# Tag Info

29

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.

10

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 ...

8

It is a very interesting question. There is a brief explanation in the book Martingale methods in financial modelling. Basically, it says that, the interest short rate $r_t$ can be modeled in any martingale measure $Q$, however, as long as the zero-coupon bond price $P(t, T)$ is defined by \begin{align*} P(t, T) = E^{Q}\Big(e^{-\int_t^T r_s ds} \mid \...

7

$$dS / S = \mu dt + \sigma dW \\ \\ dS / S -r dt= \mu dt - rdt + \sigma dW \\ \\ dS / S -r dt= [\frac{(\mu - r)}{\sigma}dt + dW]\sigma \\$$ 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 ...

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 ...

7

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 ...

7

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 ...

7

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 ...

7

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 $P(V_T > 0) > ... 6 You can find a simple proof in the discrete time case at http://kalx.net/ftapd.pdf. I'm not sure what you are trying to derive with your Ito calculus, but here is a rigourous derivation of the Black-Sholes/Merton PDE: http://kalx.net/dsS2011/bms.pdf. The Black-Scholes '73 derivation is not mathematically correct. The modern approach does not use so called ... 6 A stochastic volatility model for a single risky asset can't be complete because you have two sources of randomness. But you can easily make it complete by adding a derivative whose value depends on the volatility. For example, if you add a variance swap in the Heston model then it becomes complete. This allows you to calibrate the model. But your ... 6 Q: What does the risk-neutral price represent if the option is not replicable? In an incomplete market, there is no unique martingale measure but instead a set$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] \Big)...

6

Very simply, Ross' framework assumes a great deal to extract the true pricing kernel. Time homogeneity, additively separable state dependent utility, (discrete time Markovian structure - though these have been relaxed.) In particular, there are two schools of criticism, one is that time homogeneity makes little sense in the real market. In fact, the Recovery ...

6

The short answer is: As long as a derivative can be perfectly replicated via hedging in the underlying asset then the price of the derivative should be independent of investors' risk aversion and hence the application of risk-neutral probabilities and discounting of the future expected payoff under risk neutral probability leads to the same price of the ...

6

We first list the assumptions. \begin{align*} g_{t+1} &= \mu_g + \sigma_{g, t} z_{g, t+1}, \tag{1}\\ \sigma_{g, t+1}^2 &= a_{\sigma} + \rho_{\sigma} \sigma_{g, t}^2 + \sqrt{q_t} z_{\sigma, t+1}, \tag{2} \\ q_{t+1} &= a_{q} + \rho_q q_t + \varphi_q \sqrt{q_t} z_{q, t+1}. \tag{3} \end{align*} Moreover, \begin{align*} r_{t+1} &= -\ln \delta +\...

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

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 ...

5

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 ...

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

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 see,...

5

Yes, you may as well take this as the definition of the risk-neutral probability $Q$. I will now try to give you some intuition for that kind of construction. Assume the risk-free interest rate $r$ is constant and that the world ends at time $T$. Suppose you have a security $B=B_t$ which is riskless, i.e. which follows the dynamics $$dB/B = r \, dt$$ so ...

5

The risk-neutral probability density function $q(.)$ is indeed given by $$q(S_T=s) = \frac{1}{P(0,T)} \frac{ \partial^2 C }{\partial K^2} (K=s,T)$$ where $P(0,T)$ figures the relevant discount factor. This is known as the Breeden-Litzenberger identity. Because you do not observe a continuum of call prices in practice, you can use a finite difference ...

5

Sell 1 unit of S1,2,3 respectively, gain 3; buy 2 units of risk-free asset, cost 2. No matter which state appears, the future payoff/loss is 0 for sure, while you will gain 1 at the beginning.

5

I believe that the confusion arises because of the wrong treatment of NIG. The answer to the question you link is misleading, as it simulates under P which is not appropriate for option pricing. None of the NIG parameters under P carries over to Q in general, but especially the drift is the problem here. First use the mom gen function of NIG to find the ...

5

You should see this as a comment to @Kiwiakos answer which already hit the bull's eye. In the SE question you're referring to and to which I have answered, the idea was simply to provide you with a sound way of simulating returns out of a NIG distribution. It so happens that, for whatever your reason was, you decided to calibrate your NIG parameters based ...

4

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 $... 4 If$\mu\$ is large, then it is more likely for the call to finish in the money. Your and my intuitions suggest that this means that the option is more valuable. But this is wrong. A call option is an insurance policy. A call option is useful because it protects you in the case that the value of the stock goes down. That is why call options are valuable for ...

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, ...

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