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Figure 1: Seminar 3 In the graph I have drawn prots as a function of qA for a given level of qB : 2 Π(qA |qB = q B ) = −qA + (100 − q B ) qA If qB ∈ [a, b], then I can draw 2 Π(qA |qB = b) = −qA + (100 − b) qA 2 Π(qA |qB = a) = −qA + (100 − a) qA and notice that for every qB ∈ (a, b), the prot schedule Π(qA |qB ∈ (a, b)) will ∗ ∗ lie somewhere between the two and will peak between qA (b) and qA (a). From the graph you should try to gure out the following: ∗ ∗ • ∀qB ∈ (a, b), every qA < qA (b) is dominated by qA (b) ∗ ∗ • ∀qB ∈ (a, b), every qA > qA (a) is dominated by qA (a) Hence, by elimination of strategies, it must be the case that strictly dominated ∗ ∗ qA ∈ [qA (b) , qA (a)] = 100−b 100−a 2 , 2 • At round 0 , qi ∈ [0, 100] ≡ [a0 , b0 ] 0 100−a0 • At round 1, qi ∈ 100−b , 2 ≡ [a1 , b1 ] 2 • ..... 1 • At round k + 1, qi ∈ 100−bk 2 k , 100−a ≡ [ak+1 , bk+1 ] 2 What I was trying to explain this morning is that we can equivalently impose the following condition ∂Π > 0 or • qA is dominated if ∀qB ∈ [a, b], either ∂q A Now, ∂Π ∂qA ∂Π ∂qA = 100 − 2qA − qB and so ∂Π ∈ [100 − 2qA − b, 100 − 2qA − a] ∂qA • ∂Π ∂qA > 0 implies 0 < 100 − 2qA − b ⇒ qA < 100−b 2 • ∂Π ∂qA < 0 implies 100 − 2qA − a < 0 ⇒ qA > 100−a 2 and everything goes as before. 2 <0