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Plugging the first equation into the second yields:

( P q )

³

Pj=6

i

R j ´

Ã

1+

c d EQ,i

c) dx

P q

1+

³

Pj=6

i

R j ´

!

1

+ P q +

d θ dx(c) c E Q , i

=0

And solving for

d dx c EQ,i c)

we finally obtain:

2+ 1+ R j dx(c) j 6 = i R Pj=6 P i j P q + d θ c E Q , i = ( P q à )

Ã

(

1)

1+

Pj=6

Pj=6

i

2+

j =6 i X R j !

i R j R j

>0

+1

!

d θ dx(c) c E Q , i

=0

(29)

And solving for

d Q i ( dx ) E Q c EQ,i we obtain: ) c) d Q i E Q ( θ c E Q , i X R = j dx(c)

and

d q dx(c) E Q j

= R j > 0

(30)

j =6 i

F i n a l l y w e d e r i v e a n i m p o r t a n t p r o p e r t y o f R j ( θ c E Q , i

) which will be needed later in order

to proof the second order conditions. We obtain for all j =6

i:

d R j ( θ c E Q , i

d y c 0 i

)

=

( C j q q

P q

P q ) 2 C j q q q | { z }

>0

(·)

d q E Q j

( θ c i )

|

dx(c) {z }

>0 see (30)

<0

(31)

W e c a n t h u s c o n c l u d e , t h a t w h e n e v e r ( C j q q q

0 ) ( x 0 0 j ( c ) 0 ) , w e o b t a i n

d R j dx (

c EQ,i

c)

)

0.

( i i i ) P r o p e r t i e s o f t h e s p o t m a r k e t e q u i l i b r i u m a t θ c E Q , i

, derive

d

c EQ,i

dc

:

R e m e m b e r θ c E Q , i

is defined by the equation system given by (28).

Differentiation wrt c yields:

(c)

q +P

(c)

q + 2P

C j q q

¢

d q dc EQ j

+ P q

d q dc EQ i

=0

d q dc EQ i

1

=0

j:

d

...

i:

d

+ P q

d Q i E Q

dc

+ P q

d Q i E Q

dc

c EQ,i

dc

c EQ,i

dc

Solving for

d q dc EQ j

, then summing up and solving for

d Q dc E Q i

yields:

d q E Q j dc

d Q i E Q

dc

= ( 1 ) R j

Ã

= ( 1)

d q dc EQ i

d Q dc E Q i

+

d q dc E Q i

Pj=6

i

R j +

1 P q

d

1+

Pj=6

i

+ dc c EQ,i 1 P q d θ Pj=6 dc i c E Q , i R j !

R j

38

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