= Solution
Use the natural filtration $\mathcal F_t=\sigma(Z_1,\ldots,Z_t)$ and the usual integrable definition of the <Snell envelope>. Since the factors are positive, its terminal value is $U_T=Y_T$. If $U_{t+1}=Y_{t+1}c_{t+1}$, <independence of random variables> gives
$$
\mathbb E[U_{t+1}\mid\mathcal F_t]=Y_tc_{t+1}\mathbb E Z_{t+1}.
$$
The backward recursion therefore becomes
$$
\boxed{U_t=Y_tc_t,\qquad c_T=1,\qquad c_t=\max\{1,c_{t+1}\mathbb E Z_{t+1}\}.}
$$
This proves the assertion by backward induction. All constants are finite and at least one. More explicitly,
$$
c_t=\max_{t\le u\le T}\prod_{j=t+1}^u\mathbb E Z_j,
$$
where the empty product is one. This is the <Snell envelope of a multiplicative process>. The standard <Snell envelope> hypothesis includes integrable rewards; <independence of random variables> and positivity then give finite first moments of the factors. If that hypothesis is dropped, an infinite-mean factor gives an infinite continuation value, so finite constants need not exist.
For the <discrete-time binomial market>, write $p=(1+r/\varepsilon)/2$. The assumptions give $0<p<1$, positive <bank account> and <stock> prices, and
$$
\mathbb E R_{t+1}=p\varepsilon-(1-p)\varepsilon=r.
$$
Thus
$$
\mathbb E\left[\frac{S_{t+1}}{B_{t+1}}\mid\mathcal F_t\right]=\frac{S_t}{B_t}.
$$
The finite discrete-time <fundamental theorem of asset pricing> states that a market with a strictly positive <numéraire> has no <arbitrage> if there is an equivalent probability measure under which all numéraire-denominated asset prices are <martingales>. The original probability measure is already such an <equivalent martingale measure>, proving \b[no <arbitrage>].
The complete-market American-option pricing theorem identifies the discounted replication or superhedging value with the <Snell envelope> of the discounted exercise rewards. Here
$$
Y_t=\frac{S_t^2}{(1+r)^t}=S_0^2\prod_{j=1}^t\frac{(1+R_j)^2}{1+r}.
$$
The independent factors have common mean
$$
\lambda=\frac{\mathbb E(1+R_j)^2}{1+r}=\frac{1+2r+\varepsilon^2}{1+r}=1+\frac{r+\varepsilon^2}{1+r}>1.
$$
The <Snell envelope> recursion consequently gives $c_t=\lambda^{T-t}$. At every time before $T$, the continuation value strictly exceeds immediate exercise, so \b[optimal exercise is at maturity]. The undiscounted value process is $V_t=S_t^2\lambda^{T-t}$, and the time-zero replication cost is
$$
\boxed{V_0=S_0^2\left(\frac{1+2r+\varepsilon^2}{1+r}\right)^T.}
$$
In fact $V_t/B_t$ is a <martingale>, not just a <supermartingale>, because the <Snell envelope> has no exercise compensator before maturity.
To hedge the optimally exercised payoff, subtract the two next-state <portfolio> equations. The first-period <delta hedge> is
$$
\Delta_0=\frac{S_0^2(1+\varepsilon)^2\lambda^{T-1}-S_0^2(1-\varepsilon)^2\lambda^{T-1}}{S_0(1+\varepsilon)-S_0(1-\varepsilon)}
=\boxed{2S_0\lambda^{T-1}}.
$$
The corresponding <bank account> holding, since $B_0=1$, is $V_0-\Delta_0S_0=S_0^2\lambda^{T-1}(\lambda-2)$. Substitution into either next-state equation verifies both terminal values at time one. Repeating this two-state calculation at every node gives a <self-financing strategy> replicating the maturity payoff. It also covers earlier exercise: $V_t\ge S_t^2$ at all dates, so the seller can meet any earlier payment with the available wealth.
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