# Tag Info

9

If you think the stock is going to continue going up, just wait. If you think the stock has reached its peak, then short it in the open market. If the shorted stock continues to climb, you can always cover with your call option. If however the stock falls below your strike price, then let the option expire and cover at the market price. It's this very ...

7

I guess if your American-style option is in no-exercise region, you can use exactly the same bisection method as for European option.The implied volatility will be different, but the method is still the same. See for example, here, chapter 9.3.3. The applicability of bisection method for American-style options is discussed in the book "Binomial Models in ...

6

A stochastic volatility model for a single risky asset can't be complete because you have two sources of randomness. But you can easily make it complete by adding a derivative whose value depends on the volatility. For example, if you add a variance swap in the Heston model then it becomes complete. This allows you to calibrate the model. But your ...

6

it's a model-free result. The conditions are $d\leq 0, r\geq 0.$ The proof is that for a european $$C_t > S_t - Ke^{-r(T-t)} \geq S_T - K$$ and the American is worth at least as much so you never early exercise. So it's worth the same as European. To prove the inequality, observe if $B_T =1,$ take $K$ units of $B_t$ and one of $C_t$ to get something ...

6

Dividends do not matter for the determination of the upper bound. Indeed, the maximum profit which the holder of a put option can make (be it through a European or an American exercise feature) is exactly equal to the strike price $X$. This can be seen by simply looking at the payout function: the maximum profit is finite and located on the downside when the ...

5

Standard Options on CBOE expire on the Saturday following the third Friday of a month. Additionally to that there exist weekly options. That's why you see these two series of options.

5

American calls on a non-dividend paying stock are worth the same as European ones so there is no point to using least-squares.

4

The algorithm is the same, you just need to use appropriate (American/Exotic) pricer instead of black-scholes.

4

As OracleOfNJ said, there is never any advantage to early exercise of an American style call option unless the underlying asset offers some advantage, usually dividends, which does not apply to interest rate futures. American put options were among the biggest open problems in finance until people learned how to treat them as free boundary problems. In ...

4

Since the treasury note future does not pay coupons or dividends, and is a future as opposed to a cash instrument that you purchase, it is never optimal to exercise early.

4

For a standard American exercise option expiring at $T>0$, price is still monotically increasing in volatility under the Black-Scholes model (though obviously it is not strictly monotonic, due to early exercise rendering price insensitive to volatility in some regions of parameter space). To see this, you can use one of three techniques: Investigate ...

4

For a vanilla option, this is a very slow way to get the boundary, and it's somewhat unreliable for any option. In either a more standard grid scheme or in a LS solver, you obtain the boundary by finding two nodes such that one of them has option value equal to early exercise value, and its neighbor has option value above early exercise value. This gives ...

4

The model here is the binomial option pricing model, so the second term in the brackets represents the expected future value of the option (under riskneutral probabilities). The aim of the option holder is always to maximize the value of his option. He can at any point sell the option at the fair market price $E(V_{n+1})$ or exercise it to get $G_n$. So if ...

4

It can also be proved by Jenson's inequality. It can only be optimal to exercise the American option if the option is below its intrinsic value; but since the "max" function is convex, the European price satisfies the following inequality: $$c(S_t, t)=e^{-rT}\mathbb{E}[(S_T-K)^+]>=e^{-rT}\left(\mathbb{E}[S_T]-\mathbb{E}[K]\right)^+=S_t-Ke^{-rT}$$ The ...

3

There are several ways to choose a particular EMM. I believe that the most popular approach is to use a "distance" between $\mathbb{P}$ and $\mathbb{Q}$. Most papers use a minimal entropy approach(for example, Fujiwara and Miyahara, Esche and Schweizer, or Hubalek and Sgarra) or a relative q-entropy approach (for example, Jeanblanc, Klöppel, & Miyahara) ...

3

I am not sure (had only a quick look), but isn't it that we have $\hat{X} \leq \bar{X}$ and hence we have the same for the $sup$ and given that $p \geq 1$ we have this for the power-of-$p$.

3

Because you would make a higher profit if you sold the option on the open market at that point in time, rather than exercising it at that point in time due to the time value of money.

3

If you can dynamically hedge then you can monetize the value of your option without prematurely exercising it. Before writing about Randomness and Black Swans, Taleb wrote a book on the topic. The short version of the story is find the DV01 of your position, and take an opposite position with the same DV01. (& if you want, line up the rest of the ...

3

Because if you sell, you will get a higher value than 20USD per share. You can think of the reason behind this added value is that having a deep ITM option is better than having a stock: your downside is limited. Therefore your option is worth more on the market than it's exercise value. This is why you are better off by selling it in your case (if you know ...

3

You compare apples and oranges here. You can't possibly compare the profit generated involving S(t) on one side and S(T) on the other side. at time t you do not know what the stock will be worth at time T. Merton made the statement in the context of deciding whether to exercise the call option at any time before expiration OR to simply sell the call ...

3

Bisection method is rather fast but it has only linear convergence. Newton's method offers quadratic convergence but it requires the knowledge of Vega (which AFAIK is only accessible numerically with binomial model). However, the convergence of Newton's method can suffer from poor initial approximation. In this case Brent's method tends to perform better. ...

3

These options can be priced by adding an early exercise premium value to the intrinsic value: http://www.statistics.nus.edu.sg/~stalimtw/PDF/lb-float.pdf

3

The argument that the American and European call are worth the same is model independent. So it holds for the binomial model. So there is no need to check to see if the early exercise occurs because it won't. Of course, if you have written general purpose code, it is much easier to test for early exercise and always have the test fail than to try and deal ...

3

For American options, the longer the maturity, the more choices for the optimal exercises time, then the option value is bigger. For example, consider maturities $T_1$ and $T_2$, for the same option except for different maturities. Any optimal exercise time within $[0, T_1]$ is a possible exercise time within $[0, T_2]$, with a better time possibly falls in ...

3

Use Dynkin's formula to write the expectation: $\mathbb{E}[e^{-r\tau} \phi(S_\tau)]= g(S_0)+\mathbb{E}[\int_ 0 ^ \tau (A g -rg) dt]$ where $\phi$ is the payoff. Use the infinitismal generator $A$ to derive an ODE which describes the solution Use the fact that American options must be equal to or greater than their intrinsic value to derive boundary ...

3

I would not say there is no link to what you say but here would be my view. Intuitive explanation If you wait for a delay $h$ before exercising, you lose your exercise right between $t$ and $t+h$, this leads to a loss in value. Supermartingale property proof (to apply it in your case : $\phi_t=e^{-rt}(L-S_t)^+$) If we denote $\phi$ the obstacle, and ...

2

They'll be correlated, and generally close to one another, but rarely identical. In fact differences of 2 points in implied vol are common. The reason for the differences comes down to the portfolio construction and tracking error of the SPY ETF. While generally quite low over a long period of time, the tracking error on a 1-day or less basis can be ...

2

at the time of exercise, you don't know what the final expiry stock value is. Consider the portfolio consisting of the option and $E$ zero coupon bonds worth $B_t \leq 1.$ At expiry its value is $$\max(S,E) > S.$$ since can you exercise and get the stock if $S>E$ and have $E$ otherwise. So at all times previously, $$C_t + K B_t > S_t$$ ...

2

Let's talk about your first equation: If you exercised your option early, you got this payoff. But if you are a rational investor you'd realize that this is less than what you would get if you would just sell your option itself. i.e. the payoff at time t will be more than S(t)-K because the option is worth more than that as it also has some time value. so ...

2

Hum, that's one of the most important questions in financial engineering, that why no answer is proposed. If you have available data as option prices, you may calibrate a parametric EMM but nothing can tell that it's the best EMM (cause there is no best EMM). So make a choice and defend your choice by saying 'it's simple and allows beautiful result' like ...

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