# Tag Info

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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 ...

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 ...

5

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: $S$ ...

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 ...

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

(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$is a ... 2 I do this question to death in Concepts and ... If (discounted price of) everything is a martingale then every trading strategy is a martingale. Therefore any self-financing portfolio of initial value zero and has expectation zero. Therefore there are no arbitrages (since these have positive expectation and initial value zero). So there is no arbitrage in ... 2 Okay, this is a bit of an involved question, but the intuition is as follows: As Ross (1976) truly conceived it, being risk-neutral means being indifferent between any gamble and its mean payoff. This is equivalent to linear Von-Neumann Morgenstern preferences over all wealth levels, not just positive ones. A classic experiment to distinguish between ... 2 Unfortunately, I do not know the model you talk about. However, the law of one price is a direct implication of the no-arbitrage assumption, which is assumed in many models (if not all). I do agree that the law of one price should be stated as a theorem rather than a definition. Anyway. Consider the case in which two portfolios A and B have the same value ... 2 Regarding (1). Assume for some time$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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Show that the discounted expectation price of the new security is the same as the solution of the PDE. Once this is done all three assets have discounted price processes which are martingales so there can be no arbitrage.

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Lets look at generic markets with a single market filtration, then if $\mathbb{E}[S_t]=S_0$ then the market should be arbitrage free (absence of interest rates.) Otherwise there would be a butterfly arbitrage. But for more sophisticated markets, not at all. Consider a market where there are only is only one period and there are two agents, one who knows ...

1

this is just theory, don't take it as serious, theory it's just take on approximation of reality and in this case not good one, people trade to check that strategy is profitable or trade because they think it will profitable, besides that you have many other spaces on what people compete with each other in this game

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In our lecture, we were told to omit the proof because it was too difficult. Maybe it will help you though if you can read it here:

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