Advances in Risk Management

FRANÇOIS-SERGE LHABITAN T 201

hypotheses of the model may simply not hold true in the real world, result-
ing in a model that performs poorly. As an illustration, a model may assume
that zero-coupon bonds exist for all required maturities, while in practice
the set of available maturity dates will be restricted. One should always
remember that markets are driven by psychology, by supply and demand,
by consensus, and not by simple diffusion processes.

10.6 What if the model is wrong? a case study

When running a derivatives book, the first obvious impact of model risk is
on pricing – model-based prices will diverge from observed ones. However,
model risk also affects hedging, in a more subtle way. As an illustration,
let us consider the case of the Black and Scholes (1973) framework. In a
complete perfect market, a stock priceS(t) follows a geometric Brownian
motion with constant volatility parameter and drift parameters:^5

dS(t)

S(t)

=μdt+σdW(t) (10.1)

Equation (10.1) defines the true model, for example, the ones that rules the
world. We denote byC(t) the value at timetof a European call option with
maturityTand exercise priceKon this stock. By Ito’s lemma, we obtain:

dC(t)=

(

∂C(t)

∂S(t)

μS(t)+

∂C(t)

∂t

+

1

2

∂^2 C(t)

∂S^2 (t)

σ^2 S^2 (t)

)

dt+

∂C(t)

∂S(t)

σS(t)dW(t)

(10.2)

Furthermore, we know that the call priceC(t) must satisfy the following
partial differential equation:

∂C(t)

∂S(t)

rS(t)+

∂C(t)

∂t

+

1

2

∂^2 C(t)

∂S^2 (t)

σ^2 S^2 (t)−rC(t)= 0 (10.3)

with boundary conditionC(T)=Max [S(T)−K, 0].
Now, consider the case of a trader that is short one call option and needs
to hedge his position. In theory, if he knew the “true” model, he could hedge

perfectly in continuous-time by holding∂∂CS((tt))units of the underlying asset

and

(

C(t)−∂∂CS((tt))S(t)

)

units of a zero-coupon bond maturing at timeT.In

the absence of arbitrage opportunities, the value of his overall portfolio(t)
is equal to zero, and its instantaneous variations are defined by:

d(t)=−dC(t)+

∂C(t)

∂S(t)

dS(t)+

(

C(t)−

∂C(t)

∂S(t)

S(t)

)

rdt (10.4)

which can also be shown to equal zero.