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To recover the Black-Scholes pricing equation, you should first express the standard normal cdf in terms of its characteristic function analogous to the Heston solution: $$N(x) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi x} f(\phi)}{i\phi}] d\phi$$ where $f(\phi)$ is the characteristic function of the standard normal distribution: $$... 3 I think you need to go even one step further than vonjd went in his reply. If liquid trading of the underlying is not possible, not only the arbitrage argument underlying risk neutral pricing breaks down. In that case there is simply no reason why the prices of those two assets (the option and its underlying) should be related in any way at all. So in my ... 2 The consensus nowadays is that stable distributions are not a well fit, although they do possess heavy tails. In particular Cauchy has too fat tails. The reasons for this are disparate, however the first that comes to mind is that empirically longer horizons show a decrease in tail thickness, approaching normality for 1-year returns (although this has been ... 2 Behavioral Finance is a wide topic, which I believe is still today underestimated by many financial professionals. How can it be used by quants? Well, in portfolio optimization it can be used "as an overlay" in the form of constraints where the optimal portfolio can not be too different from the current portfolio, because clients have behavioral biases ... 2 the answer for calculating the prices can be found here - see chapter: Black–Scholes valuation ;) The put-call parity in that case is pretty straight forward: P=Se^{-qT}-C. Using the results presented on the Wikipedia page in the aforementioned section this can be proved as follows P=Se^{-qT}-C =Se^{-qT}-Se^{-qT}\Phi(d_1) ... 2 No the discounting factor that you use for backward induction won't change. (confer here Chapter IV) This is only seems confusiong due to the mathematical formulation. Introducing continuous dividends basically adjusts your stock price (down) by discoutning the divididend (for it is paid out and thus dicreases the stock value). Your "risk-free" stock value ... 2 There is no difference in information, though the fitting algorithm may increase in complexity. First note that in practice you never have an entire curve or surface of prices C(K,T) of any kind of option. You only have a finite number of observations and even those typically have a bid and an offer. I would therefore argue that the correct picture of ... 2 In effect, you are wondering whether to price this option on risk-free probability distributions (B-S drift r_f), or real-world ones (B-S drift \mu, however calibrated) One cannot short the mutual fund, so the argument for using risk-free is weakened. But, there are various economic equilibrium arguments why using it may still be OK. If you use the ... 1 Look the first answer of this thread: How to derive the implied probability distribution from B-S volatilities? Also many papers in Dupire volatility have your formula derivation. For example, look at (10) in http://www.javaquant.net/papers/DupireLocalVolatility.pdf 1 The Price of an American option may contain information on the expected behaviour of it's holder. When might he/she exercise the option ? Contrary to European options that don't. Thus when you are primarily interested in "reconstructing" the transition density - I would stick with the European-Option-Prices. If however you were to price path dependant ... 1 Hint The future world has 4 states: (0.5,0.5), (2,0.5), (0.5,2), (2,2). You have 4 instruments - cash, each stock, and an option they are both \2 which is traded. Take x,y,z,w of each and match the portfolio to the price of the option in each market state. You get 4 equations and 4 unknowns, solve, and supposedly you get a unique solution, which ... 1 I would put it differently. Modelling variance in an additive way (an OU process is in some regard additive) is more natural than e.g. a gemetric Brownian motion model (which on the other hand does not model mean reversion). Volatility as it is a square-root is by no means additive. Let (B_t)_{t \ge 0} be Brownian motion then we have$$ VAR(B_t) = t = ...

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I would define the weights $w_1,\ldots,w_n$ as whatever number you want and the basket given by $$B_t = \sum_{i=1}^n \frac{w_i}{W}S_t^{(i)}\ , \qquad W = \sum_{i=1}^nw_i$$ so the weights always sum to one. This doesn't make much sense, however, because you are changing the product, not a market variable. This meaning that when the weights change, the ...

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you get a volatility skew by imposing a neumann-like barrier if market makers think a stock won't surpass a certain threshold, a skew is inevitable if one were to match the pricing under a barrier with the BS formula https://en.wikipedia.org/wiki/User:Barrieroption/sandbox

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