In class we analyzed a $2-$period bargaining game in which player $1$ makes a proposal x

$($player $1$s proposed share$)$ in the first period t $=1($in which the pie is of size $1),$ and

player $2$ makes a proposal y $($player $2$s proposed share$)$ if the second period t $=2$ is

reached $($in which the pie is of size $0<$ $<1).$ Rejection of the second proposal results

in each player getting $0.$

Now consider a variant in which the only change is that, if player $1$ rejects the proposal

in the second period, then player $2$ still gets $0$ but now player $1$ gets a $($where $0<$ a $<$ $),$

i$.$e$.,$ player $1$ has a positive $$outside option$.$

$($a$)(10$ points$)$ Draw the game$-$tree.

$($b$)(20$ points$)$ Use backwards induction to find the SPE. $($The SPE includes all of

the acceptance rules, as well as the actual proposals each player would make.$)$

$($c$)(10$ points$)$ As a becomes larger, what happens to player $1$s share of the pie in

the SPE outcome? Explain.

$3$