Feasibility of Layer-by-Layer Calibration in Dynamic SABR

# Dynamic SABR layer-by-layer calibration fails at third layer: implementation issue or no admissible solution?


# Dynamic SABR layer-by-layer calibration fails at third layer: implementation issue or no admissible solution?












I am implementing the layer-by-layer calibration of the time-dependent SABR model described in the paper “Managing Vol Surfaces” by Hagan et al in 2018. In particular, I am using the effective-parameter formulas around equations (3.7)–(3.8) and the piecewise-constant integral update formulas in equation (3.11) of the paper.

For each expiry $T_j$, I first calibrate a standard SABR smile and obtain target effective SABR parameters

$$ (\alpha_j^*, \rho_j^*, \nu_j^*). $$

Then I assume piecewise-constant dynamic SABR parameters on each interval

$$ (T_{j-1},T_j): \qquad \sigma(T)=\sigma_j,\quad \rho(T)=\rho_j,\quad \nu(T)=\nu_j, $$

and solve for

$$ x_j = (\sigma_j,\rho_j,\nu_j) $$

such that the effective SABR parameters at $T_j$ match the static SABR parameters calibrated from the smile.

The global settings are:





- $\varepsilon = 1.0$.

All times are expressed in years and all volatilities/parameters are annualized. The static SABR parameters are obtained using the same backbone/beta convention for all expiries.

The static SABR target parameters used as inputs are:

| Layer | $T_{\text{start}}$ | $T_{\text{end}}$ | $d_j$ | $\alpha_j^*$ | $\rho_j^*$ | $\nu_j^*$ |
| 1 | 0.00 | 0.25 | 0.25 | 0.01863 | 0.41786 | 1.21964 |
| 2 | 0.25 | 1.00 | 0.75 | 0.01666 | 0.31270 | 0.76510 |
| 3 | 1.00 | 5.00 | 4.00 | 0.01406 | 0.13788 | 0.35115 |
| 4 | 5.00 | 10.00 | 5.00 | 0.01300 | 0.12279 | 0.21804 |
| 5 | 10.00 | 20.00 | 10.00 | 0.01248 | -0.00542 | 0.18992 |
| 6 | 20.00 | 30.00 | 10.00 | 0.01213 | -0.10955 | 0.21490 |

The first two layers are matched, but the third layer does not seem to reach the target effective parameters. I would like to understand whether this is a numerical issue, an implementation mistake, or whether the sequence of independently calibrated static SABR slices may simply be dynamically inconsistent under the time-dependent SABR effective-parameter mapping.

The update formulas I use are the following. Let

$$ d_j = T_j - T_{j-1}. $$

Given the cumulative integrals from the previous layer, I update:

$$ I_0^j = I_0^{j-1} + \sigma_j^2 d_j, $$

$$ I_1^j = I_1^{j-1} + \rho_j \nu_j \sigma_j d_j, $$

$$ I_2^j = I_2^{j-1} + \sigma_j^2 d_j I_1^{j-1} + \frac{1}{2}\rho_j\nu_j\sigma_j^3 d_j^2, $$

$$ I_3^j = I_3^{j-1} + \sigma_j^2 d_j \left[ \left( I_1^{j-1} + \frac{1}{2}\rho_j\nu_j\sigma_j d_j \right)^2 + \frac{1}{12} \left( \rho_j\nu_j\sigma_j d_j \right)^2 \right], $$

$$ I_4^j = I_4^{j-1} + \nu_j^2 d_j, $$

$$ I_5^j = I_5^{j-1} + \sigma_j^2 d_j I_4^{j-1} + \frac{1}{2}\nu_j^2\sigma_j^2 d_j^2, $$

$$ I_6^j = I_6^{j-1} + 2\sigma_j^2 d_j I_5^{j-1} + \sigma_j^4 d_j^2 I_4^{j-1} + \frac{1}{3}\nu_j^2\sigma_j^4 d_j^3. $$

Then I compute

$$\Delta_j^2 = \frac{I_0^j}{T_j},$$

$$\bar b_j = \frac{2I_2^j}{(I_0^j)^2},$$

$$\bar c_j = \frac{3I_6^j + 9I_3^j}{(I_0^j)^3} - 3\bar b_j^2,$$

$$G_j^* = \frac{2I_5^j}{(I_0^j)^2} - \bar c_j.$$

Finally, the effective SABR parameters are

$$\alpha_{\mathrm{eff},j} = \sqrt{\Delta_j^2}\exp\left(\frac{1}{4}\varepsilon^2 I_0^j G_j^*\right),$$

$$\rho_{\mathrm{eff},j} = \frac{\bar b_j}{\sqrt{\bar c_j}},$$

$$\nu_{\mathrm{eff},j} = \sqrt{\Delta_j^2}\sqrt{\bar c_j}.$$

So the nonlinear system I solve at each layer is

$$ F(\sigma_j,\rho_j,\nu_j) = \begin{pmatrix} \alpha_{\mathrm{eff},j}-\alpha_j^* \\ \rho_{\mathrm{eff},j}-\rho_j^* \\ \nu_{\mathrm{eff},j}-\nu_j^* \end{pmatrix} = 0. $$

with constraints

$$ \sigma_j > 0, \qquad \nu_j > 0, \qquad -1 < \rho_j < 1. $$

The objective minimized by the 3D solver is the sum of squared residuals:

$$ SSR = (\alpha_{\mathrm{eff},j}-\alpha_j^*)^2 + (\rho_{\mathrm{eff},j}-\rho_j^*)^2 + (\nu_{\mathrm{eff},j}-\nu_j^*)^2. $$

The 3D calibration method is a multi-start Levenberg-Marquardt scheme:

- 27 initial guesses on a $3\times 3\times 3$ grid;

- $\sigma_j$ starting values: ${0.2\alpha_j^*, 1.0\alpha_j^*, 5.0\alpha_j^*}$;

- $\rho_j$ starting values: ${-0.7, \rho_j^*, 0.7}$;

- $\nu_j$ starting values: ${0.2\nu_j^*, 1.0\nu_j^*, 5.0\nu_j^*}$;

- first scan: 50 LM iterations for each starting point;

- final polish: 400 LM iterations from the best candidate;

- finite-difference Jacobian;

- bounds enforced numerically: $\sigma_j\geq 10^{-8}$, $\nu_j\geq 10^{-8}$, and $-0.9999\leq \rho_j\leq 0.9999$.

For the failing third layer, the previous cumulative integrals after layer 2 are:

| Quantity | Value |
| $I_0^{2}$ | $9.386031997022536\times 10^{-5}$ |
| $I_1^{2}$ | $-7.383657219878922\times 10^{-4}$ |
| $I_2^{2}$ | $1.0877751497615332\times 10^{-7}$ |
| $I_3^{2}$ | $1.7342127739522748\times 10^{-10}$ |
| $I_4^{2}$ | $31.528024105319403$ |
| $I_5^{2}$ | $1.2923651476505206\times 10^{-4}$ |
| $I_6^{2}$ | $1.7030113443892636\times 10^{-9}$ |

The third layer data are:

| Quantity | Value |
| $T_3$ | $5.0$ |
| $d_3=T_3-T_2$ | $4.0$ |
| target $\alpha_3^*$ | $0.01406$ |
| target $\rho_3^*$ | $0.13788$ |
| target $\nu_3^*$ | $0.35115$ |

Equivalently, for the third layer I am asking whether the target point

$$ (\alpha_3^*,\rho_3^*,\nu_3^*)=(0.01406,0.13788,0.35115) $$

belongs to the image of the map

$$ (\sigma_3,\rho_3,\nu_3) \mapsto (\alpha_{\mathrm{eff},3},\rho_{\mathrm{eff},3},\nu_{\mathrm{eff},3}) $$

given the cumulative integrals after layer 2.

The best 3D candidate found by the multi-start Levenberg-Marquardt routine is:

| Quantity | Value |
| $\sigma_3$ | $6.732644144691606\times 10^{-4}$ |
| $\rho_3$ | $-0.9999$ |
| $\nu_3$ | $4.172091357712332$ |

Using this candidate, the updated integrals become:

| Quantity | Value |
| $I_0^3$ | $9.567345985738738\times 10^{-5}$ |
| $I_1^3$ | $-1.1972924733978022\times 10^{-2}$ |
| $I_2^3$ | $9.725384110464655\times 10^{-8}$ |
| $I_3^3$ | $2.6573216225899337\times 10^{-10}$ |
| $I_4^3$ | $101.15340929371112$ |
| $I_5^3$ | $2.495215143558578\times 10^{-4}$ |
| $I_6^3$ | $2.351604001199082\times 10^{-9}$ |

The reduced/effective quantities are:

| Quantity | Value |
| $\Delta_3^2=I_0^3/T_3$ | $1.9134691971477476\times 10^{-5}$ |
| $\bar b_3$ | $21.249748742436793$ |
| $\bar c_3$ | $9432.127971997032$ |
| $G_3^*$ | $45087.769912437114$ |

The produced effective SABR parameters are:

| Parameter | Produced | Target | Relative error |
| $\alpha_{\mathrm{eff},3}$ | $0.012860714109057685$ | $0.01406$ | $-8.5297716283%$ |
| $\rho_{\mathrm{eff},3}$ | $0.2188008259409324$ | $0.13788$ | $+58.6893138533%$ |
| $\nu_{\mathrm{eff},3}$ | $0.4248303936628354$ | $0.35115$ | $+20.9825982238%$ |

The corresponding unscaled sum of squared residuals is approximately

$$ SSR \approx 0.0119784187679213. $$

The first two layers matched the target effective parameters essentially exactly. The resulting dynamic parameters for the first two layers were:

| Layer | $\sigma_j$ | $\rho_j$ | $\nu_j$ |
| 1 | $0.01863$ | $0.41786$ | $1.21964$ |
| 2 | $0.0030748647602185392$ | $-0.2093689360371161$ | $6.44527151462418$ |

For layer 2, the produced effective parameters were equal to the static targets within numerical precision:

| Parameter | Produced | Target |
| $\alpha_{\mathrm{eff},2}$ | $0.01666$ | $0.01666$ |
| $\rho_{\mathrm{eff},2}$ | $0.31270$ | $0.31270$ |
| $\nu_{\mathrm{eff},2}$ | $0.76510$ | $0.76510$ |

I am not primarily asking for help with the VBA/Excel implementation. I am trying to understand whether, for the numerical data above, the third-layer inverse problem should theoretically admit an admissible solution, or whether the failure may indicate that the independently calibrated static SABR slices are not dynamically consistent under this time-dependent SABR construction.

My questions are:

- Are the formulas above the correct interpretation of the layer-by-layer calibration equations in the Hagan-Lesniewski-Woodward dynamic SABR paper?

- Given the previous cumulative integrals after layer 2 and the target static SABR parameters at $T_3=5$, should an admissible solution $(\sigma_3,\rho_3,\nu_3)$ always exist?

- If no admissible solution exists, is this an expected indication that independently calibrated static SABR slices are not dynamically consistent under the time-dependent SABR effective-parameter mapping?

- Is the fact that the best candidate pushes $\rho_3$ to the lower bound $-0.9999$ a sign of a numerical optimizer issue, or is it evidence that the target effective parameters at layer 3 may lie outside the image of the layer map generated by the previous integrals?

- Are there known necessary conditions on the sequence of static effective SABR parameters $(\alpha_j^*,\rho_j^*,\nu_j^*)$ for the layer-by-layer dynamic SABR calibration to be feasible?