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01 — Article
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Quiz
Quiz: Newton-Raphson and Brent for Implied Volatility
Module 1 of 4 · Hard
Quick Quiz
1.
The BS vega is
V
=
S
e
−
q
T
T
ϕ
(
d
1
)
\mathcal{V}=Se^{-qT}\sqrt{T}\,\phi(d_1)
V
=
S
e
−
q
T
T
ϕ
(
d
1
)
. Why does plain Newton-Raphson misbehave for deep OTM options where
d
1
≫
1
d_1\gg 1
d
1
≫
1
?
The objective
f
(
σ
)
f(\sigma)
f
(
σ
)
acquires several roots deep OTM
The BS price becomes non-monotone in
σ
\sigma
σ
for deep OTM strikes
Vega is exponentially small, so the step
−
f
/
V
-f/\mathcal{V}
−
f
/
V
overshoots wildly
The Brenner-Subrahmanyam initial guess is undefined deep OTM
2.
Brent needs a bracket
[
σ
l
o
,
σ
h
i
]
[\sigma_{\mathrm{lo}},\sigma_{\mathrm{hi}}]
[
σ
lo
,
σ
hi
]
with
f
(
σ
l
o
)
f
(
σ
h
i
)
<
0
f(\sigma_{\mathrm{lo}})\,f(\sigma_{\mathrm{hi}})<0
f
(
σ
lo
)
f
(
σ
hi
)
<
0
. A robust universal bracket for a valid European-call quote is:
[
10
−
8
,
10.0
]
[10^{-8},\,10.0]
[
1
0
−
8
,
10.0
]
—
f
f
f
is monotone and the no-arbitrage bounds place the root inside
[
0
,
1
]
[0,\,1]
[
0
,
1
]
— implied vols above 100% never occur in practice
[
σ
A
T
M
−
0.5
,
σ
A
T
M
+
0.5
]
[\sigma_{\mathrm{ATM}}-0.5,\ \sigma_{\mathrm{ATM}}+0.5]
[
σ
ATM
−
0.5
,
σ
ATM
+
0.5
]
— centred on ATM vol
No fixed bracket exists; it must be found by trial and error
3.
Near the root, Newton-Raphson converges quadratically. In terms of correct decimal digits, this means the digit count:
Grows linearly with the iteration number
Grows by a fixed additive constant each iteration
Depends on the initial guess but not on
f
f
f
Roughly doubles each iteration
4.
Because the BS call price is strictly increasing in
σ
\sigma
σ
at every strike, the implied-vol root-finding problem has a solution for any positive market price.
True
False
5.
In a hybrid Newton/Brent solver the bracket must stay valid after each iterate
σ
n
e
w
\sigma_{\mathrm{new}}
σ
new
. The correct update is:
Replace the bracket midpoint regardless of the sign of
f
f
f
Replace the endpoint whose
f
f
f
-value shares the sign of
f
(
σ
n
e
w
)
f(\sigma_{\mathrm{new}})
f
(
σ
new
)
Reset the bracket to
[
10
−
8
,
10
]
[10^{-8},10]
[
1
0
−
8
,
10
]
whenever Newton diverges
Always overwrite
σ
h
i
\sigma_{\mathrm{hi}}
σ
hi
with
σ
n
e
w
\sigma_{\mathrm{new}}
σ
new
6.
The Brenner-Subrahmanyam guess
σ
0
≈
2
π
/
T
C
m
k
t
/
(
S
e
−
q
T
)
\sigma_0\approx\sqrt{2\pi/T}\,C_{\mathrm{mkt}}/(Se^{-qT})
σ
0
≈
2
π
/
T
C
mkt
/
(
S
e
−
q
T
)
comes from which ATM approximation?
Linearising the ATM price as
C
≈
S
e
−
q
T
σ
T
/
(
2
π
)
C\approx Se^{-qT}\sigma\sqrt{T/(2\pi)}
C
≈
S
e
−
q
T
σ
T
/
(
2
π
)
The Bachelier (normal-model) ATM option price formula
Differentiating the exact BS ATM price with respect to
σ
\sigma
σ
Put-call parity applied to the at-the-money straddle position
Submit
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