# Arbitrage from ATM option trading?

So I was testing out a collar options strategy (long put, short call, and long shares of the underlying stock) in a backtest for a school finance project, and the profits & losses are given by the formula

$$\text{PnL}_{t+1} = (S_{t+1} - S_t) + [C_t(K_1) - P_t(K_2)] *(1 + R_t) + \max[K_1 - S_{t+1}, 0] - \max[S_{t+1} - K_2, 0]$$.

where $$S_t$$ is the price of the underlying asset at time $$t$$, $$C_t(K_1)$$ is premium of the call option at time $$t$$ given strike price $$K_1$$, $$P_t(K_2)$$ is premium of the put option at time $$t$$ given strike price $$K_2$$.

The first strategy I have tested involves using at-the-money (ATM) options, where the strike price equals the current spot price, i.e. $$S_t = K_1 = K_2$$. So the $$\text{PnL}_{t+1}$$ is reduced to just $$[C_t(K_1) - P_t(K_2)] *(1 + R_t)$$. Then using the put-call parity formula we also have $$C_t - P_t = S_t - \frac{S_t}{1+R_t},$$ then $$[C_t(K_1) - P_t(K_2)] *(1 + R_t) = S_t \cdot R_t,$$ where $$R_t$$ is the risk free rate for time $$t$$.

So the value of \$1 following this strategy at period $$t = 1 + S_0\cdot R_0 + \dots + S_{t-1} \cdot R_{t-1}$$ It is noted that the growth of \$1 under this strategy has an extremely high Sharpe ratio of approximately $$2.7$$ and the period to period returns is more superior than the risk-free rate of return, doesn't this mean that this is an arbitrage opportunity?

• When $K_1=K_2=S$ the profit from this strategy is going to be essentially zero, in dollar terms. It is a stock position fully hedged with a "syntetic short" position. No investment, no profit, return $=\frac{0}{0}$ – noob2 Nov 6 '18 at 19:23
• @noob2 but I still get some form of arbitrage profits from investing in the difference of the call and the put in a risk free asset like t-bills isn't it? So how do we explain the profits from there? – Weichen Christopher Xu Nov 6 '18 at 19:44
• its directly offset by the financing cost of buying the stock which u left out of the PnL equation – hjw Nov 7 '18 at 8:36