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You said you've used the one-step binomial method to calculate 11.8 for one share. I'm not sure if that's right, since the call option $C_{0}$ $$ C_{0} = \left(\frac{S^+X^--S^-X^+}{S^+-S^-}\right)e^{-rT}+\left(\frac{X^+-X^-}{S^+-S^-}\right)S_{0}. $$ In your case $S^+=240, S^-=180 \implies X^+=S^+-E=240-205, X^-=0.$ using these I have $13.75 per a share.


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Even if you are sure that the option is misvalued, you can't say that there is arbitrage gain. Why?? because the volatilty you used to price your option is almost sure different from the one used from the other side. So if you are sure of your volatility, you can buy a variance or a volatility swap/option to make gain of your information. Regarding your ...


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You almost got it... but you are missing one important piece the "every" HFT has. They use what is called "prime brokers", where they hold their account and have the ability to open/close trades in multiple venues/ecn's That means, that all their positions will be held by the prime broker. So, let's say you have an account with these Prime Brokers and you ...


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The only missing point is that, by NA, if an asset has zero volatility, it is riskless and must therefore grow at the risk-free interest rate: $\mu \equiv r$ (Else, you buy the highest yielding asset and sell the lowest yielding). In such situation, the valuation of an option is straightforward: it is the discounted payoff $e^{-r\left(T - t\right)} \left[...


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I don't know where to begin with the potential problems calling this an "arbitrage", but here then below is the kind of "answer" I suspect your lecturer is looking for: If you're supposed to use the 2y or 3y in any way, the "arb" becomes a "carry trade", as played by many a carry-and-roll junkie ;-)


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