# Tag Info

18

In general there are two basic ways to make money out of your option pricing models: Sell side (market maker, risk neutral): You use these models to calculate your greeks to hedge your portfolio, so that you live on the spread. Buy side (market/risk taker): You use your model to find mispriced options in the market and buy/sell accordingly. (A third ...

13

Here couple pointers that may make it clearer: Drift can be replaced by the risk-free rate through a mathematical construct called risk-neutral probability pricing. Why can we get away with that without introducing errors? The reason lies in the ability to setup a hedge portfolio, thus the market will not compensate us for the drift above and beyond the ...

8

It is not possible for what most people think of as options, but there are classes of options for which an ODE is used. For a nontrivial example, think of perpetual American-exercise options. Because of perpetual exercise, the option value is independent of time. In place of the Black-Scholes PDE $$\frac{\partial f}{\partial t} = \frac12 \sigma^2 x^2 ... 7 This is in fact a tricky matter. As you say one way is to calculate delta by an analytic formula, i.e. calculate the first derivative of the option pricing formula you are using with respect to the underlying's spot price. The second way is to do it numerically, i.e. change the spot price by a small value dS, calculate the value of the option and then ... 6 Agree with all of vonjd's points though I like to add the following: First of all, market practitioners do not read options prices or set options prices in the market, they price the option through models primarily on the basis of implied volatility. Im plied volatility is actually traded, options prices is what comes out on the other side. I know there ... 6 A simple intuitive answer why the OTM Call is more expensive than the OTM Put is because of the skewness of the log-normal distribution. Think about it, what is the probability that the stock price is above 110 at expiration and what is the probability it is below 90? This should answer your question. Written in probability terms: The median of the ... 5 While the translation between implied volatility (iVol) and options prices is of a strictly mathematical nature (when you feed 10 market makers with the same iVol you most likely get 10 identical or close to identical option prices in vanilla structures). What is on the other side more of an art than science is how to assess whether iVols/prices trade ... 5 You may need to differentiate between the use of options and the pricing of options. How options are used has no bearing on the price of such options. Options can be used as leveraged investments or as insurance or as hedges. Any such use does not change the fair value derived for the option. By the way you are in fact compensated the risk premium but it is ... 4 The put call parity is given as follows:$$c_t-p_t = S_t - \frac{X}{e^{r(T-t)}}$$If you assume r=0, you get$$c_t-p_t = S_t - X$$So, c_t \neq p_t. The rationale behind it is much more financial than mathematical. You have to look at the payoff on both side of the equation, and you see that both portfolio will give the same payoff at time T (the ... 4 He is approximating C(S_{t+1}) around t:$$C(S_{t+1})=C(S_{t}) + \frac{\partial C(S_{t})}{\partial S_{t}}(S_{t+1}-S_{t})+\frac{(S_{t+1}-S_t)²}{2}\frac{\partial^{2}C(S_{t})}{\partial S_{t}^{2}} + ...$$In addition, he takes the time value of C(S_t) into account (and I look only at the time contribution here):$$C(S_{t+1})-C(S_t)=\Delta ...

4

Being on the sell side and selling options you can intuitively think of it like this: An option is like any other product that is being produced out of ingredients and because of the competitive situation of the producer this is done by the cheapest possible production process. The ingredients are in a simple (Black Scholes) setting a stock and and a risk ...

4

Recently I came across an interesting intuitive explanation: Suppose driftless market. Market price is 105, strike price is 100. Call option costs 8, put option 3. (intrinsic value of call is 5, time value of both is 3) Now the market starts drifting upwards massively. You say, that you would probably price call higher, e.g. at 10. Would you also price put ...

3

On a single-option basis, there is this paper comparing methods by Mark Joshi. It doesn't specifically examine portfolios, but there's a reason for that. Portfolio and scenario computations are embarrassingly parallel, so once you have achieved your most efficient available option pricer, the rest is simply about wise distribution of your computational ...

3

I don't have the original reference here but what about Merton's Jump-Diffusion model? Stochastic Calculus for Finance II: Continuous-Time Models by Steven Shreve has some chapters about modelling with jump-processes. I think it is a slightly easier introduction to the topic than Cont/Tankov.

3

Yes. The risk neutral and the real path share the same volatility, so the difference is in the drift rate, where the risk-neutral path drifts with the risk-free rate r. You may want to check out Paul Willmots book, esp. ch. 26, for applications.

3

I found that sometimes going back to the source gets me the farthest. Here is what you are probably looking for: http://www.math.ku.dk/kurser/2005-1/finmathtowork/ODD.pdf Discrete dividends that have not been declared yet need to be estimated. Estimating and updating dividend expectations is part of the job of every single stock vol trader. You are asking ...

3

Fourier Transform seems a good method for option pricing by take advantage of Fast Fourier Transform technique, such as the following paper written by Peter Carr and Dilip B. Madan: http://portal.tugraz.at/portal/page/portal/Files/i5060/files/staff/mueller/FinanzSeminar2012/CarrMadan_OptionValuationUsingtheFastFourierTransform_1999.pdf

3

There are two excellent choices for implementing prediction markets: (1) Use book orders that stand until filled, just as intrade.com does. (2) Use an automated market maker (like Robin Hanson's) that stands ready to make trades. The book orders model is very simple to implement, but can suffer from very wide Bid/Ask spreads. And, it can be tough to bet ...

3

I think you should not just ask what the implied vol is of a basket of equity derivatives but you should aim to generate a volatility surface. A spot implied vol gives you nothing to work with. What you need is an implied vol surface in order to understand the smile and skew effects when you quote basket options in the market and/or as price taker. Take a ...

3

a. is correct, but you should derive it using appropriate logic, not just guessing the answer. Ie the drift of discounted stock should be 0. Define a bond dB = rBdt. d(S/B) should have no drift. This can help you find the correct mu. You can find the sde for S/B using two dimensional ito b. don't really know about market price of risk. c. In this case the ...

3

$$dS / S = \mu dt + \sigma dW \\ \\ dS / S -r dt= \mu dt - rdt + \sigma dW \\ \\ dS / S -r dt= [\frac{(\mu - r)}{\sigma}dt + dW]\sigma \\$$ Then, Girsanov tells us that, as long as the risk premium is bounded from below, we can write $[\frac{(\mu - r)}{\sigma}dt + dW]\sigma$ as $\sigma d\tilde{W}$ where $\tilde{W}$ is simply another brownian motion with ...

3

As you correctly pointed out volume has no place in the pricing models of most any option(Unless of course you create an option whose underlying or is volume in some way or if volume is used as some sort of barrier). The reason is simple: The contingent payoff and hence the probability of ending up in the money is not a function of volume. Why the market ...

3

The first Google result seems clear enough: A seagull option is structured through the purchase of a call spread and the sale of a put option (or vice versa)....This structure is appropriate when volatility is high but expected to fall, and the price is expected to trade with a lack of certainty on direction. So, for example, you might buy the 105% ...

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

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

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 ...

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 ...

2

Once you have slogged through all the relatively useless theoretical literature, this paper is a rediscovery (and pretty good write-up) of how basket option pricing is really done in serious quant packages at the big banks.

2

Let $\beta$ denote the relative machine precision, usually $\beta = 1E-16$. Assume the you can evaluate the value V up to precision $\alpha$. The best you can get is $\alpha = \beta \cdot V$ if $V$ is not underflow or overflow. Then you can calculate the finite difference up to precision $4 \alpha / \epsilon$ (the 4 might be a rough estimate, but it comes ...

2

Jim Gatherals Book deals with the models you mention and gives an intuitive understanding about calibration and issues that arise. Mostly basic stuff, but very useful if you're just starting out. Also very understandable without an extensive math background.

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