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PSOR for American Options
Hard
·
22 min read
Numerical Methods
American Options
PSOR
Linear Complementarity
Free Boundary
1
Article
2
Notebook
3
Quiz
Quick Quiz
1.
Why can an American put on a non-dividend stock be optimal to exercise early, while an American call on the same stock never is?
Exercising the put earns interest on the strike; the call would only sacrifice time value
The put's delta is negative while the call's delta stays positive
Calls have unlimited upside whereas puts have payoffs capped at the strike
Black-Scholes assumes American puts should always be exercised early
2.
For the American-put LCP, which conditions hold at a point in the holding (continuation) region
S
>
S
∗
(
t
)
S>S^*(t)
S
>
S
∗
(
t
)
?
L
V
=
0
\mathcal{L}V=0
L
V
=
0
and
V
>
g
V>g
V
>
g
— the PDE holds, value exceeds intrinsic
L
V
>
0
\mathcal{L}V>0
L
V
>
0
and
V
>
g
V>g
V
>
g
— both strict inequalities active
L
V
>
0
\mathcal{L}V>0
L
V
>
0
and
V
=
g
V=g
V
=
g
— early exercise is forced
L
V
=
0
\mathcal{L}V=0
L
V
=
0
,
V
=
g
V=g
V
=
g
, and
L
V
(
V
−
g
)
=
0
\mathcal{L}V\,(V-g)=0
L
V
(
V
−
g
)
=
0
simultaneously
3.
In PSOR, what does the projection step
V
i
new
=
max
(
g
i
,
V
~
i
)
V_i^{\text{new}}=\max(g_i,\tilde V_i)
V
i
new
=
max
(
g
i
,
V
~
i
)
accomplish?
It projects the iterate onto the space of smooth functions
It over-relaxes the update by the factor
ω
\omega
ω
to speed convergence
It removes Crank-Nicolson oscillations near the payoff kink
It enforces
V
i
≥
g
i
V_i \geq g_i
V
i
≥
g
i
node-by-node, with no free-boundary tracking
4.
PSOR for the American-option LCP converges for every relaxation parameter
ω
∈
(
0
,
2
)
\omega\in(0,2)
ω
∈
(
0
,
2
)
when the discretisation matrix
A
A
A
is an M-matrix.
True
False
5.
At matched cost, a Crank-Nicolson + PSOR scheme typically beats a binomial tree for the American put mainly because:
CN is second-order in time; the tree is low-order and oscillatory
PSOR eliminates all numerical error while the tree only approximates
The binomial tree cannot price American puts
The binomial tree needs more memory than a finite-difference grid
6.
For the American put, what is the terminal free boundary
S
∗
(
T
)
S^*(T)
S
∗
(
T
)
and its behaviour just before expiry?
S
∗
(
T
)
=
K
S^*(T)=K
S
∗
(
T
)
=
K
, and
∂
S
∗
/
∂
t
→
0
\partial S^*/\partial t\to 0
∂
S
∗
/
∂
t
→
0
as
t
→
T
t\to T
t
→
T
S
∗
(
T
)
=
K
e
−
r
T
S^*(T)=Ke^{-rT}
S
∗
(
T
)
=
K
e
−
r
T
(the discounted strike)
S
∗
(
T
)
=
K
S^*(T)=K
S
∗
(
T
)
=
K
, approached like
O
(
(
T
−
t
)
∣
ln
(
T
−
t
)
∣
)
O(\sqrt{(T-t)\,|\ln(T-t)|})
O
(
(
T
−
t
)
∣
ln
(
T
−
t
)
∣
)
S
∗
(
T
)
=
0
S^*(T)=0
S
∗
(
T
)
=
0
, rising monotonically from 0 to
K
K
K
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