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Delphine LAUTIER and Alain GALLI Delphine LAUTIER(*) is ...

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As the storage theory showed, the two state variables are correlated because both<br />

the spot price <strong>and</strong> the convenience yield are an inverse function of the inventory levels.<br />

Nevertheless, as Gibson <strong>and</strong> Schwartz (1990) demonstrated, the correlation between these<br />

two variables <strong>is</strong> not perfect:<br />

[ dz × dz ] = ρdt<br />

E S C<br />

where ρ <strong>is</strong> the correlation between the two Brownian motions associated with S <strong>and</strong> C.<br />

The convenience yield <strong>is</strong> mean reverting <strong>and</strong> <strong>is</strong> involved in the spot price<br />

dynamics. Mean reversion relies on the hypothes<strong>is</strong> that there <strong>is</strong> a level of stocks which<br />

sat<strong>is</strong>fies the needs of industry under normal conditions. The behaviour of the operators in<br />

the physical market guarantees the ex<strong>is</strong>tence of th<strong>is</strong> normal level of stocks. When the<br />

convenience yield <strong>is</strong> low, the stocks are abundant <strong>and</strong> the operators sustain a high storage<br />

cost compared with the benefits related to holding the raw materials. So, if they are<br />

rational, they try to reduce these surplus stocks. Conversely, when the stocks are rare, the<br />

operators tend to reconstitute them.<br />

The solution of the term structure model can be expressed in a r<strong>is</strong>k neutral<br />

framework, using a Feynman-Kac solution. Therefore, the value of the futures prices can<br />

be written as:<br />

F<br />

Qλ ( t,<br />

T ) = E [ S(<br />

T ) ]<br />

where F(t,T) <strong>is</strong> the futures price at t for delivery at T, <strong>and</strong> Q λ denotes the r<strong>is</strong>k neutral<br />

probability 6 , which <strong>is</strong> dependent of an unknown value λ. The latter <strong>is</strong> the market price of<br />

convenience yield r<strong>is</strong>k. The solution <strong>is</strong>:<br />

with: B(<br />

τ )<br />

−<br />

⎡ 1−<br />

e<br />

F(<br />

S,<br />

C,<br />

t,<br />

T ) = S(<br />

t)<br />

× exp⎢−<br />

C(<br />

t)<br />

⎣ κ<br />

6<br />

κτ<br />

⎤<br />

+ B(<br />

τ )⎥<br />

⎦<br />

⎡<br />

2<br />

⎤ 2 −2κτ<br />

⎡<br />

2<br />

−<br />

⎛<br />

⎞ ⎡ − ⎤ ⎛<br />

⎞ ⎛ −<br />

⎢⎜<br />

) σC<br />

σSσC<br />

ρ σ<br />

⎟<br />

C 1 e<br />

× ⎥ + ⎢ × ⎥ + ⎢⎜<br />

) σC<br />

1 e<br />

=<br />

⎟×<br />

⎜<br />

⎜<br />

r −α<br />

+ −<br />

⎟<br />

τ<br />

⎜<br />

ακ<br />

+ σ σ ρ −<br />

2<br />

3<br />

S C ⎟ ⎜ 2<br />

⎢⎣<br />

⎝ 2κ κ ⎠ ⎥⎦<br />

⎣ 4 κ ⎦ ⎢⎣<br />

⎝<br />

κ ⎠ ⎝ κ<br />

)<br />

α = α − /<br />

( λ κ )<br />

κτ<br />

⎞⎤<br />

⎟<br />

⎥ ,<br />

⎠⎥⎦

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