_i} \left ( \delta_k^j j(1+Z_j)^{-j-1} + \alpha_k^j S_j d_k k(1+Z_k)^{-k-1} \right ) $$

where $a_k^j=1$ if $k<=j$ or 0 otherwise. This equation has a scalar, a first term and a second term. All of which are visible in the calculations shown below in Excel, all based off the know discount factors.

This is replicable numerically as follows:

```
from rateslib import *

defaults.convention="ACt365F"
par_curve = Curve(
    nodes={
        dt(2022, 1, 1): 1.0,
        dt(2023, 1, 1): 1.0,
        dt(2024, 1, 1): 1.0,
        dt(2025, 1, 1): 1.0,
        dt(2026, 1, 1): 1.0,
        dt(2027, 1, 1): 1.0,
    },
    id="curve"
)
zero_curve = Curve(
    nodes={
        dt(2022, 1, 1): 1.0,
        dt(2023, 1, 1): 1.0,
        dt(2024, 1, 1): 1.0,
        dt(2025, 1, 1): 1.0,
        dt(2026, 1, 1): 1.0,
        dt(2027, 1, 1): 1.0,
    },
    id="curve"
)
par_inst = [
    IRS(dt(2022, 1, 1), "1Y", "A", curves="curve"),
    IRS(dt(2022, 1, 1), "2Y", "A", curves="curve"),
    IRS(dt(2022, 1, 1), "3Y", "A", curves="curve"),
    IRS(dt(2022, 1, 1), "4Y", "A", curves="curve"),
    IRS(dt(2022, 1, 1), "5Y", "A", curves="curve"),
]
zero_inst = [
    ZCS(dt(2022, 1, 1), "1Y", "A", curves="curve"),
    ZCS(dt(2022, 1, 1), "2Y", "A", curves="curve"),
    ZCS(dt(2022, 1, 1), "3Y", "A", curves="curve"),
    ZCS(dt(2022, 1, 1), "4Y", "A", curves="curve"),
    ZCS(dt(2022, 1, 1), "5Y", "A", curves="curve"),
]
par_solver=Solver(
    curves=[par_curve],
    instruments=par_inst,
    s=[3.5, 3.25, 2.75, 2.9, 3.1],
    instrument_labels=["1Yp", "2Yp", "3Yp", "4Yp", "5Yp"],
    id="Par IRS",
)
zero_solver=Solver(
    curves=[zero_curve],
    instruments=zero_inst,
    s=[float(_.rate(solver=par_solver)) for _ in zero_inst],
    instrument_labels=["1Yz", "2Yz", "3Yz", "4Yz", "5Yz"],
    id="Zero IRS",
)
par_solver.jacobian(zero_solver).style.format(precision=3)
```

## Answer by Attack68 (score 1)

https://quant.stackexchange.com/a/76944

You could try this and see how closely it aligns with the numerical Jacobian.

The continuously compounded zero rate is defined by the formula:

$$v_n = e^{-D_nr_z} $$

where $v_n$ is the discount factor at maturity and $D_n$ is the DCF til maturity.

On the other hand in a self discounted curve a par swap rate is defined as,

$$r_p = \frac{1-v_n}{\sum_i^n d_iv_i} \implies v_n = 1 - r_p \sum_i^n d_i v_i$$.

Now the par rate sensitivities, and zero rate sensitivities, respectively of your portfolio are defined as,

$$ \frac{\partial P}{\partial r_p}, \quad \frac{\partial P}{\partial r_z} = \frac{\partial P}{\partial r_p} \frac{\partial r_p}{\partial r_z}$$

where,

$$-D_n e^{-D_nr_z} = - \frac{\partial r_p}{\partial r_z} \sum_i^n d_i v_i \quad \implies \quad \frac{\partial r_p}{\partial r_z} = \frac{D_n v_n}{\sum_i^n d_i v_i}$$

I would like to see how well this approximates by doing the numerical analysis myself but dont have the time right now.

## Answer by Kermittfrog (score 1)

https://quant.stackexchange.com/a/76961

Just an addendum to @Attack68's answer:

Mathematically, we have that for a vector of par rates $\mathbf{s}$, consisting of par rates $s_i, i=1\ldots N$ in above answer, the corresponding vector of zero rates of equal length, $\mathbf{z}$, $z_i,i=1\ldots N$ is set such that the vector of reference instruments prices $\mathbb{f_0}$ (conveniently of length $N$) is met with equality (curves have been calibrated).

$$\mathbf{f}(\mathbf{s},\mathbf{z})\stackrel{!}{=}\mathbf{f_0}$$

When our curves are calibrated, we have the relationship in Attack's answer regarding the Jacobian:

$$ \mathbf{df}=\frac{\partial \mathbf{f}}{\partial \mathbf{s}}\mathbf{ds} + \frac{\partial \mathbf{f}}{\partial \mathbf{z}}\mathbb{dz}\stackrel{!}{=}0 $$

And thence

$$ \mathbf{dz}=-\left(\frac{\partial \mathbf{f}}{\partial \mathbf{z}}\right)^{-1}\frac{\partial \mathbf{f}}{\partial \mathbf{s}}\mathbb{ds} $$

Where the derivatives are understood matrices with typical component

$$ \left(\frac{\partial \mathbf{f}}{\partial \mathbf{z}}\right)_{i,j}=\frac{\partial f_i}{\partial z_j} $$ i.e. the derivative of swap $i$ with resepct to zero rate $z_j$.