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5

Setting aside, that it's not pure riskless arbitrage, but rather statistical arbitrage: You can extract the profit by performing continuous delta hedging. If you constantly adjust your hedge position you gain/lose money by delta hedging. Being long option (gamma long), you sell at higher prices and buy at lower ones. Over the course of time you realize ...

1

What if the actual volatility during the following period is lower, so my bet was correct, but the implied volatility stays the same for the whole period anyway? If under those circumstances I liquidate the position, wouldn't the profit be 0? I think I know where your confusion comes from. 1) this isn't arb - it is not a risk free strategy ...

0

I know that this question is quite old, but I uploaded a matlab implementation of the method to fileexchange: http://www.mathworks.com/matlabcentral/fileexchange/46253-arbitrage-free-smoothing-of-the-implied-volatility-surface

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you never trade spot volatility. you exchange it for something else. if you want to exchange it for spot implied volatility, you buy a volatility swap. if you want to exchange it for forward implied volatility you get options.

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They are not referring to any implied volatility but actual volatility, i.e. statistical standard deviation. The price volatility is the annualized standard deviation of bond price changes and the yield volatility the annualized standard deviation of bond yield changes. These quantities are usually estimated using a historical estimator. If you have n ...

2

we use implied vol for similar reasons why we use duration. we know that security prices are not linear functions of rates, yet we look at the duration, because it gives us an idea of sensitivity to a rate. implied vol gives you a measure of volatility, it doesn't perfectly describe it, but as long as we know this, it's still a valuable metric.

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I'll try to anwswer too. 1) You seem to try to interpret implied volatility as having a statistical nature. In fact implied volatility is nothing but (today's) market prices except that you look at them through Black and Scholes "glasses". Why the Black-Scholes model? There are many reasons for that. this is the simplest sensible model: basically you ...

3

My try to answer this question with some other questions: Is the BS model right? No. Is it useful: yes. Taking a traded price and the BS Model there is only one input factor that is not given by the market: the implied volatility. It is a measure to compare options across time and strike. Are there better models? yes. Those that you mention: The local vol ...

1

I am not a trader and will pass on question 1 for now. To answer your second question. You want you model to be able to reproduce market prices of certain vanilla instruments. This way you achieve consistency with the market. Thus if you want to price an exotict call option you will calibrate your model to liquid call option prices. If the option has more ...

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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 ...

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 ...

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 ...

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