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2

As barrycarter stated in the comment - the value of a set of [European!] options is the sum of the values of the individual options. This is simply follows from integral of a sum being a sum of integrals. $$butterfly\,option\,price = \\ \int_0^\infty butterfly\,payoff(S) dS = \\ \int_0^\infty (call\,payoff(S,K)+call\,payoff(S,K')+call\,payoff(S,K'')) dS ...


0

You are onto something, it is inconsistent to be calculating vega with Black-Scholes considering it assumes that volatility is constant. Black-Scholes is not a good for modeling option prices/implied volatility. It's a very good intuitive model (like the CAPM), and a good way of organizing thoughts, but it is not an accurate depiction of reality. If it ...


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On the topic of your second paragraph, the author below is the authority on precisely that topic. Start at page 19 https://www8.gsb.columbia.edu/leadership/sites/leadership/files/Is%20economics.pdf


2

As your code works for the short maturity case, I assume that it is correct. The volatility of $80 \%$ is simply huge. Thus the area covered by the paths is huge too. As you can read e.g. here the sampling error is proportional to the variance of the process, which is huge in your case. As a brute force solution you can just enlarge the number of samples. ...


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Increase the number of paths in your simulation for the getting the terminal prices, and at some point your monte carlo option price will finally converge to Black scholes option price as you are using a very longer maturity call option i.e. 10 year call option.


1

Yes. If you know what the drift is.


4

Note that \begin{align*} (K-S_T)^+ \ge K-S_T. \end{align*} Then \begin{align*} p &\equiv E\Big(e^{-rT} (K-S_T)^+ \Big)\\ &\ge E\Big(e^{-rT} (K-S_T) \Big)\\ &=K\, e^{-rT} - S_0\\ &= 670 \times e^{-0.05 \times 55/365} - 563.48\\ &=102.49. \end{align*} However, the option price is 101.375, which is smaller. This is the reason that you have ...


0

"So how come traders actually use that information in trading if itstems from an 'arbitrary' Blackbox?". You make a bet on realized vol via delta-neutral option position. if you believe that realized vol will be different from IV of BS (or your another model) than do exactly what you mention in question 2. if IV is too high: sell option, delta-hedge with ...


0

The Black-Scholes PDE holds in the continuation region : $$ u_t = - \frac{1}{2} \sigma^2 u_{ss} $$ (ignoring interest rate). This says that theta is as smooth as gamma. The "smooth pasting" literature you mention shows that delta is continuous at the exercise boundary, but gamma has a jump. So theta has a jump as well. In other words, in the time ...


2

For a call option with price given by \begin{align*} c = S_0 \Phi(d_1) - K e^{-rT}\Phi(d_2), \end{align*} the delta hedge ratio $\Phi(d_1)$ is the number of shares to hold. That is, $S_0 \Phi(d_1)$ is the total holding share value for hedging, while $K e^{-rT}\Phi(d_2)$ is the total cash amount in short. In the question, it says that, for $N$ options, ...


0

if you have all options same strike then you have no vol smile risk (at the first glance). vol smile is about different IVs for different strikes in series. the only possible effect on you from vol smile is how vol surface evolves when underlying moves (read about sticky strike and sticky delta). if IV surface behaves according to sticky delta rule then ...


1

A 3x leveraged fund that experiences a drop of more than 33% will lose all its money and close down when the value hits 0. Most funds, however, set a lower loss limit, usually around 5%, 10% or 20%, and will try to exit the market if those values are triggered. Of course, it is not that easy to exit efficiently in a day where the market drops that much. ...



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