5 Choosing to minimize the variance
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Part Five advanced topics
and
2
∂ 2v
∂v 1 2 2 ∂ 2v
1 2∂ v
+ 2σ S
+
+
ρσ
Sq
q
2
∂t
∂S 2
∂S∂σ
∂σ 2
+ µS
∂v
∂m
∂v
+p
+ q 2 (1 − ρ 2 )
∂S
∂σ
∂σ
2
− 2rv = 0.
(54.5)
The ﬁnal conditions for these are obviously the payoff, for m(S, σ , T ), and zero for
v(S, σ , T ). If the portfolio contains options with different maturities, the equations must satisfy
the corresponding jump conditions as well.
Since the ﬁnal condition for v is zero and the only ‘forcing term’ in (54.5) is (∂m/∂σ )2 ,
equation (54.5) shows that the only way we can have a perfect hedge is for either q to be zero,
i.e. deterministic volatility, or to have ρ = ±1. In the latter case the asset and volatility (changes)
are perfectly correlated. The solution of (54.4) is then different from the Black–Scholes solution.
Equation (54.4) is very much like the pricing equation for stochastic volatility in a risk-neutral
setting. It’s rather like having a market price of volatility risk of (µ − r)ρ/σ . But, of course,
the reasoning and model are completely different in our case.
The system of equations is non linear (actually two linear equations, coupled by a non-linear
forcing term). We are going to exploit this fact shortly.
54.7 HOW TO INTERPRET AND USE THE MEAN
AND VARIANCE
Take an option position in a world with stochastic volatility, and delta hedge as proposed above.
Because we cannot eliminate all the risk we cannot be certain how accurate our hedging will
be. Think of the ﬁnal value of the portfolio together with accumulated hedging as being the
‘outcome.’ The distribution of the outcome will generally not be Normal. The shape will depend
very much on the option position we are hedging. But we have calculated both the mean and
the variance of the hedged portfolio. If we made the assumption that the distribution was not
too far from Normal then the mean and the variance are sufﬁcient to describe the probabilities
of any outcome. If we wanted to be 95% certain that we would make money then we would
have to sell the option for
m + 1.644853v 1/2
or buy it for
m − 1.644853v 1/2 .
The 1.644853 comes from the position of the 95th percentile assuming a Normal distribution.
We’ll use this idea below, but with a requirement that we are within one standard deviation
of the mean, i.e. we make money 84% of the time.
More generally we would price at
m ± ξ v 1/2 ,
where the ξ is a personal choice.
Clearly the larger ξ the greater the potential for proﬁt from a single trade (see Figure 54.1).
stochastic volatility and mean-variance analysis Chapter 54
0.6
Expected
profit
0.5
0.4
0.3
0.2
0.1
ξ
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure 54.1 Expected proﬁt from a single trade versus ξ .
Number of
trades
ξ
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure 54.2 Number of trades versus ξ .
However, the larger ξ the fewer trades (see Figure 54.2).
The net result is that the total proﬁt potential, being a product of the number of trades and
the proﬁt from each trade, is of the form shown in Figure 54.3. Don’t be too greedy or too
generous.
We’ll use this idea in the example below, but we will insist that we are within one standard
deviation of the mean so that ξ = 1. This is simply so that we have fewer parameters to carry
around.
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Part Five advanced topics
Profitability
ξ
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure 54.3 Total proﬁt potential versus ξ .
54.8 STATIC HEDGING AND PORTFOLIO OPTIMIZATION
If we use as our option (portfolio) ‘price’ the following
mean − (variance)1/2 = m − v 1/2
then we have a non-linear model. Everything that will be said in Chapter 60 about non-linear
pricing models applies here, in particular the possibility of optimal static hedging.
54.9 EXAMPLE: VALUING AND HEDGING AN UP-AND-OUT
CALL
In this section, we consider the pricing and hedging of a short up-and-out call. Throughout this
section, our choice of mean-variance combination is:
m − v 1/2 .
(54.6)
First consider a single up-and-out call with barrier located at Su . In this case, we solve the
equations (54.4) and (54.5) subject to:
(a) m(Su , σ , t) = v(Su , σ , t) = 0 for each (σ , t) ∈ (0, ∞) × (0, T ) where T is maturity;
(b) m(S, σ , T ) = − max(S − E, 0) for each (S, σ ) ∈ (0, X) × (0, ∞) where E is the strike;
(c) v(S, σ , T ) = 0 for each (S, σ ).
The discontinuity of the payoff at the knock-out barrier makes this position particularly difﬁcult
to hedge. In fact this can be easily seen from our equations. Figures 54.4 and 54.5 are the
pictures of calculated mean and variance respectively with the following speciﬁcations:
−8
−6
−4
−2
0
stochastic volatility and mean-variance analysis Chapter 54
0.4
120
0.3
110
vo
lat 0.2
ilit
y
0.1
90
100
t
spo
80
0 1
2
3
4
5
6
Figure 54.4 Mean for a single up-and-out call.
0.4
120
0.3
110
vo
lat 0.2
ilit
y
0.1
90
100
t
spo
80
Figure 54.5 Variance for a single up-and-out call.
• Strike at 100, barrier at 110, and expiry 30 days;
• p(σ ) = 0.8(σ −1 − 0.2) and q(σ ) = 0.5 .
Near the barrier, (∂m/∂σ )2 is huge (see Figure 54.4) and this feeds the variance, being the
source term in (54.5). If the spot S is 100, and the current spot volatility σ is 20% per annum,
the mean is −1.1101 and the variance is 0.3290. Thus if there is no other instrument available
in the market, one would price this option at $1.6836 to match with Equation (54.6).
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