# Tag Info

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The text of your question doesn't actually match the question title. The answer to your title is of course yes binary options make sense. And as others have pointed out with binary options your reward is limited, and conversely the risk involved in writing them is less. To answer your additional question you can replicate a binary option with a tight call ...

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The price of a binary option, ignoring interest rates, is basically the same as the CDF $\phi(S)$ (or $1-\phi(S)$ ) of the terminal probability distribution. Generally that terminal distribution will be lognormal from the Black-Scholes model, or close to it. Option price is $$C = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$ for calls and $$P = e^{-rT} ... 5 The key difference besides the cap is that there is nothing in between: its 100 or nothing (binary!) - with traditional options you have S-K as long as S>K (for the call). You can find out more here: http://en.wikipedia.org/wiki/Binary_option 3 They would make sense in certain narrow applications; one can perhaps think about scenarios where binary option might be the most efficient, quickest or easiest way to either benefit from a particular insight OR to hedge against some sort of event ... the real question is whether the volume in the markets for binary options will continue to sufficient to ... 2 For a long or short position in a vanilla put, the maximum risk/reward is known and capped. For a long position in a vanilla call, the maximum risk is also always known and capped. The maximum reward is therefore known and capped for a short vanilla call position. However, the reward is unbounded for a long vanilla call position, and therefore the risk is ... 2 5 minutes is a very short time period! If you have access to real time data of Implied Volatility and transaction Volume of the underlying of your option than you can take a look to the following article: Volatility Forecasts, Trading Volume, and the ARCH versus Option-Implied Volatility Trade-off In this article, the authors use the information from ... 2 First of all, don't forget that there are two different probability measures at play here: the frequentist market measure that reflects actual observation of the market "in the long run" and the market-neutral martingale measure which is pertinent for pricing options. More or less, we can take the frequentist measure, and "back out" the effects of market ... 2 No, there is an upper limit to a binary option's value, based on the interest rate and how much of the distribution can be packed under the payoff region. Essentially$$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$$for puts. Neither of the integrals can ever exceed 1.0 and often they take on a ... 1 Ofcourse, It is always possible to find the implied volatility. The value of binary call is$$ {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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Black–Scholes usually assumes your time and volatility are annualised. Accordingly, when you calculate the volatility term you would usually annualise it to 252 or 260 (or however many trading days a year are applicable to your situation). Accordingly, the time remaining term of the Binary Option must also be expressed as a fraction of a year (again, 252, or ...

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A really simple and arbitrage free solution is to extrapolate flat volatility on the same moneyness. Let's say that you want an implied volatility for strike $K$ at time $t<t_1$, and $t_1$ is the first pillar on the surface. You look at the moneyness level $k=K/F_t$, then look for $K'$ to get the volatility at the same moneyness level of the first ...

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