# Tag Info

1

This question is extremely interesting and not that straightforward. See answer here. From a financial perspective this is very much like pricing an American call (stopping rule = intrinsic value from exercice (i.e. current cash earned) > continuation value (i.e. what you can expect to gain). Note that you can never win more than 13 nor lose (at worst you ...

0

You're half way there. When $R < D \ (< U)$, the return of the stock dominates the risk-free return in all states of the world. To benefit from that, just borrow cash and invest in the stock. At $t=0$ this requires no net investment: borrowing cash means your account is credited $S_0$, while subsequently buying the stock suggests it is debited ...

0

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, & ...

1

Exploiting an arbitrage is straightforward. Constructing and noticing one is the hard part. In your case if you know that Swptn(K,T1,T2)+Swptn(K,T2,T3) >= Swptn(K,T1,T3), Simply sell Swptn(K,T1,T2)+Swptn(K,T2,T3) and buy Swptn(K,T1,T3). Sell the most expensive and buy the cheapest. L.

Top 50 recent answers are included