Seminar 3

```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 &lt; qA
(b) is dominated by qA
(b)
∗
∗
• ∀qB ∈ (a, b), every qA &gt; 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
&gt; 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
&gt; 0 implies 0 &lt; 100 − 2qA − b ⇒ qA &lt;
100−b
2
•
∂Π
∂qA
&lt; 0 implies 100 − 2qA − a &lt; 0 ⇒ qA &gt;
100−a
2
and everything goes as before.
2
&lt;0
```