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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 ...


10

I think you are interpreting too much into the matter. The $-\frac12\sigma^2$ is just a correction term that comes from Jensen's inequality. You need this when switching from supposedly symmetric returns (normal distribution) to the skewed price process (log-normal distribution). I think there are no deeper truths to be found here.


7

Note first that this key equation is only assumed to hold true under some extra assumptions. Typically those assumptions are taken to be about absence of arbitrage, though it is possible to weaken them somewhat if you are willing to consider portfolio arguments or collectively agreeable objective function. Anyway, the argument is this: if all the risk can ...


6

The classic argument using risk-neutral pricing is to assume that discounted stock prices are $\tilde{P}$-martingales where $\tilde{P}$ is the risk-neutral probability measure. Then, you know that $$\frac{S_t}{(1+r)^t}=\tilde{E}[\frac{S_T}{(1+r)^T} | \mathcal{F}_t]$$ by definition of a martingale process. As the discounts are non-stochastic, you can ...


6

You can find a simple proof in the discrete time case at http://kalx.net/ftapd.pdf. I'm not sure what you are trying to derive with your Ito calculus, but here is a rigourous derivation of the Black-Sholes/Merton PDE: http://kalx.net/dsS2011/bms.pdf. The Black-Scholes '73 derivation is not mathematically correct. The modern approach does not use so called ...


5

It depends on the purpose of your simulation. If you want to model the asset price path for pricing some derivative then you need the risk-neutral measure (thus you take the risk-less rate as drift). Why? Because the risk-neutral measure makes your pricing compatible with the pricing of other contracts in the market. It makes the prices consistent. If ...


5

Not all binomial trees take $u=e^{\sigma\sqrt{\Delta t}}$. Thinking of the binomial tree as a discrete approximation (on a grid) to a continuous process, it makes sense that a variety of choices for where to place grid points will work. For a listing of a few different choices of $u$, see the Tian Tree settings and others. From this Sitmo page you can ...


4

No, you obtain a risk-neutral measure by any change of measure; invariance is far more restrictive. Because in your formula $\mu\circ f^{-1} (A)=\mu(A)$, it has to be for any $A$. Risk-neutrality can be seen as a way to inject into your model a list of market prices you really want to not be exposed to: once they are taken into account (i.e. once you made ...


4

The use of risk-neutral measure is based on the ability to arbitrage away the instantaneous risk of contingent claims. Although for forward contracts the hedge quantity is 1.0, in the general contingent claims case we must assume it varies instantaneously with the market state. The Girsanov Theorem tells us what the difference is, instantaneously, between ...


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 ...


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

Risk-neutrality isn't really a property of a model. Instead, it describes a certain calibration of a model (almost always represented by an SDE). We say a model has been calibrated to risk-neutral probabilities if model parameters can be inferred from traded security prices, and there's some anti-arbitrage assumption and hedging scheme available for ...


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 ...


3

What a great question -- it touches on many issues at the core of quantitative finance. This answer might be a lot more than you bargained for, but it's too interesting to pass up. References Mostly, this subject falls somewhere at the intersection of these three highly-interrelated topics: risk-neutral valuation, rational pricing and the fundamental ...


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

One thing to keep in mind here is that the world of risk-free/arbitrage-free models is not necessarily the real world. Specifically, this equation $$ \mu = r - \frac{1}{2}\sigma^2 $$ occurs not because this is the way stocks behave in reality (they don't! For S&P 500, long-run $\mu$ is closer to 6-9%, if I recall correctly), but because using any ...


3

You may want to consider splitting two important, yet very different concepts: Pricing a derivative security with contingent payoff and forecasting an asset. Pricing a derivative can be achieved through setting up a hedge portfolio and track its evolution and "value" at any point in time before the derivative security pays off. Risk-neutral pricing is a ...


3

First of all, I must say that it's a very general question, and the answer can vary depending on type of assets you model. In quant finance real world probabilities are generally used for risk management. It can be said, that in order to use real-world probabilities you have to calibrate your models to history. In order to obtain risk-neutral probabilities, ...


3

This is indeed one of the most difficult tasks to do (if not next to impossible). I would say the standard reference is the following: Expected Returns: An Investor's Guide to Harvesting Market Rewards by Antti Ilmanen An abridged, yet more current - and free (!) version in different formats can be found here:Expected Returns on Major Asset Classes by ...


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

Just following Musiela Rutkowski (the link redirects to Amazon). The risk neutral measure is derived form imposing that the present value of a self financed portfolio (i.e.; no infusion or withdraw of money) is a martingale. A portfolio can be seen as a stochastic process where its value at time $t$ is given by $$ V_t = \phi^0_tP_t + \phi^1_tS_t\ , $$ ...


3

Since I did not get any comments to my latest update, and since I find it quite convincing, I hereby post my solution as an answer. maybe I can prove that Q exists assuming a lognormal distribution of $S_t$. Assuming $dS_t = \mu S_t dt + \sigma S_t dW_t$ By Itô, $d(e^{-rt} S_t) = -r e^{-rt} S_t dt + e^{-rt} S_t dS_t$. Replacing with the definition of ...


3

The only requirement if you are risk neutral is the property of martingale on your discounted stock price $M_t=e^{-rt} S_t$. But if you apply Itô $d( S_t\cdot e^{-rt} e^{rt})=d(M_t\cdot e^{rt})=r_tM_te^{rt}dt + ..dW_t=r_tS_tdt+..dW_t$ you see see that under the risk free probability, the asset price must have $r_t$ as yield and to answer to your question, ...


2

The first think you have to ask is ¿¿What price??? Monetary price or equity price?? All answers,the ones I read, related to monetary price, but are equity price really risk free???? One of the biggest problem with Black Scholes (personal opinion) is that they consider the behave of equity price as monetary price: Solve this ODE: S(t)'/dt= r*S(0), this tell ...


2

ad) "Is it normal to assume no other drift?" Under measure P you might have drift. You could use it as a working assumption, but in general indices drift every now and then. So, no, usually you do not assume away the drift. "The index is described as "following a geometric Brownian motion", which to me says that the there is no other drift going on" ...


2

This depends. I am not aware of a general risk neutral pricing framework applying to all asset classes and/or stochastic processes. In order to reach more general statements about risk neutral pricing you need to consider jumps and autocorrelation depending on the asset regarded, maybe stochastic interest rates and/or volatility. If I remember correctly, in ...


2

Have you looked at using Laplace in a Monte Carlo simulation? Here is how you price American style options within a MC framework: http://www2.math.uu.se/research/pub/Jia1.pdf and the Longstaff, Schwartz paper: http://escholarship.org/uc/item/43n1k4jb#page-1 Regarding the discretization of a process that draws its random variables from a Laplace ...


2

In my mind you are simply right: you arrive at $$ f(t,S) = S(t) - K e^{-r(T-t)}. $$ Assume that $t=0$, so we are at the inception of the contract, then $$ f(0,S) = S(0) - Ke^{-r T}. $$ If you choose $K = S(0) e^{r T}$ then the contract value at inception is zero. This simply means that the fair price for the forward is given by $K= S(0) e^{r T}$ which is ...


2

Hum, that's one of the most important questions in financial engineering, that why no answer is proposed. If you have available data as option prices, you may calibrate a parametric EMM but nothing can tell that it's the best EMM (cause there is no best EMM). So make a choice and defend your choice by saying 'it's simple and allows beautiful result' like ...


2

There are several ways to choose a particular EMM. I believe that the most popular approach is to use a "distance" between $\mathbb{P}$ and $\mathbb{Q}$. Most papers use a minimal entropy approach(for example, Fujiwara and Miyahara, Esche and Schweizer, or Hubalek and Sgarra) or a relative q-entropy approach (for example, Jeanblanc, Klöppel, & Miyahara) ...



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