Hot answers tagged binary-options
10
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 ...
10
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
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
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 ...
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