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Regularisation and Stability in Calibration
Hard
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25 min read
Calibration
Regularisation
Tikhonov
Inverse Problems
Numerical Stability
1
Article
2
Notebook
3
Quiz
Quick Quiz
1.
With
J
=
U
Σ
V
⊤
J=U\Sigma V^\top
J
=
U
Σ
V
⊤
, the least-squares step is
δ
θ
L
S
=
∑
j
u
j
⊤
δ
σ
^
σ
j
v
j
\delta\theta_{\mathrm{LS}}=\sum_j \frac{u_j^\top\delta\hat\sigma}{\sigma_j}v_j
δ
θ
LS
=
∑
j
σ
j
u
j
⊤
δ
σ
^
v
j
. Why do near-zero singular values
σ
j
\sigma_j
σ
j
destabilise calibration?
They mean the model ignores the data, so the objective is trivially satisfied
They make
J
J
J
non-square, so the normal equations cannot be solved
They force the LM damping parameter
λ
\lambda
λ
to diverge
The noise term
u
j
⊤
δ
σ
^
u_j^\top\delta\hat\sigma
u
j
⊤
δ
σ
^
is amplified by
1
/
σ
j
1/\sigma_j
1/
σ
j
, giving huge
δ
θ
\delta\theta
δ
θ
2.
Tikhonov regularisation has SVD filter factors
f
j
(
λ
)
=
σ
j
2
/
(
σ
j
2
+
λ
)
f_j(\lambda)=\sigma_j^2/(\sigma_j^2+\lambda)
f
j
(
λ
)
=
σ
j
2
/
(
σ
j
2
+
λ
)
. What happens to directions with large singular values (
σ
j
≫
λ
\sigma_j\gg\sqrt\lambda
σ
j
≫
λ
)?
Heavily suppressed by the penalty toward the prior
Nearly unaffected:
f
j
≈
1
f_j\approx 1
f
j
≈
1
Set exactly to zero, as in truncated SVD
Pulled strongly toward the prior
θ
0
\theta_0
θ
0
3.
Morozov's discrepancy principle sets
∥
r
(
θ
λ
)
∥
≈
ϵ
N
\|r(\theta_\lambda)\|\approx\epsilon\sqrt N
∥
r
(
θ
λ
)
∥
≈
ϵ
N
. With
N
=
50
N=50
N
=
50
instruments and half-spread
ϵ
=
0.002
\epsilon=0.002
ϵ
=
0.002
vol points, the target residual norm is:
0.002
×
50
≈
0.0141
0.002\times\sqrt{50}\approx 0.0141
0.002
×
50
≈
0.0141
0.002
2
×
50
=
0.0002
0.002^2\times 50=0.0002
0.00
2
2
×
50
=
0.0002
0.002
×
50
=
0.1
0.002\times 50=0.1
0.002
×
50
=
0.1
0.002
/
50
≈
2.8
×
10
−
4
0.002/\sqrt{50}\approx 2.8\times10^{-4}
0.002/
50
≈
2.8
×
1
0
−
4
4.
Adding a Tikhonov penalty
λ
∥
θ
−
θ
p
r
e
v
∥
2
\lambda\|\theta-\theta_{\mathrm{prev}}\|^2
λ
∥
θ
−
θ
prev
∥
2
(with
θ
p
r
e
v
\theta_{\mathrm{prev}}
θ
prev
= yesterday's calibration) removes calibration bias on days when the market moves sharply.
True
False
5.
On an L-curve (residual norm vs regularisation norm), what characterises the 'horizontal arm' at small
λ
\lambda
λ
?
Large residual (poor fit) with small regularisation norm (parameters near the prior)
Small residual (good fit) with large regularisation norm (far from prior, unstable)
The calibration has reached the global minimum of the objective
Both the residual and the regularisation norm are small (the ideal point)
6.
Penalising surface curvature
λ
1
∥
∂
k
k
σ
l
o
c
∥
2
\lambda_1\|\partial_{kk}\sigma_{\mathrm{loc}}\|^2
λ
1
∥
∂
k
k
σ
loc
∥
2
in local-vol calibration. What is the main failure when
λ
1
\lambda_1
λ
1
is too large?
Genuine sharp smile features get smoothed away, mispricing liquid near-ATM options
The objective becomes non-convex and LM stops converging
The surface can no longer reprice any European option
The local-vol surface violates the no-butterfly-arbitrage condition
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