25

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*} ...


19

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-...


10

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}{...


7

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

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

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

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 ...


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 ...


5

What you need to note is the following: \begin{align*} S_T - \max(S_T-K, \,0) &= S_T + \min(K-S_T, \,0)\\ &=\min(K, \, S_T) >0. \end{align*}


5

Firstly, there is nothing wrong with the fast Fourier transform approach from Carr and Madan (1999). However, there is a whole range of reasons why there is research about other numerical approaches. You may have a model where you do not know the characteristic function (e.g. local volatility). Then, the Carr Madan method does not apply at all and you got ...


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

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 ...


4

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 ...


4

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 ...


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 $K\gg S_t$,...


4

Your code looks fine and it is encouraging that both MC simulations yield similar results. Please look at this simplified code for the analytical part of the Monte Carlo simulation. As you know, $$S_T=S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)T+\sigma W_T\right).$$ A call is path-independent, so there is no need to simulate the entire path. I guess you ...


4

You can use the standard black-scholes formula to price an european option. The only parameter you do not know to use the formula is the volatility. If you have the price of an american option then you can use the Cox-Ross-Rubinstein (CRR) model to backout the implied volatility. Then just use black scholes. The CRR model: In the framework of the CRR model, ...


4

SPY pays dividends ~1.8%, and the expiry is ~3y (as of date was 2018, 2021 expiry), so the it looks like there is a discount Assuming $0 time value $$OptionValue=Intrinsic Value+Time Value $$ $$OptionValue= (S-K)-Dividend$$ $$OptionValue=267x(1-0.02)-267x1.8\%x3=\\\$247$$


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

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 ...


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