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Say we have the following binary option $B$ on asset $S$ with strike K and expiration time T, assume also that the following relation holds at time $0$:

$B > N*C(K,T)-N*C(K+1/N,T)$

Where $N$ is some natural number and $C(K,T)$ is the call option on asset $S$ with strike K and expiration time T

How is it possible to find an arbitrage strategy in that case , under the assumption that we can buy call options at any strike?

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  • $\begingroup$ 1) is B an up or a down binary option ? 2) right hand side is negative $\endgroup$ – MJ73550 Apr 12 '16 at 15:08
  • $\begingroup$ 1) up binary option 2) edited $\endgroup$ – pointiy Apr 12 '16 at 15:12
  • $\begingroup$ Your inequality is always true, and then it is difficult to find an arbitrage opportunity, unless you change your inequality to $B < N*C(K,T)-N*C(K+1/N,T)$ or $B > N*C(K-1/N,T)-N*C(K,T)$ $\endgroup$ – Gordon Apr 12 '16 at 15:36
  • $\begingroup$ Thanks @Gordon is it possible in case $B < N*C(K,T)-N*C(K+1/N,T)$? $\endgroup$ – pointiy Apr 12 '16 at 15:38
  • $\begingroup$ The argument in that case is similar. $\endgroup$ – Gordon Apr 12 '16 at 15:40
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We assume that the inequality is given by \begin{align*} B > N C(K-1/N, T) - N C(K, T).\tag{1} \end{align*} The argument for the case with the inequality \begin{align*} B < N C(K, T) - N C(K+1/N, T) \end{align*} is similar. $$$$ For the binary option, \begin{align*} \pmb{1}_{\{S_T \ge K\}} = \begin{cases} 1, & \textrm{if } S_T \ge K,\\ 0, & \textrm{otherwise}, \end{cases} \end{align*} while for the portfolio with payoff \begin{align*} X_T &= N\bigg[S_T-\Big(K-\frac{1}{N}\Big)\bigg]^+ - N (S_T-K)^+\\ &= \begin{cases} 1, & \textrm{if } S_T \ge K,\\ N\bigg[S_T-\Big(K-\frac{1}{N}\Big)\bigg], & \textrm{if } K-\frac{1}{N} \le S_T \le K,\\ 0, & \textrm{otherwise}. \end{cases} \end{align*} Then, it is obvious that \begin{align*} \pmb{1}_{\{S_T \ge K\}} \le X_T,\tag{2} \end{align*} and, consequently, \begin{align*} B \le N C(K-1/N, T) -N C(K, T). \end{align*}

$$$$ However, if (1) holds, we can then short the binary option, short $N$ units call option with strike $K$, and long $N$ units call option with strike $K-1/N$. We have a net profit \begin{align*} B - \big[N C(K-1/N, T) - N C(K, T)\big]. \end{align*} Moreover, at maturity $T$, we have the portfolio payoff \begin{align*} X_T - \pmb{1}_{\{S_T \ge K\}} \ge 0, \end{align*} as per (2) above.

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  • $\begingroup$ In case $B < N*C(K,T)-N*C(K+1/N,T)$ and doing a similar strategy we will not earn at time $T$ in case $K<S_T<K+1/N$ since we short the binary option and loss 1 while the profit from the rest is smaller then 1 $\endgroup$ – pointiy Apr 12 '16 at 15:47
  • $\begingroup$ @pointiy: In this case, we short $N$ units call options with strike $K$ and long $N$ units call options with strike $K+1/N$, and short the binary option. $\endgroup$ – Gordon Apr 12 '16 at 15:57
  • $\begingroup$ I still don't understand how in that case we have profit in case $K<S_T<K+1/N$. $\endgroup$ – pointiy Apr 12 '16 at 16:07
  • $\begingroup$ Sorry @pointiy: In this case, we short $N$ units call options with strike, long $N$ units call options with strike $K+1/N$, and long (NOT short) the binary option. $\endgroup$ – Gordon Apr 12 '16 at 17:16

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