Tag Info

Hot answers tagged

17

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 ...


6

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 ...


5

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 (but still about 170 pages long), yet more current - and free (!) version in different formats (pdf, mobi for the Kindle and ...


5

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 ...


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

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 ...


5

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 ...


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 ...


5

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] ...


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

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

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 ...


4

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, ...


4

In general these are the two basic approaches to QuantFinance: Sell side (market maker, risk neutral): You use risk-neutral probabilities ("$\mathbb{Q}$") e.g. in option pricing (to e.g. calculate your greeks and hedge your portfolio), so that you live on the spread. Buy side (market/risk taker): You use real-world probabilites ("$\mathbb{P}$") for e.g. ...


4

In the derivatives context, "arbitrage free" means almost surely for the probability measure under consideration. This is in opposition with statistical arbitrage used at high frequencies for example. More precisely the assumption is that there is no $T\geq 0$ and self-financed portfolio $V$ such that $V_0 = 0$, $P(V_T < 0) = 0$ and $P(V_T > 0) > ...


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. ...


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

When using Monte Carlo for option pricing you numerically approximate expectation under a risk-neutral probability measure $Q$. Your undiscounted stock price process in GBM framework has as a drift equal to risk free rate under $Q$. So the answer to your question is affirmative.


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, ...


3

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) ...


3

If you are interested in the proof of the Baye's Rule for conditional expectations you can find it here The sake of completeness: The Baye's rule for conditional expectations states $$ E^Q[X|\mathcal{F}]E^P[f|\mathcal{F}]=E^P[Xf|\mathcal{F}] $$ With $f=dQ/dP$ - thus being the Radon-Nikodyn derivative and $X$ being some random variable and ...



Only top voted, non community-wiki answers of a minimum length are eligible