How to prove Rubinstein's Bargaining Game with Random (instead of Alternating) Offers? Ariel Rubinstein introduced in Econometrica, 1982 the canonical model of bargaining, which I describe below.

Two Players $i,j\in N$ bargain over how to split \$1. Both Players have linear (discounted) utility functions $u_i=\delta^{t-1}x_i$ and $u_j=\delta^{t-1}x_j$. The bargaining procedure is as follows: one Player is chosen to make an offer $(x,1-x)$ to Player $j$, who then accepts or rejects. If he accepts, the game ends and Player $i$ gets $u_i=x$ and Player $j$ gets $u_j=\delta^{t-1}(1-x)$. If he rejects, he makes an offer $(1-y,y)$ to Player $i$, who then accepts or rejects. If he accepts, the game ends and Player $i$ gets $u_i=\delta^{t-1}(1-y)$ and Player $j$ gets $\delta^{t-1}y$. If he rejects, he makes an offer $(x^{\prime},1-x^{\prime})$ to Player $i$ and so on. The game ends whenever Players reach an agreement; if they never do so, they get $u_i=0$ and $u_j=0$.

The unique Stationary Subgame Perfect Equilibrium (SSPE) of this game is well-known and several explicit proofs can be found out there. Because of that, let me just remark that, in the unique SSPE of the game described above, an agreement is reached at $t=1$ and has the following form:
\begin{gather*}
x=\frac{1}{1+\delta}\\
1-x=\frac{\delta}{1+\delta}
\end{gather*}
Consider now the canonical Rubinstein bargaining model described above with a small tweak:

Following any rejection, Nature randomly chooses the identity of the next proponent. 

Hoel analyses the game I am considering in Economics Letters, 1987 and claims that the unique SSPE is almost identical to the one of the Rubinstein canonical bargaining game. However, he does not provide an explicit form of the Equilibrium proposals nor he provides an explicit proof of his claim (I'm not blaming him; it's probably a fairly easy claim to show that I just can't see). Instead, he argues that his claim can be derived following the results in Shaked & Sutton in Econometrica, 1984 and in Hoel in Scandinavian Journal of Economics, 1986. Unfortunately, I have thoroughly tried to derive an explicit proof of his claim, but I haven't succeeded at all so far. 
Therefore, my question is: could anyone please derive the Stationary Subgame Perfect Equilibrium of the tweaked Rubinstein game I am considering? (note: you don't need to check whether there exist non-stationary Subgame Perfect Equilibria).  
EDIT: given the null success of my question so far, I link here a clarifying video of 15 minutes that explains how to solve the canonical Rubinstein model presented above. Then, my question could be re-phrased as follows: how do I need to change the proof in this video for my modified version of the Rubinstein Bargaining Game?
Thank you all very much in advanced for your time.
 A: Suppose that the two players have the same discount factor and use stationary strategies.
At time $t$, when it's $i$'s turn to propose, she proposes $m_i$ to the receiver $j$. If the receiver accepts, she keeps $1-m_i$. 
The receiver $j$ accepts $m_i$ if this value is not smaller than what she may get waiting one period: with probability $1/2$, she will be proposer and get a value $(1-m_j)$; with probability $1/2$, she will be again receiver and will be offered $m_i$ again. Discounting the expected value from the next period, she accepts an offer $m_i$ when
$$m_i \ge \delta \left( \frac{1}{2} (1-m_j) + \frac{1}{2} m_i \right)$$ 
Since the proposer $i$ offers the minimum amount $m_i$ necessary for acceptance, we get
$$m_i = \delta \left( \frac{1}{2} (1-m_j) + \frac{1}{2} m_i \right)$$ 
Switching identities, we obtain the second equation
$$m_j = \delta \left( \frac{1}{2} (1-m_i) + \frac{1}{2} m_j \right)$$ 
Solving the system, we find
$$m_i = \frac{\delta}{2} \quad \mbox{and} \quad m_j = \frac{2 - \delta}{2}$$
as equilibrium offers. The offers are accepted in the first period and the two players obtain
$$u_1 = 1 - m_i = \frac{2 - \delta}{2} \quad \mbox{and} \quad u_2 = \frac{\delta}{2}$$
Incidentally, you can check that these are the same values you obtain if you replace $\delta$ with $\delta/(2 - \delta)$ as claimed (without proof) in Hoel (1987).
