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2

I would argue that there is some path-dependency involved. The BS model is considered the big breakthrough and it presented the world with some kind of tractable toy model. After that people saw that you had to adjust the model to account for all kinds of stylized facts (e.g. non-constant volatility for different strikes, over time and so on). Yet finite ...


0

On a pure technical aspect, a model does not need to have a finite variance. In the context of option pricing, what you need it a way to replicate the behaviour of the stock price. Once you have it you need to find a corresponding risk-neutral measure. There you will have the first difficulty, with infinite variance, the corresponding hedging strategy is ...


3

Answering my own question as it could be useful for others. Actually package fOptions is vectorized. The only constraint (and that make sense) is that you can't compute at the same time 2 different greeks, or mix up calls and puts. So assuming that you want to compute the delta of a set of puts, the code will be the following: ...


0

if you put all your option objects into a list then you can use lapply. Read the documentation or just thist post for details.


0

Employee stock options (ESOs) have a couple of disadvantages, but the high volatility of small tech company stocks is NOT one of them. It is actually an advantage, as the higher the volatility of the underlying, the higher the option value. Disadvantages for ESO holders: Withholding fees for early exercise. Income tax treatment instead of capital gains. ...


0

The typical approach is: you only use option data from the last day. Furthermore, you only include those points that are liquid enough. One approach to this is to weigh the modelling error of an option by its bid-ask spread and vega. Using data from multiple days is not a good approach, because you might have options with the same strike but different ...


2

One does not estimate the local volatility at a given $T$ and $K$. Instead, Dupire's formula actually gives $\sigma(T,K)$ for all $T$ and $K$. $$ \sigma^2(t_0,S_0;T,K)= \frac{\frac{\partial C}{\partial T} + (r - q)K \frac{\partial C}{\partial K} + qC}{\frac{1}{2} K^2 \frac{\partial^2C}{\partial K^2}} $$ where $C(t_0,S_0;T,K)$ are the call prices for ...


2

you can do the bounds without using a model or martingales. At maturity $$ 0 \leq C \leq S_T $$ with positive probability of strict inequalities. So before maturity, $$ 0 < C < S_t. $$ Since if these are violated, you can make an arbitrage. eg if $C \geq S_t$ hold $S_t - C$ to get a profit with positive probability and no chance of loss. Similarly, ...


3

For a call option, the payoff is given by $(S_T-K)^+$. Note that the function $x^+$ is convex, then, by Jensen's inequality, the price $c$ satisfies \begin{align*} c &= e^{-rT}E\big((S_T-K)^+\big) \\ & \geq e^{-rT}\big(E(S_T-K)\big)^+\\ &=\big(S_0 - K \, e^{-rT}\big)^+. \end{align*} For the upper bound, note that \begin{align*} c &= ...


0

The PDE only holds in t he continuation region, in the excerise region, P is just the pay off of the function. Let $\tau$ be the first time you enter the stopping region, then by the martingale property of the option price up to the first stopping time ...


3

The No-Arbitrage bounds for a European put are: $$ (Ke^{-rT}-S)^+ \leq P \leq K e^{-rT}$$ This is because the maximum payoff at maturity is $K$ (discounted) and the minimum value is the discounted intrinsic value (since $E(e^{-rT}S_T)=S_t$ by the martingale condition and the payoff being always semi-positive).


4

I know two papers explaining how to calibrate this kind of models, and one of them explain the impact of the quality of the fit on a pricing model: Aït-Sahalia, Y. (2002, January). Maximum likelihood estimation of discretely sampled diffusions: A closed-form approximation approach. Econometrica 70 (1), 223-262. Azencott, R., Y. Gadhyan, and R. Glowinski ...


1

Given two representations: $$ C = E_f[\varphi(X)] = \int \varphi(x) f(x)dx = \int \varphi(x) \frac{f(x)}{g(x)}g(x)dx = E_g[\varphi(X)\frac{f(X)}{g(X)}] $$ The difference of the variances of the MC estimators associated with the two expression is $$ Var[\widehat{C}^f_N] - Var[\widehat{C}^g_N] = \frac{1}{N}\int \varphi(x)^2 \left(1 - ...


5

importance sampling is well known to be tricky. See the extensive discussion in Glasserman's book. I presume that you are simply meanshifting and multiply by the ratio of normal densities. For this sort of problem, I'd use a more stratified algorithm instead and force every path to end in the money. To do this I'd compute the uniform that goes to the ...


1

This is the Black Scholes Call Price: \begin{align} C(S, t) &= N(d_1)S - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T - ...


1

A convex function is when the line between two points on the graph always lies above the graph. And this does hold for the put, its also sometimes called a sublinear function. Also see http://en.wikipedia.org/wiki/Convex_function So the author is correct in saying that $(K-s)^+$ is convex.


0

We assume that the short interest rate $r_t$ follows the Hull-White model, that is, the short rate $r$ and the stock price $S$ satisfies a system of SDEs of the form \begin{align*} dr_t &= \lambda(\theta_t -r_t)dt + \sigma_0 dW_t,\\ dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t + \sqrt{1-\rho^2} dB_t\Big)\Big], \end{align*} where $\lambda$, ...


3

This is a bit of an old question, but I thought I'd contribute to add more weight to to what some people have been saying. A CSO (calendar spread option) is NOT a calendar spread of options. If you read it carefully, you can see the Hull quote Max Li posted is talking about a calendar spread, not a CSO. A CSO needs to be priced the same way as a spread ...


3

Fact 1: if you are not good at pricing options, of course you can create a lot of arbitrage opportunities for the rest of the market. It does not matter whether the reason is in dividends or anything else. Fact 2: if you are good in pricing options, you price the dividend effect in advance. Consider the situation of the European calls, and suppose that both ...


2

Generally no, because 'dividends' are already 'priced into' the options. Which means, if an ATM call cost 0.50, and stock price drops by 1.00(amount of dividend), the ATM becomes OTM, but it may still cost 0.50, because the initial price of 0.50 already factored in the dividend.


0

consider adjoint algorithmic differentiation to get an exact derivative here. Works especially well for monte carlo. Here is an example paper: http://luca-capriotti.net/pdfs/Finance/jcf_capriotti_press_web.pdf


6

the problem is that the pay-off has discontinuous first derivative. Try a contract with pay-off that is twice differentiable and it will probably work. The problem is that all the value comes from the tiny number of paths within $\Delta S$ of the strike, and these paths have huge value. This is a well-known problem. As the bump size goes to zero, the ...


0

http://finance.bi.no/~bernt/gcc_prog/recipes/recipes/img169.png let $q t$ be big (t goes to infinity where q is the yield) and you will see why . The first part of the BS formula becomes zero. Also in accordance to put call parity, the call must be worth zero if the entire stock price has been paid out in dividends: ...


1

It's a combination of too few sample paths and/or too small an increment. Your estimation error on the price is magnified by the $dS^2$. Try using a larger sample or a larger increment. Alternatively, you can use a multiplier instead of a fixed increment; in my experience, it usually yields better results.



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