# Tag Info

21

A similar question for put option has been discussed in this question: Finding Arbitrage in two Puts. Basically, the call option payoff is a convex function of the strike. Then the call option price is also a convex function of the strike. Specifically, let $C(K)$ denote the call option price with strike $K$. Then for $0 < K_1 < K_2$, \begin{align*} ...

16

For a sufficiently smooth function $f$, positive constant $a$, and $x>0$, Note that, \begin{align*} f(x) -f(a) &= \int_a^{x} f'(v) dv \\ &= \int_a^{x} \big[f'(v) -f'(a) + f'(a) \big] dv \\ &= f'(a) (x-a) + \int_a^{x}\!\! \int_a^v f''(u)du dv\\ &= f'(a) (x-a) + \int_a^{x}\!\! \int_u^{x} f''(u)dv du\\ &= f'(a) (x-a) + \int_a^{x}f''(u)(x-...

8

You generally can't simply subtract two inequalities as you did in your attempt. Here are two approaches to solve your problem: No-Arbitrage Argument Assume that the initial value of the Butterfly spread was strictly negative $V_0 < 0$. Buying the butterfly spread would thus yield a strictly positive cash-flow at time $t = 0$. Next note that the ...

7

From the SDE \begin{align*} \frac{dS_t}{S_t}= k(\theta-\ln S_t) dt + \sigma dW_t, \end{align*} where $\{W_t,\, t\ge 0\}$ is a standard Brownian motion, we obtain that \begin{align*} d(e^{kt}\ln S_t) = ke^{kt} \Big(\theta -\frac{1}{2k}\sigma^2\Big) dt + \sigma e^{kt} dW_t. \end{align*} Then, \begin{align*} \ln S_T = e^{-k(T-t)} \ln S_t + \Big(\theta -\frac{1}{...

6

You've tagged this with 'black-scholes' but you don't have to make the assumptions of the Black-Scholes-Merton model to understand why the option price with time to expiry. Consider this example: Consider 2 ATM put options on a stock with a time to expiry of one month and one year with some strike price. The maximum pay-out is achieved when the company goes ...

4

Practitioners tend to wear Black-Scholes glasses when dealing with European options: to them, quoting a certain option price today $V(S_0;T,K)$ is equivalent to quoting the forward price of the underlying $F(0,T)$ along with a relevant Black-Scholes volatility figure $\sigma(T,K)$(*) That being said, when you are asked to price a European option on a stock $... 4 You are wrong. Using the maximum of the prices of the European options is equivalent to choosing (and making that choice final) on$t=0$the date$t_i$on which you will exercise. As such a choice would be sub-optimal, you would be giving up value. Therefore the Bermuda option is worth more than the maximum of the prices of the European options. 4 It seems like he is assuming that the shorter term volatilities change more than the longer term ones and the relatively sensitivity is proportional to$1 / \sqrt{T}$. Thus, this hedge is not against a parallel shift of the surface. This is not an uncommon assumption and the corresponding vegas are often referred to as "time weighted vegas". 4 Gamma and vega have the same general shape , peaking at ATM and tapering to the tails. But gamma concentrate as the option gets closer to expiry (when vega is small). For options a long way from maturity, vega increases and gamma is small. Consequently for short dated options, if the price is close to strike, the option will have to be rehedged often (... 3 Below is an example of how you could plot a "call" option value with RQuantLib: library(RQuantLib) library(ggplot2) call_price <- sapply(seq(365,0,-1), function(x) AmericanOption("call", 100, 100, 0.2, 0.03, x/365, 0.4)$value) qplot(day, call_price, data=data.frame(day=0:365, call_price=call_price), geom="line") The code output: Another useful package ...

3

how to construct the portfolio (St−K)+ or how to make this arbitrage If you have this scenario on your hands then you construct the portfolio by putting as much capital as you can into the trade. It's an all reward and no risk scenario. Max it out! You "make" the arb by buying the call, shorting the equivalent amount the underlying at the current price ...

3

The intrinsic value of a call is the price of the underlying minus the strike (S0-K), so if you find a european call whose value is less that that you would: Sell (or short) the underlying at S0 Use the proceeds to buy the call at C and wait. At maturity, the price of the underlying is Sm, and you will make a profit in either case: If Sm < K, the call ...

3

You should go back to the derivation of the Black-Scholes equation (see this answer for example). The main point is that you can cancel the risk of the derivative over an infinitesimal time period $dt$ by holding a certain amount $\Delta$ of the asset. When applying this hedging strategy, in this continuous limit, the variance of your PnL is zero. So ...

3

For the US market nearly all options on securities are american whereas the options on indexes are european. What you can do is to use a database such as OptionMetrics which adjusts the stock american options to european options by taking into account the early exercise premium.

3

First let's note that in practice exercise notice (of US equity options) is given after the end of the trading day, when we may have a bid and offer coming in for after-hours trading with very wide spread. That makes your example fairly important. In the situation you cite, where the bid and ask are $S^B=90\$$and S^A=110\$$, the true "fair" mid-market ... 3 The main interest of the formula is that it allows you, at least theoretically, to replicate any European option with payoff$f(\cdot)$using only Call and Put options. As simple examples, consider$f(S)=S$and$f(S)=(S-K)^+$. The formula also implies that knowing all Puts and Calls for all strikes for a given maturity gives you the price of any European ... 3 There is a logical fallacy in your argument. The price of a European call expiring 1 day before a dividend payment may well be greater than that of a call expiring after it. In other words, claiming that $$C_E (S_0,K,t_D-1\text {day}; D, t_D) < C_E (S_0,K,T; D, t_D)$$ is not necessarily true. Try the above inequality with a huge dividend (e.g.$D =...

3

To keep notations uncluttered, consider that $r=q=0$ in what follows, while focusing on the particular case of an ATM option i.e. $K=S$ (otherwise use the same reasoning with $K=F(0,T)=Se^{(r-q)T}$ i.e. an ATMF option, the conclusion won't change that much). In your first question, you're looking for the sign of the derivative of Vega with respect to the ...

3

We know that $-1\le\rho_{imp}\le 1$ so perhaps the simplest approach is to try the possible values $\rho_{imp}=\{-1,-0.9,-0.8,\cdots,0.8,0.9,+1\}$, to calculate resulting $\sigma$ values, d± values, and $M_{quote}$ values, then see which of these is closest to the observed market price. If desired you can then search a finer grid between two adjacent assumed ...

2

For homogenous diffusion models (i.e. models such that the distribution of $\ln(S_t)-\ln(S_0)$ is level-independent, e.g. Black-Scholes, Heston, Bates etc.), this would indeed hold. To illustrate this, consider an exponential Lévy model for the spot price under the risk-neutral measure $\mathbb{Q}$ $$S_t = S_0 e^{X_t},\ \forall t \in [0,T]$$ The price ...

2

Besides the code's problem, I highly recommend the Brownian Bridge correction method which can compensate the pricing error resulting from discretization of the continuous path.

2

Basically it boils down to this: You either use a descriptive or a prescriptive (normative) model, i.e. you either think that the market is always right or you think that you alone know how to determine the "true" price of an option. The original idea of BS was to build a prescriptive model but most modern models try to take the market prices as given and ...

2

First, you need to specify your working modelling assumptions by selecting a (jump)-diffusion framework, or more exactly, by postulating the risk-neutral dynamics of the underlying (e.g. Black-Scholes). Then, you choose a pricing method, which should allow you to price both American-style and European-style vanillas (e.g. a binomial lattice) At this point,...

2

Carr-Madan formula tells you that the European-style payoff $f(F_T)$ can be decomposed as: $$f(F_T)=f(\kappa) + f'(\kappa) [(F_T - \kappa)^+ - (\kappa - F_T)^+] + \int_0^{\kappa} f''(K) (K-F_T)^+ \ d K + \int_{\kappa}^{\infty} f''(K) (F_T-K)^+ \ d K$$ for any positive $\kappa$ of your choice. Assuming deterministic rates w.l.o.g. as well as a complete ...

2

From the book mentioned in the question, the short hedging option portfolio is used to replicate the option value with investment in the stock and money market account so that the portfolio value $X(t)$ at each time $t \in [0, T]$ agrees with the option value $c(t, S(t))$. That is, \begin{align*} c(t, S(t)) = \Delta(t) S_t + \Delta_t^1 B_t, \end{align*} ...

2

It's really simpler then it might sound at first. The concept at the root of all of this is "arbitrage free pricing". That means that there is a price at which the buyer and the seller of the future each are at break even and no one is making an arbitrage. For futures it's pretty simple. If I can buy an ounce of gold for \$1000 now and interest rates are ... 2 Just to be clear: please provide the 2 dates you are pricing your package at and the values of the spot and IV on each of those pricings. Also you are pricing very short dated options immediately before earnings and there was very sharpe move (unsurprisingly). So it seems pretty clear that your greeks will move! It is obvious you need to perform some ... 2 Let's consider the following 2 portfolios: Portfolio A: one European put option plus one share, Portfolio B: a zero coupon bond paying off$K$at time$T$. If$S_T<K$then the option in portfolio A is exercised at$T$and the portfolio is worth$K$. If$S_T>K$, then the put option expires worthless and the portfolio is worth$S_T\$ at this time. ...

2

You compare the result of an analytical solution (european call) with the numerical solution for the american option. It seems as if you use to few steps to calculate your American option price. Just try to increase the number of steps and see what happens. Or just compare the european price based on the same binomial tree with the american one and you ...

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