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I have this exercice, and for the last question, i tried to say that with lower bound, $C > S_0 - Ke^{-rT}$ which is $-8$ something but it doesn't make sense so i don't know what to do. Could we just say that under risk-neutral probabilities, price of call is $0.5*5*e^{-rT}$ , as $S_0$ should be the expected value of $110$ and $90$ in $t_1$ with $p = 0.5$ for example ?


A call struck at $100$ costs $2.97$, therefore a call with a strike higher than $100$ must cost less than $2.97$.

  • $\begingroup$ It makes sense, i was looking for an answer with some calculs but can't find one ! $\endgroup$ – TmSmth Nov 27 '19 at 18:33
  • $\begingroup$ The 110 Call pays less than the 100 Call in every situation in which $S_T>100$, and they both pay nothing for $<=100$. So 100 call is worth more by AOA. $\endgroup$ – Alex C Nov 27 '19 at 19:12

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