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An out-of-the money binary call option will have two implied volatilities. After the first implied volatility keep looking at increasingly higher implied volatilities. After a while the binary call option price doesn't rise further, i.e. the binary call vega falls to zero, and then the binary call option price starts falling as implied volatility continues ...


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Ofcourse, It is always possible to find the implied volatility. The value of binary call is $$ {e}^{-r(T-t)}N(d_2) $$ where $$ d_2=\frac{ln(\frac{S}{E})+(r-D-\frac{\sigma^2}{2})\tau}{\sigma\sqrt\tau} $$ Now, there is nothing that can ever ever stop the newton raphson method to find a $\sigma$ for which the value of binary call is given and is positive ...


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No, there is an upper limit to a binary option's value, based on the interest rate and how much of the distribution can be packed under the payoff region. Essentially $$C = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$ for calls and $$ P = e^{-rT} \int_0^K \psi(S_T) dS_T$$ for puts. Neither of the integrals can ever exceed 1.0 and often they take on a ...



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