In the second period, the buyer accepts any offer $s_{2}$ $\leq$ $0.7$ if he has rejected the first offer and any offer $s_{2}$ $\leq$ $0.3$ if he has accepted the first offer.
Given this, there are only two offers that may be optimal for the second seller: either $s_{2}$ $=$ $0.3$
or $s_{2}$ $=$ $0.7$. Let $\mu$ denote the probability that the second seller assigns to the fact that the buyer has rejected the first offer. The optimal offer in the second period is:
\begin{equation}
s_{2}=\left\{
\begin{array}{@{}ll@{}}
0.3 & \text{if}\ \mu < \frac{3}{7} \\
0.7 & \text{if}\ \mu > \frac{3}{7} \\
\end{array}\right.
\end{equation}
any randomization between $0.3$ and $0.7$ $\text{if}\ \mu = \frac{3}{7} $
Suppose that the equilibrium is such that the buyer rejects the first offer. In the second
period, the second seller will offer $s_{2} = 0.7$ and the buyer's payoff is equal to 0. Because of this, the buyer should accept any offer smaller than $0.7$ (the buyer can guarantee a positive payoff by accepting the first offer and rejecting the second one). But then the first buyer
should offer $s_{1} < 0.7$ and get a positive payoff. In other words, we have shown that there is no PBE in which the buyer rejects the first offer.
Let us now see whether we can construct an equilibrium in which the buyer accepts the first offer. In this case $s_{2} = 0.3$. This implies that the buyer will accept the first offer only
if $s_{1} < 0.3$ (in fact, by rejecting the first offer and accepting the second one, the buyer
guarantees a payoff equal to $0.4$). Finally, given this, it is optimal for the first seller to offer $s_{1} = 0.3$
To sum up, we have the following pure-strategy PBE.
The first seller offers $s_{1} = 0.3$ In the first period, the buyer accepts an o¤er if and only if $s_{1} = 0.3$
In the second period, the second seller assigns probability $\mu = 0$ to the fact that the buyer has rejected the first offer.
Thus, the second seller offers $s_{2} = 0.3$
Finally, the buyer's strategy in the second period is described above.
