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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: $$... 5 Aleš Černý has very simple examples in his book. Alternatively, this paper seems to recap part of the chapter on Fourier series: Introduction to Fast Fourier Transform in Finance - Aleš Černý 5 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 ... 5 Q: What does the risk-neutral price represent if the option is not replicable? In an incomplete market, there is no unique martingale measure but instead a set Q of equivalent martingale measures. Consequently, there is an interval of arbitrage-free prices:  \Big( inf_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX], sup_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX] ... 5 The following paper gives you really all of the missing steps in a very detailed form: Black-Scholes Option Pricing Formula by Michael Tomas and Ravi Shukla From the paper: "This presentation is purely for pedagogical purposes. In the course of doing work on option pricing, we found no complete solution for the Black-Scholes model. By complete, we mean ... 4 In binomial tree models, there is no such a thing as a path. The binomial tree represents information about the distribution of the zero-curve at a given time and preserve enough information between different times to let you compute conditional expectations. Generally, you can not price path-dependant instruments in a model based on trees—because there is ... 4 A stochastic volatility model for a single risky asset can't be complete because you have two sources of randomness. But you can easily make it complete by adding a derivative whose value depends on the volatility. For example, if you add a variance swap in the Heston model then it becomes complete. This allows you to calibrate the model. But your ... 4 The dynamics of the underlying stock process are obviously crucial to the derivative's price. Thus if you don't necessarily assume S_t to be log normally distributed (B&S-Model) you won't get the same price even if the market is arbitrage free. Example: Assume S_t=C  \forall t \in \mathbb{R}^+ and r=0. Thus S_t is constant and the interest ... 4 Most of the time, when you have a simple SDE without a drift, it's a martingale because the Wiener process itself is a martingale. In your example, you have a constant with the Wiener process, therefore the whole process must also be a martingale because the expectation is clearly X(t). However, we can't conclude a driftless SDE is always a martingale. ... 4 I would use the following arguments: If the option were on the first throw of the dice, then we would price it using the expectation, which is 3.5 (= (1+2+\cdots+6)/6. Now we have a 2 stage game: First throw : if the player throws more than 3.5 points, i.e. 4,5,6, then there is no sense in throwing again. If he throws 1-3 then it makes sense to ... 3 Two parts Real world vs risk neutral: Can we even estimate risk neutral volatility using historical data? There is a difference in distribution of the underlying stock price under the real world and risk neutral measures. Luckily, changing to the risk neutral measure does not affect volatility, only the drift. Thus, a real world measure of volatility will ... 3 The theoretical idea of Delta hedging is placed in a setting of infinitesimally small time steps. In such small time steps and if no jumps occur (eg. in the diffusion case) the underlying can not move that much and Delta makes sense (at least theoretically). In continuous time things work out. As in practice (or in simulations) the Delta hedge can only be ... 3 Maybe this could also be a comment but I think an it is not possible to answer this question with a 'yes and here is how you do it'. It has been tried, e.g. by me for a university research project. In this research we focused primarily on aggregation of returns and the main problem was the tractability of the resulting distributions and expressions, also ... 3 First find minimum value of j that assures the option is in the money. This would be function of u,d,n,S,K,p,q Any particular value of j has a probability associated with it. You gave formula above. Then you need sum probabilities from j=min j needed to n. See wikipedia binomial distribution to find formula for sum. Sum is based on cdf of binomial ... 3 The following is a standard exercise that will help you answer your own question. Consider a one-period binomial lattice for a stock with a constant risk-free rate. Determine the initial cost of a portfolio that perfectly hedges a contingent claim with payoff xu in the upstate and xd in the downstate (you can do this so long as the up and down price are ... 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 ... 3 Assume the stock pays no dividends before 100 dollars is hit. Interest rates can be arbitrary. Buy 1/100 of a share for 75 cents. Hold until 100 is hit then sell. The payoff of 1 dollar is replicated for an upfront cost of 75 cents. The arbitrage-free value of the option is 75 cents. 3 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 ... 3 Implied volatility cannot be calculated analytically with a closed formula. Instead, you have to approximate it numerically. There are multiple methods to compute IV on an option: Bi-section method Newton-Raphson method Secant method A quick google search came up with the following code for C++ using bi-section and newton methods: Implied ... 3 Mersenne Twister is currently the most used PRNG in the quant world. It was even incorporated in C++11 so it can be considered standard nowadays. Any PRNG with reasonable statistical quality shall perform well (equivalently) for pricing, so that differences relate more to convenience (speed, parallelizability etc..). If the statistical quality is poor then ... 3 Asian options: strike is average of underlying over tenor. Underlying is stochastic. Options with kock-ins/knock-outs: Underlying is stochastic and may cross the kock threshold as it evolves. Option value depends on this cross or lack thereof (boolean). Options on Options, too. Motivations for Asian options you can google. Kock-ins and knock-outs ... 3 Benoit Mandelbrot applied fractals and self-similarity to financial markets and the hurst exponent has its roots in chaos theory. Look at this article from Wilmott magazine. Just a personal note: I have not worked that much with this kind of theory so far but I also have not seen any of my peers being exceptionally sucessfull with these methods. 3 The Black-Scholes price of this option is approximately 14.8. When I run a Monte Carlo simulation with 10000 paths and "exact" time stepping, I get results very close to this value. You are simulating the terminal asset price with the first-order Euler approximation over multiple time steps:$$S(t+\Delta t)= S(t) + rS(t)\Delta t + \sigma ...

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The error is, you are not storing the random numbers for the same path at the end: xbefore = x + c*tau + sigma*sqrt(tau)*randn() A = muA + sigmaA*randn(); xafter = xbefore + A; But then at end you set a different path here by creating a new random number: xT = log(S0)+(c+muA*lambda)*T+sqrt((sigma^2+(muA^2+sigmaA^2)*lambda)*T)*randn(); randn() ...

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These options can be priced by adding an early exercise premium value to the intrinsic value: http://www.statistics.nus.edu.sg/~stalimtw/PDF/lb-float.pdf

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The condition $$ud=1\text{, or equivalently }u=1/d$$ is necessary to ensure convergence of the Binomial tree's mean $\mu$ and standard deviation $\sigma$ to nonfinite values when $n$ (number of steps) goes to infinity. Cox-Rubinstein-Ross showed in their famous paper, that to achieve this, we must have: $$u=e^{\sigma\sqrt{t/n}}\text{, ... 2 You're missing the point of the risk-neutral framework. The idea is as follows: assume the real probability measure called \mathbb{P}. The thing is, because investors are not risk-neutral, you cannot write that v_0 = E_\mathbb{P} [ e^{-rT} V_T]. Using the Fundamental Theorem of Asset Pricing, you know that if the market is arbitrage-free, then there ... 2 Risk neutral probability differs from the actual probability by removing any trend component from the security apart from one given to it by the risk free rate of growth. If you think that the price of the security is to go up, you have a probability different from risk neutral probability. In very layman terms, the expectation is taken with respect to the ... 2 I think the classic explanation (any other measure costs money) may not be the most intuitive explanation but it is also the most clear in some sense and therefore does not really require a intuitive explanation. That is to say: you could use any measure you want, measures that make sense, measures that don't but if the measure you choose is a measure ... 2 When the pdf of a distribution is not known analytically, it's common to compute by taking the inverse Fourier transform of its characteristic function. The same idea applies here. Consider the discounted expectation formula of a European option$$V (S,\tau) = e^{-r\tau} \mathbb{E}_{x_0} [\theta(x_T)] . for log prices $x$ and time to expiry $\tau=T-t$. In ...

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