# Tag Info

1

it's the liquidity of the underlying that matters not the market in which the option was bought.

4

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 ...

0

Let me provide an intuitive answer that I just thought of (correct me if I am wrong). So starting with two Stochastic Differential Equations (SDE) $\frac{dS_t}{S_t}=μdt+σdW_t$ $\frac{dD_t}{D_t}=-rdt$ (I am assuming our risk-free rate to be constant as is done in most introductory financial math courses) Notice: $D_t = e^{-rt}$ is the solution to the ODE ...

3

As Degustaf mentioned above, one of the keys to the exponential dynamics is just the fact that you first model (relative/log) returns and then describe the dynamics of the stock price itself. I am not sure whether the arbitrage argument of experquisite is realistic, though. Regarding the estimation of the drift: if you know the drift, you can just trade ...

4

Some option prices can't be converted to volatility. E.g. A bid for an in-the-money call which is below its intrinsic value. So sometimes NaN is a valid answer. Best way to handle it is to do precursory checks before going down to the search loop.

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.NET doubles return double.NaN when you do things like divide zero by zero. With doubles, anything less than double.Epsilon is "zero" for the purpose of this result. I suggest that your vega is less than double.Epsilon What happens if you run the same method using decimal instead?

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I like Financial Calculus: An Introduction to Derivative Pricing by Baxter and Rennie. It's less technical than Shreve or Bjork, whether that's an advantage or disadvantage is up to you.

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Have a look at the following titles: http://www.worldscientific.com/worldscibooks/10.1142/3856 http://books.google.de/books?id=z42E_VIaQbIC&lpg=PR2&pg=PR2#v=onepage&q&f=false http://www.markjoshi.com/concepts/ http://www.worldscientific.com/worldscibooks/10.1142/8495

1

I suggest you Arbitrage Theory in Continuous Time by Tomas Bjork. It is a standard reference introducing Stochastic Calculus, then Black-Scholes both from a hedging portfolio perspective and a martingale point of view. It has also some nice chapters on American Options, exotic options and Fixed income derivatives.

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http://www.iaqf.org/bookstore Here you can find the list of books recommended by the professionals in quantitative finance. You can search the recommended books for each topic.

2

This is based on observations of historical data. If you looked at a histogram of daily changes, you would notice that the distribution is heavily skewed. Whereas if you looked at a histogram of daily returns, you would see that it is much closer to normally distributed. As for how to find $\mu$, you don't. The beauty of the Black-Scholes model is that ...

4

You derivation here is flawed because you are deriving with respect to two processes and you do not take into account that the variable $W_t$ is stochastic and hence $S_t$ is as well. So, to derive $S_t$ from $dS_t$, you have to apply Ito's Lemma, see this question for details. This is the "classic" way you see it. If you want to do it the other way ...

0

A martingale must have constant expectation, such that adding a deterministic finite variation process $(b-r)dt$ would break the martingale property (except for when its a constant, which it is not by multiplication with $dt$). Hence the finite variation process must be eliminated under $Q$ for LRS to be an (equivalent) martingale measure, and as shown the ...

1

I saw a quote from Brigo & Mercurio "IR models" (page 26, 2.1 No-Arbitrage in Continuous Time) . May be it will help you to find answer: Harrison and Pliska (1983) proved the following fundamental result. A financial market is (arbitrage free and) complete if and only if there exists a unique equivalent martingale measure.

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