# Tag Info

## Hot answers tagged no-arbitrage-theory

8

I teach Derivative Securities in the mathematical finance program at NYU and was rather surprised to learn that there is no proof of the FTAP that is accessible to masters level students. So I wrote this. It is a simple proof for the discrete time case. One bonus of the proof of the one period case is that it tells you how to find the arbitrage if one ...

8

This question requires a comprehensive answer, perhaps beyond the confines of my input box :) Suffices here to state the following: The First Fundamental Theorem of Asset Pricing states that in an arbitrage-free market, there exists a ("net") present value function, that is, a linear valuation rule whose value is zero when evaluated in any traded cash-flow. ...

6

Making money is not the only reasonable objective to trading. Another common reason is to manage/reallocate risk. For example, this is exactly the objective of liability-driven-investors, such as pension funds. They're specifically trying to match durations of their liabilities. It doesn't matter if pension fund managers believe there are no inefficiencies ...

6

This option is a perpetual one touch option. Its price depends on the model used; additional assumptions are required to get a model-independent price. Let us first consider 3 important example models for stock price $S$. Constant: $S(t) \equiv 1.$ There is $0$ probability that the perpetual one touch pays off, so its price is $0.$ Black-Scholes: $S$ ...

5

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 ...

5

This is not an arbitrage because the transaction costs of the basket of goods is too high. Ever try to sell an item on eBay? I doubt you'll get 2-3% more for it next year, even new in box. Some of the items in the basket are current consumption goods. Good luck selling those fresh fruits and vegetables next year for 2-3% more than you paid. Others are ...

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 $P(V_T > 0) > ... 4 The dynamics of the underlying stock process are obviously crucial to the derivative's price. Thus if you don't necessarily assume$S_t$to be log normally distributed (B&S-Model) you won't get the same price even if the market is arbitrage free. Example: Assume$S_t=C \forall t \in \mathbb{R}^+$and$r=0$. Thus$S_t$is constant and the interest ... 3 The formation of asset price bubbles, such as the recent US housing market bubble, is perhaps the clearest indication that markets are not efficient. Hundreds of bubbles have been documented for all kinds of traded assets; see the tulip mania for an extreme case. Many practitioners also routinely use trading strategies such as momentum or reversion to the ... 3 No this is not a risk free arbitrage. What you are talking about is modeling a stock price with GBM and it has nothing to do with Black-Scholes. Black-Scholes is an option pricing formula that assumes that stocks follow GBM (which is a bad assumption to begin with but we won't get into that). What you are talking about doing is taking on leverage.$ ...

3

A very good book covering such fundamentals with no or only a minimal amount of maths — highly recommended! Puzzles of Finance: Six Practical Problems and Their Remarkable Solutions by Mark P. Kritzman The topics that are covered here are: Siegel's Paradox Likelihood of Loss Time Diversification Why the Expected Return Is Not To Be Expected Half Stocks ...

3

I think that you are missing one key condition on the call prices that I would say is standard, namely that the call prices should be bounded below by an "intrinsic" value. Specifically, we would expect $C(K) \ge (S-e^{-rT}K)_+$, and this can easily be seen to yield a static arbitrage if violated. This condition (in a slightly different form) can be found ...

3

Consider a portfolio where I sell $\frac{1}{H}$ in stock and use that to buy an option. This is a 0 cost portfolio. When I hit the barrier the price of this portfolio is also 0. Law of one price would suggest that this portfolio should be zero cost at all times. So the price of the option at any time must be $$C_t = \frac{1}{H}*S_t$$ Also, the option ...

3

Fact 1: if you are not good at pricing options, of course you can create a lot of arbitrage opportunities for the rest of the market. It does not matter whether the reason is in dividends or anything else. Fact 2: if you are good in pricing options, you price the dividend effect in advance. Consider the situation of the European calls, and suppose that both ...

2

Here's an answer from a purely statistical point of view: http://www.duke.edu/~rnau/regnotes.htm#constant And another from Cross Validated: http://stats.stackexchange.com/questions/7948/when-is-it-ok-to-remove-the-intercept-in-lm The lean in both cases is to include the intercept unless there is a strong theoretical reason. A more satisfying answer would ...

2

Time-series regression is not a great method for determining betas on individual securities. Rather, the most common method used by the commercial risk model providers is called "predicted beta" or "fundamental beta." The leader in this area is Barra. The way they define the predicted beta, it appears that they include the constant in the regression.

2

(If I remember well,) the local nature of the equivalent measure in the NFLVR theory comes from the fact that the market $S$ is a locally bounded semi-martingale. If it is bounded, you obtain an equivalent martingale measure. Should be in A general version of the fundamental theorem of asset pricing, by Freddy Delbaen and Walter Schachermayer (thanks to ...

2

For the first one absurd reasoning allows you to construct an arbitrage (as r=0) by investing (or short selling according to the sign of $\mu$) at the time where $\sigma$ is null, or if you prefer as soon as $t$ is in $B$ (which is not a Lebesgue negligible set by hypothesis) which is absurd as no-arbitrage holds. The details that remain to be proved is that ...

2

It does not seem you feel the question is answered so I will try to elaborate over what I think seems to bother you. Let $S_t = e^{(\mu -\sigma^2/2) t + \sigma W_t}$ be the stock price process and $B_t=e^{rt}$ be the risk free. The arbitrage you describe is then choosing a nice $\varepsilon >0$ and setting $\tilde{T}=\inf \{t>0 : (\mu -r ... 2 It's unclear what type of trading you are referring to (day trading sort of?). Also I'm not familiar with the aforementioned paradox. However, I think it's weird to say that you can't make money from trading, the semi-strong (strong) from of the EMH only states that the current share price incorporates all publicly (and non-publicly) available information. ... 2 Let$X$be endowed with the following partial order:$y \geq x $means that$\Bbb P(y\geq x) = 1$. The AOA condition in your case states that the pricing law$p$is strictly inctreasing with respect to$\geq$, whereas LOP says that$p$is a linear function. Neither if the two implies another one in general. For example, if$X = \Bbb R$then$p(x) = x^3$is a ... 2 I do this question to death in Concepts and ... If (discounted price of) everything is a martingale then every trading strategy is a martingale. Therefore any self-financing portfolio of initial value zero and has expectation zero. Therefore there are no arbitrages (since these have positive expectation and initial value zero). So there is no arbitrage in ... 2 Okay, this is a bit of an involved question, but the intuition is as follows: As Ross (1976) truly conceived it, being risk-neutral means being indifferent between any gamble and its mean payoff. This is equivalent to linear Von-Neumann Morgenstern preferences over all wealth levels, not just positive ones. A classic experiment to distinguish between ... 2 Let$T= \inf\{t>0: S_t = H\}$. Then the option payoff is given by$\mathbb{1}_{\{T < \infty\}}$, and the value of the option is given by$\mathbb{P}(T< \infty)$. We assume that the stock price process is a geometric Brownian motion, that is, for$t>0$$$S_t = \exp\big(-\frac{1}{2}\sigma^2 t + \sigma W_t\big),$$ where$\{W_t, t \geq 0\}$is a ... 2 Important assumptions: - we have zero interest rate, - option is perpetual, EDIT: with probability 1, share price will hit the barrier$H$(in fact this is a hidden assumption that price changes continuously or we can at least trade at the very moment when$S_t = H$). No, we can't assume that, because , as @q.t.f noted, it would imply arbitrage. ... 2 Unfortunately, I do not know the model you talk about. However, the law of one price is a direct implication of the no-arbitrage assumption, which is assumed in many models (if not all). I do agree that the law of one price should be stated as a theorem rather than a definition. Anyway. Consider the case in which two portfolios A and B have the same value ... 2 Regarding (1). Assume for some time$k$,$\mathcal{V}_k(\Phi_1) > \mathcal{V}_k(\Phi_2)$(w.l.o.g.) with full knowledge that these strategies have equal value at$T$,($\mathcal{V}_T(\Phi_1)=\mathcal{V}_T(\Phi_2)$). I claim that this situation admits arbitrage. I can sell$\mathcal{V}_k(\Phi_1)$and buy$\mathcal{V}_k(\Phi_2)$and pocket the ... 2 (1) To get an arbitrage, buy low and sell high. Consider the following strategy: at$k$, in the event$V_k(\Phi_1) < V_k(\Phi_2)$, buy$\Phi_1$and sell$\Phi_2$invest the difference at the risk free rate. At maturity, your portfolio is worth what the you put in the bank plus interest. Formally, if$V_t(\Phi_\alpha) = \sum_{i=0}^d \Phi^i_{\alpha,t} ...

2

The claim payoff you describe, $g(M)$, looks to me like a tight butterfly spread that pays off only in one state of the world. Can't you just replicate that by short two calls with strike $K_0$ and long two calls, with strikes one either side at $K_0\pm 1$? Then the price of your option would be $C(K_0+1)+C(K_0-1)-2\cdot C(K_0)$. This is effectively the ...

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