28

Cases where exercising a call early makes sense: There is a dividend payment the next day that is >= the interest on the strike price + a put with the same strike and expiration. Exercise the call, buy the put, and sell a zero-coupon bond for (strike price + put price - dividend). The transaction has a net 0 cashflow at time 0 and at expiry you will be ...


15

No, you should not expect such a relationship to hold in general. The reason is that American options have an "exercise barrier" which European options don't, and this results in different prices and greeks. In the case of put options (with interest rate $r>0$) as the spot price falls, at some point it becomes optimal to exercise early and take the cash. ...


13

It is indeed. The price of an American option is the Bermuda option in the limit that the exercising interval approaches zero. The Bermuda option at any exercising time can be evaluated inductively via the dynamic programming principle as the maximum of the payoff and the risk-neutral expected value of the Bermuda option price at the next exercise time. The ...


11

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 ...


11

To compute the price of an American option or a callable instrument in general, at each potential exercise date, one is required to compare its continuation value (discounted risk-neutral expectation of what the option would pay off if it was not exercised) to the relevant exercise value/early redemption price. By construction, lattice and finite difference ...


9

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


9

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 ...


9

Having traded these options for a number of years I have some insight. It’s my belief that those that make a living specifically out of these options do have tree-style models that take into account early exercise. On the other hand , those that have occasional use of these options (such as interest rate derivatives dealers who might use them to hedge otc ...


9

Here is a much more straightforward proof of the convexity of the American option with respect to a parameter, if it is independent of time and sample, than my previous one, though I am happy to have made the connection amongst the dynamics programming principle, the discrete time process and the continuous time process there. Let $g(t,\omega,x)$ be the ...


8

If implemented properly, least-squares Monte Carlo as originally suggested by Longstaff-Schwartz should allow you to identify sub-optimal exercise dates and a lower bound of the true option price. There are many articles out there discussing this non trivial topic. @MarkJoshi can probably shed some more light, see this nice paper. You claim that your LSM ...


8

There are 2 ways to do it. The good-enough way, and the complete and complex way. The Good-Enough Way Here you will convert to a situation where you can apply put-call parity. Begin by finding the strike $K$ where put and call prices are closest to each other. This might not end up being the closest-to-the-money strike, but it will do. Now run the ...


8

European Contracts It's a really important question and as @noob2 commented, the FTAP is normally applied to European-style derivatives, even if they are (strongly) path-dependent, including barrier options and Asian options. The idea is always the same, $V_t=B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi_T}{B_T}\Big|\mathcal{F}_t\right]$, that is the derivative's ...


7

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 ...


7

Probably because your risk-free rate is 0.3070664 (30%) Try 0.3%


7

So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value. Yes indeed However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to indicate that it would be an unwise decision to ever exercise ...


7

Let’s forget about dividends (actually assume there are no dividends). By Put Call parity $C^E(K)= P^E(K) + S - Ke^{-rt}$. Suppose that $S>K$ [otherwise you don’t even think about exercising!], if you exercise the American Call now you get $S - K$ that for sure is less than the intrinsic value of the European call, i.e. when the American Call is still ...


7

The reason is that, as shown in Proposition 2.1 of that paper, in order to exclude static calendar arbitrage, the total variance has to be strictly increasing in forward moneyness. See also the below to links for details on this result. The intuition is that for European options, only the distribution of the terminal spot price is relevant. Furthermore, $F_t^...


7

You're not setting the global evaluation date. If you don't, you're in December 2017 and your option has expired a good while ago. Adding ql.Settings.instance().evaluationDate = valuation_date before the calculations will give you the expected results.


7

The Black-Scholes differential equation is a second-order PDE in two dimensions and reads as \begin{align*} \frac{\partial f}{\partial t} + rx\frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2 x^2 \frac{\partial^2 f}{\partial x^2}-rf&=0, \\ \Theta+rx\Delta+ \frac{1}{2}\sigma^2 x^2 \Gamma-rf&= 0, \end{align*} assuming that $f\in \mathcal{C}^{1,2}([0,...


7

Theoretically, this is a more difficult problem than it looks like at first glance. Unfortunately, existing literature taking into account a proper dividend consideration is rare (at least from a practical viewpoint). There are several options: 1) Use what is called "De-Americanization": In this case, based on your input dividends (maybe based on other ...


6

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) ...


6

The classic result is never early exercise an American call if $r \geq 0, d \leq 0.$ If we think in terms of FX, calls and puts are really the same thing and by switching currency, we get never early exercise an American put if $ r \leq 0, d \geq 0.$ If one of these is violated it may be worth early exercising.


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 ...


6

The exercise boundary $B_t$ for a finite maturity American put option is not a constant function of time as in your plot. As mentioned in the excerpt, $B_T = K$ at maturity. But for $t < T$, we have $B_t < K$ as you would never pre-maturely exercise to receive a zero payoff. Below is a plot of the early exercise boundary that I once produced for a ...


6

Ikonen and Toivanen don't say that the LCP is solved exactly, they simply say that the modified back-substitution is a valid algorithm to solve the LCP. A numerical error may arise around the location of optimal exercise, since it does not fall directly on the finite difference grid. I think that however, the error is of the same order as the discretization ...


6

Let $\mathscr{T}$ be the set of stopping times with values in $[0, T]$. Note that, for any $\tau \in \mathscr{T}$, $\lambda_1\ge 0$, $\lambda_2 \ge 0$, and $\lambda_1+\lambda_2 =1$, \begin{align*} &\ \max(\lambda_1 K_1+\lambda_2 K_2 -S_{\tau}, 0) \\ =&\ \max\big(\lambda_1 (K_1-S_{\tau})+\lambda_2 (K_2 -S_{\tau}), 0\big)\\ =&\ \lambda_1\max(K_1-...


5

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 ...


5

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


5

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 ...


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.


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