15

I generally agree with @dm63's answer: A convex (concave) smile around the forward usually indicates and leptokurtic (platykurtic) implied risk-neutral probability density. Both situations can or cannot admit arbitrage. I provide you with two counterexamples to your statements. A volatility smile that is concave around the forward does not necessarily ...


10

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) > ...


9

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: $S$ ...


9

This is an interesting question that I have asked myself. Below is my take. Let us consider an economy $(\Omega,\mathcal{F},P)$ equipped with a filtration $(\mathcal{F})_{t \geq 0}$ consisting on a traded asset $S_t$ and a numéraire $N_t$ specified by the following stochastic differential equations: $$\begin{align} \text{d}S_t&=\alpha(t,S_t)\text{d}t+\...


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 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 ...


6

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

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 ...


4

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 ...


4

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 ...


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 ...


4

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 ...


4

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 ...


4

I believe there is not a unique price if you can't short. Say, instead of buying the option you spent 0.5 on a half a unit of the asset $S^2_1$ This asset pays out $[0.4, 0.6, 0.8]$ which first order stochastically dominates the option. So, no matter your probability beliefs about the states, in that setting you'd never pay $0.5$ for the option which pays ...


4

Suppose that the given condition is true. You want to construct an arbitrage portfolio to take advantage of this. Now, $d$ is an interest rate, and the condition suggests that $d$ is too high. So you will want to receive $d$ in order to profit. If you could, you would borrow money at $r$ and lend it to the stock broker or exchange to collect the interest ...


4

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.


4

a) From the no arbitrage condition, and without ressorting to a specific model $$ PV[S(T)|S(T_0)] = S(T_0) $$ $$ S(T_0) = (1-\delta) S(T_0^-) $$ $$ PV[S(T_0^-)|S(0)] = S(0) $$ Therefore the PV of $X$ at time $0$ is $$ PV[S(T)|S(0)] = PV[S(T_0)|S(0)] = PV[(1-\delta) S(T_0^-)|S(0)] = (1-\delta) S(0) $$ b) on $t=0$ you buy $1-\delta$ units of the stock ...


4

You'll find here that in terms of European option prices, the absence of calendar arbitrage writes $$ \frac{\tilde{C}(k\, F(0,t_2),t_2)}{F(0,t_2)} \geq \frac{\tilde{C}(k \, F(0,t_1),t_1)}{F(0,t_1)}, \forall k \in \Bbb{R}, \forall \, 0 < t_1 < t_2 \tag{1} $$ where $\tilde{C}(K,t)$ denotes the undiscounted European call price for strike $K$ and time to ...


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. $ E[S_T]=...


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

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 ...


3

(This answer is broadly in line with the comment of Amsh. I added it because Amsh his 1 line solution says substract the PV (present value) of the div; However, the example below shows that one should substract the FV (future value) of the div). edit: to be more precise future value from moment you receive div, until you deliver the stock. Assume $S(0) = ...


3

You cannot use negative probabilities in this context. When there is no unique probability measure, there can be no unique price. You only know that it is in [0, 0.6] range, if you want to tighten this interval you need to make further assumptions/tweak inputs I agree with your conclusion that there no suitable probability measure. But I am not sure about ...


3

Neither situation is necessarily an arbitrage. Negative smile is consistent with a 'thin-tailed' density function , just as positive smile is consistent with a fat tailed density function . It's true that an extreme amount of negative smile could cause the implied density to be negative in places I.e an arbitrage.


3

Consider a random variable $X$ that has a probability density function (PDF) $f(x)$. $X$ being non-negative means that $f(x) = 0$ for $x < 0$. The expectation of $X$ is thus \begin{equation} \int_{-\infty}^\infty x f(x) \mathrm{d}x = \int_0^\infty x f(x) \mathrm{d}x. \end{equation} Since $f(x) \geq 0$ for it to be a valid PDF, it follows that \begin{...


3

Let's focus on a European call option for the sake of the argument. Assume deterministic rates to keep notations uncluttered. Define $\Bbb{Q}$ as the probability measure associated to the money market numéraire $B_t$. $$ C(K,T) = \frac{1}{B_T} \Bbb{E}^\Bbb{Q} \left[ (S_T-K)^+ \right] = \frac{1}{B_T} \int_K^\infty (S - K) q(S) dS $$ Whence (Leibniz rule) $$ \...


3

Forward contract: exchange is done at maturity. Future contract: margin call is paid/received every day throughout the life of the contract, thus resetting the NPV of the position to zero every day. This explains why $F^{\text{forward}}_t=E^{Q^T}_t[S_T]$ and $F^{\text{future}}_t=E^{P}_t[S_T]$ where $Q^T$ is the $T$-forward measure and $P$ is the savings ...


3

An obvious example is using the maturity $T$ zero coupon as numeraire, and a European option with premium paid at time $T$ hedged with maturity $T$ forward contracts. You do not need to trade the zero coupon, in fact you don't even really need to know its value in terms of \$ prior to $T$, because all settlements will occur on $T$. As stated by @Matthew ...


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