I am currently revising for my exam in Financial Mathematics, and I could not solve this question:
For $T > 1$, consider a $T$-period model with a single risky asset and a bank account which pays zero interest, $r = 0$. The price process of the asset is given by \begin{align*} S(0) = 1, \quad S(t) = \begin{cases} S(t-1) e^{X(t-1)}, & S(t-1) > \varepsilon \\ S(t-1) e^{X(t-1)+2}, & S(t-1) \le \varepsilon \end{cases}, \quad t = 1, \ldots, T, \end{align*} where $\varepsilon \in (0,1)$ and $X = (X(t), t \ge 0)$ is a sequence of iid random variables such that each $X(t)$ takes the value $1$ with probability $1/2$ and $-1$ with probability $1/2$.
Construct an arbitrage opportunity for this model and verify all the conditions which define such an opportunity. Hint: for a complete answer, you will need to state appropriate assumptions on $\varepsilon$ and $T$.
What I have done so far:
We search a self-financing trading strategy $H$ such that $V_0 = 0$, $V_T \ge 0$ and $\mathbb EV_T > 0$. So if \begin{align*} 0 = V_0 = H_0(1)+H_1(1) \underbrace{S(0)}_{= 1} = H_0(1)+H_1(1), \end{align*} we must have \begin{align} H_0(1) = -H_1(1). \end{align} Next, since $H$ needs to be self-financing and $r = 0$, we obtain \begin{align} V_t = V_t^* = V_0^* + G_t^* = V_0 + G_t = 0 + G_t = G_t. \end{align} Write \begin{align*} S(t) &= S(t-1)e^{X(t-1)} 1_{\{S(t-1) > \varepsilon\}} + S(t-1)e^{X(t-1)+2} 1_{\{S(t-1) \le \varepsilon\}} \\ &= S(t-1)e^{X(t-1)} (1_{\{S(t-1) > \varepsilon\}} + e^2 1_{\{S(t-1) \le \varepsilon\}}). \end{align*} We have that \begin{align*} V_T &= G_T = \sum_{u = 1}^T H_1(u) \Delta S(u) = \sum_{u = 1}^T H_1(u) (S(u) - S(u-1)) \\ &= \sum_{u = 1}^T H_1(u) (S(u-1)e^{X(u-1)} (1_{\{S(u-1) > \varepsilon\}} + e^2 1_{\{S(u-1) \le \varepsilon\}}) - S(u-1)) \\ &= \sum_{u = 1}^T H_1(u) S(u-1) (e^{X(u-1)} (1_{\{S(u-1) > \varepsilon\}} + e^2 1_{\{S(u-1) \le \varepsilon\}}) - 1) \end{align*}
Unfortunately, I don't really know how to go on and to derive some conditions. Please help.