= Solution
Let $P_k$ be the initial proposer's share, measured as a fraction of the pie currently available, with $k$ rounds remaining. The last-round argument in part (e) gives $P_1=1$. Rejecting an offer leaves the responder as next proposer with a pie smaller by factor $\alpha$, so its continuation value is $\alpha P_{k-1}$. The current proposer offers exactly that amount, which is accepted by the <continuous-offer bargaining acceptance lemma>. Hence
$$
P_k=1-\alpha P_{k-1},\qquad\boxed{P_k=1-\alpha+\alpha^2-\cdots+(-\alpha)^{k-1}=\frac{1-(-\alpha)^k}{1+\alpha}.}
$$
The formula follows by summing a finite <geometric series>, or by induction in the recurrence. Agreement is immediate in every subgame, so this share is player 1's actual initial payoff.
For odd $k$, $P_k=(1+\alpha^k)/(1+\alpha)$ lies above $1/(1+\alpha)$ and decreases with $k$. For even $k$, $P_k=(1-\alpha^k)/(1+\alpha)$ lies below that value and increases with $k$. Thus prefer odd to even, fewer rounds among odd choices, and more rounds among even choices. For the given $k_1<k_2$, this simplifies to \b[choose $k_1$ if $k_1$ is odd, and choose $k_2$ if $k_1$ is even]. All comparisons are strict because $0<\alpha<1$. This is the <finite-horizon bargaining parity effect>.
Back to article page
