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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 ... 5 In three bullet points: Efficiency: the obtained prices maximize assumed utilities of different agents. In their paper "The Valuation of Option Contracts and a Test of Market Efficiency", Cohen, Black and Scholes compare the theoretical value of options to their market price. The efficiency is in this sense: can agents obtain more or less in practice than ... 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 ... 4 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: $$u=e^{\sigma\sqrt{t/n}}\text{, ... 4 Very simply, Ross' framework assumes a great deal to extract the true pricing kernel. Time homogeneity, additively separable state dependent utility, (discrete time Markovian structure - though these have been relaxed.) In particular, there are two schools of criticism, one is that time homogeneity makes little sense in the real market. In fact, the Recovery ... 4 it doesn;t imply \ln S_T=\ln S_0+rT+σW^Q_T it implies \ln S_T=\ln S_0+(r-0.5\sigma^2)T+σW^Q_T look up Ito's lemma. This is covered in just about any book on financial maths including my own Concepts etc 4 Your characterisation is correct but incomplete. 1) The most important part of Black-Scholes is not the model but the more general framework of dynamic hedging: you can replicate your payoff by continuously trading the underlying and the amount (delta) you should hold is the derivative of the current premium with respect to the current spot. This is a much ... 3 The error is, you are not storing the random numbers for the same path at the end: xbefore = x + c*tau + sigma*sqrt(tau)*randn() A = muA + sigmaA*randn(); xafter = xbefore + A; But then at end you set a different path here by creating a new random number: xT = log(S0)+(c+muA*lambda)*T+sqrt((sigma^2+(muA^2+sigmaA^2)*lambda)*T)*randn(); randn() ... 3 The Black-Scholes price of this option is approximately 14.8. When I run a Monte Carlo simulation with 10000 paths and "exact" time stepping, I get results very close to this value. You are simulating the terminal asset price with the first-order Euler approximation over multiple time steps:$$S(t+\Delta t)= S(t) + rS(t)\Delta t + \sigma ... 3 Assume the price follows a lognormal process. We can convert it into a problem of finding the probability of a standard Brownian motion particle starting from$0$and hitting$x$before time$t$, or its first passage time$\tau_x$being less than$t$. This can be derived through the reflection principle. The paths crossing$x$are exactly paired up by the ... 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 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 3 you don't need$ud=1.$In fact, there are now about 30 binomial trees which converge to Black--Scholes in the large step limit. Most of them do not have$ud=1.$All you need is $$d < e^{r \Delta t} < u$$ The tree recombines provided$u$and$d$don't change from step to step. See my book More Mathematical Finance for a comprehensive review and ... 3 There are lots of papers online and here are a few I would suggest math.umn riskworx G. Dimitroff, J. de Kock Nowak, Sibetz I you have matlab there is an step step example to calibrate SABR model. Since it uses the financial toolbox of matlab for a few functions I dont think you can replicate it in any other language. There must be C++ code available ... 3 You can view the price of an option as the cost to dynamically replicate it. The more volatility, the more costs you will have trading the underlying to keep your delta equal to 0 (I'm assuming you sold the option, hence a negative gamma position). So, if at any spot, any date your local vol is above 0.194, rebalancing the portfolio will be constantly more ... 3 Think of moving volatility in the other direction. As volatility approaches zero, any call strike strictly smaller than the ATM strike,$K<K_{ATM}$, will have zero probability of ending in the money, and the corresponding option value will be zero. An infinitesimally small change in stock price will not move$K$past$K_{ATM}$, so the option value ... 3 I found and answer to my own question. So, I post it here for people who maybe have the same problem. The answer, however, is quite intuitive. The last observation used for the estimation of the physical density is also the time point where the investors know the most about the physical density because at this point the most possible historical observations ... 3 Simply put, no. Vega depends on a variety of factors (including the level/price of the underlying asset). However, vomma/volga/vega convexity (whatever you want to call dVega/dIV) is always positive. So as IV increases, the vega of an option increases - I think this might have been what you were getting at. It's important to understand that IV is an input ... 2 LSM is very fiddly. The most important things in my view are 1) don't believe anyone who says that the choice of basis functions doesn't matter. 2) implement an upper bounder, eg Andersen--Broadie (2003) or Joshi-Tang (2014) so you can tell if your prices are good 3) do two passes, one to build the strategy, one to price, if they give very different ... 2 There is a whole family of GARCH option pricing models; ones with complex distributions, leverage effects, skewness parameters etc. For an example see Christoffersen and Jacobs (2004). Some example GARCH models: NGARCH EGARCH TGARCH Some distributions that can be used: Hyperbolic Normal Inverse Gaussian Variance Gamma and the generalized form ... 2 The problem with your formula is the equation sign$=$. The second order finite difference is only an approximation to the true gamma: $$f^{\prime \prime}(x) \approx \frac{f(x+h)-2f(x)+f(x-h)}{h^2}.$$$h$can not be a result. Ideally, it should be small (whatever that means), so your original choice of$1\text{bp}$seems appropriate for this ... 2 Historical returns are not to be used 'untreated' for the calculation of option prices. The expectation that you will be using in Monte Carlo will take the form $$C(K,T) = E^Q\{D(T)\ \max[0, S_T-K, 0]\}$$ where$T$is the maturity,$K$is the strike price,$S$is the stock price and$D$is the discount factor. But the expectation is taken under the 'risk ... 2 I think you are right. Now when I check papers I've used for my thesis I don't see almost any with empirical data section. Maybe this one will be helpful: Roswell E. Mathis, III, Gerald O., Bierwag Pricing Eurodollar Futures Options with Ho and Lee and Black, Derman, and Toy Models: An Empirical Comparison 2 Under the Black-Scholes framework, you can calculate the implied volatility, given the option's price, underlying's price, time to maturity and the risk free rate. To calculate the implied volatility you have to use a root finding method, since there is not a closed form of the inverse of the B-S option pricing equation for volatility. In the real world ... 2 CRR is just a numerical approximation to Black--Scholes. Its main use is in getting American option price. There is no real difference other than slight inaccuracy when using it for Europeans. So no it wouldn't do what you ask. Your questions are philosophical. What is the purpose of the model? if you estimate the volatility from a time series then you can ... 2 Victor123, let's start from$\Delta. This is the expected change in the price of an option if the underlying asset moves by a currency unit, say 1 USD. For the case of a call option, the Delta varies between 0 and 1. Everything else been equal, the Delta of OTM calls will approach to 0 as the price moves out of the target barrier. Conversely for the case ... 2 Here is to continue the above answer of Emcor to make it more explicit. Note that the fact given in the question should instead be \begin{align*} P(\inf \big\{t \in [0, T], B_t +ct = a \big\} \geq T) = 1- \Phi\Big(\frac{a-cT}{\sqrt{T}}\Big) + e^{2ac}\Phi\Big(\frac{-a-cT}{\sqrt{T}}\Big). \end{align*} Then, for0<t_0\leq T\$, \begin{align*} P(\tau \leq ...