ABSTRACT
Some simple calculations then show that pt = p~ = 2ad + e while PI(Pt, pn = P2(Pr. pn = ad + e/2. Given p~, we now study the best reply PI of firm 1 in
_(p *) _p~ - PI + 2ad + 2ae s I, P2 - 4ad •
and in
with
_(P *) -P~ - PI + 2ad - 2ae s u pz - 4ad '
respectively. Consider first si l . The maximum of PIS(PU pn over [O,oo[ is reached at PI = 2ad + ae + e/2. Then we have S(PI' pn > ~ - a, from which it follows that there is no best reply against p~ in si l . Let us now come to siz. It is easy to check that the maximum of p)s(p),pn over [O,oo[ is reached at P; =2ad+e/2-ac. For .e/d ~ 4a/(l-2a), s(p;, pD ~ 1 so that PI = p;. On the other hand, for e/d > 4a/(1 - 2a), s(p;, pn > 1 which implies that P; is not the best reply of firm 1 against pr Actually, in this case, PI corresponds to the price for which 1 - S(PI' pn = 0, i.e., p) = e(l - 2a).
