# Tag Info

23

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*} ...

18

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 ...

6

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 (...

6

If you plot the function $f$, you see that you have a bear spread. You can build such vertical spreads either with call or put options. For example consider a portfolio selling one put option with strike price $K_1=30$ and purchasing one European-style put option with strike price $K_2=35$. Then, you obtain the payoff \begin{align*} \max\{35-S_T,0\}-\max\{30-...

5

It would be much easier to start by writing the payoff using indicator functions. For example, \begin{align*} f(S_T) &= 3 \mathbb{I}_{S_T \le 30} + (33-S_T) \mathbb{I}_{30<S_T < 35} -2 \mathbb{I}_{S_T \ge 35}\\ &=3\big(1-\mathbb{I}_{S_T > 30}\big) + (33-S_T) \big(\mathbb{I}_{S_T > 30} - \mathbb{I}_{S_T \ge 35}\big) -2 \mathbb{I}_{S_T \ge ...

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 $... 4 The main interest of the formula is that it allows you, at least theoretically, to replicate any European option with payoff$f(\cdot)$using only Call and Put options. As simple examples, consider$f(S)=S$and$f(S)=(S-K)^+$. The formula also implies that knowing all Puts and Calls for all strikes for a given maturity gives you the price of any European ... 4 You are wrong. Using the maximum of the prices of the European options is equivalent to choosing (and making that choice final) on$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 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 ... 4 If you hold an option, you're always vega long, i.e. if volatility increases, your position increases as well - regardless of moneyness and the option type (put or call). Note firstly that by the model-free put-call parity, put and call options have the same vega (i.e. changes in volatility affect put and call prices in an identical way). Let now$K\gg S_t$,... 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 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

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

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 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 ... 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

Let $\rho\triangleq\rho_{imp}$. Note that: $$\frac{\partial \sigma}{\partial \rho}(\rho)=-\frac{\sigma_0\sigma_1}{\sigma(\rho)}<0$$ Therefore $\sigma$ is monotonic in implied correlation. In addition, the Margrabe pricing function $M(\cdot)$ is also monotonic in volatility $\sigma$ thus you can find an unique solution to the equation: $$\tag{1}M_{\text{... 3 Maybe it will help your intuition if you think in terms of log-moneyness \ln S/K instead of S/K. Let's look at a `deep' in the money put K=100, S=10. That sounds really deep in the money, but the value of log-moneyness for this situation is only -2.3, which is not that much if you consider the possible range of \ln S/K is (-\infty,\infty). So ... 3 Here's another way to do it, that I think is useful if you don't recognize/have knowledge of specific option spreads/techniques. This might help you on exams or other problems, although recognizing the different option plays is probably easier. First you start from the left of the payoff graph, and split the graph into segments, just like how the payoff ... 3 First, you'd rather simulate \log(X) rather than X; thus, there is no level dependency in your discretisation scheme, making it more accurate.$$Z_t = \log(S_t)dZ_t = \left(r - \frac{\sigma^2}{2}\right)dt + \sigma dW_t$$You can even run one single time step, and the distribution of your final price will still be as accurate! Second, the different ... 2 For homogenous diffusion models (i.e. models such that the distribution of \ln(S_t)-\ln(S_0) is level-independent, e.g. Black-Scholes, Heston, Bates etc.), this would indeed hold. To illustrate this, consider an exponential Lévy model for the spot price under the risk-neutral measure \mathbb{Q}$$ 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

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,...

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