# Isn't this modified stop-loss strategy an arbitrage?

In John Hull's The Book, section 18.3 he briefly discussed a stop-loss strategy for writing a call option: buy one share of stock whenever $S_t>K$ and sell it otherwise (except at time $0$: if $S_0\le K$ we do nothing), and hence the call writer will own the stock in case $S_T>K$ and not otherwise, making an overall profit of $c_0 - \max(S_0 - K, 0)\ge 0$. However, he pointed out two reasons why this isn't the case: 1). it ignores the time value of money; 2). it is impossible to buy/sell exactly at $K$; also, if the stock price follows the Wiener process it can happen that $\#\{t\mid S_t=K, t\in[0,T]\}=\infty$.

Now, let's temporarily ignore the problem 2) and assume we have a super trader who can carry out the order at the exact price we want. The main problem is then the first one. However, I think it can easily be circumvented by "discounting" our stop-loss rule towards time $T$: whenever $S_t> Ke^{-r(T-t)}$, buy one share and sell it otherwise (except at time $0$: if $S_0\le Ke^{-rT}$ we do nothing). Thus, in the absence of transaction costs, it can be verified that our net profit discounted towards $T$ is $$e^{rT}\cdot\text{BS Price}-\max(S_0e^{rT}-K,0)=S_0e^{rT}\Phi(d_1)-K\Phi(d_2)-\max(S_0e^{rT}-K,0).$$ Although its sign is not obvious, by experimenting with several sets of parameters I become fairly convinced that it is non-negative. But then such a strategy would in effect become an arbitrage in this BS world. What goes wrong?

• If you require the quantity you trade to be a continuous function of time and price it is not possible to trade at an exact price, as doing so would require acquiring a non-zero quantity over an infinitesimal period of time. – yong Apr 3 '17 at 8:41
• @yong I don't think this is the essence of this problem. If you want a practical strategy, consider buy if stock price rises above $K_t + \epsilon$ and sell if it falls below $K_t-\epsilon$, and we can approach our ideal stop loss order as close as possible by letting $\epsilon$ tend to zero. (assuming sufficient computer power, etc) – Vim Apr 3 '17 at 8:45
• @yong more importantly, if my strategy really yields an arbitrage, then it would violate the arbitrage-freeness of the BS model. For the time being, I guess the problem lies in the non-predictability (in the stochastic process sense) of the stop loss order, whose stock position is expressible as $$1\cdot \mathbb{1}_{S_t>Ke^{r(T-t)}}$$. But I'm not sure though. – Vim Apr 3 '17 at 8:51
• It is a pretty huge assumption to ignore point (2), the infinite-crossing problem. The $K_t \pm \epsilon$ band will have finite trade counts for strictly positive epsilon, while the trade count will be either zero or infinite with probability 1 when $\epsilon=0$. – Brian B Apr 3 '17 at 16:58
• i think you would run into losses if the stock starts to move around K. When it is above K, you buy but when it falls you sell. If you keep doing this long enough, you will run into losses, this is a short gamma trade – nimbus3000 Apr 3 '17 at 20:26