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0

This could contain the relevant bibliography. Isn't it? http://www.yats.com/doc/stochastic-processes-en.pdf


3

Answering my own question as it could be useful for others. Actually package fOptions is vectorized. The only constraint (and that make sense) is that you can't compute at the same time 2 different greeks, or mix up calls and puts. So assuming that you want to compute the delta of a set of puts, the code will be the following: ...


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if you put all your option objects into a list then you can use lapply. Read the documentation or just thist post for details.


3

First, my notation. $K$ is the strike price, $S$ is the stock price, $r$ is the continuously compounded risk-free rate, $T$ is time at expiration, $t$ is time at issue, $\sigma$ is volatility, $\delta$ is continuously compounded dividend rate. The Black-Scholes formula for a European call is $C = Se^{-\delta (T-t)} N(d_1) - Ke^{-r(T-t)} N(d_2)$ $d_1 = ...


2

you can do the bounds without using a model or martingales. At maturity $$ 0 \leq C \leq S_T $$ with positive probability of strict inequalities. So before maturity, $$ 0 < C < S_t. $$ Since if these are violated, you can make an arbitrage. eg if $C \geq S_t$ hold $S_t - C$ to get a profit with positive probability and no chance of loss. Similarly, ...


3

For a call option, the payoff is given by $(S_T-K)^+$. Note that the function $x^+$ is convex, then, by Jensen's inequality, the price $c$ satisfies \begin{align*} c &= e^{-rT}E\big((S_T-K)^+\big) \\ & \geq e^{-rT}\big(E(S_T-K)\big)^+\\ &=\big(S_0 - K \, e^{-rT}\big)^+. \end{align*} For the upper bound, note that \begin{align*} c &= ...


3

The No-Arbitrage bounds for a European put are: $$ (Ke^{-rT}-S)^+ \leq P \leq K e^{-rT}$$ This is because the maximum payoff at maturity is $K$ (discounted) and the minimum value is the discounted intrinsic value (since $E(e^{-rT}S_T)=S_t$ by the martingale condition and the payoff being always semi-positive).


1

This is the Black Scholes Call Price: \begin{align} C(S, t) &= N(d_1)S - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T - ...


0

Long options have a positive gamma because as price increases, call Delta approaches 1 from 0 put Delta approaches 0 from -1 (think of $S=0\to+\infty$). Based on below numerical example, theta and gamma can have equal signs for both put and call. If we set $r=0$, they have different signs.


0

We assume that the short interest rate $r_t$ follows the Hull-White model, that is, the short rate $r$ and the stock price $S$ satisfies a system of SDEs of the form \begin{align*} dr_t &= \lambda(\theta_t -r_t)dt + \sigma_0 dW_t,\\ dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t + \sqrt{1-\rho^2} dB_t\Big)\Big], \end{align*} where $\lambda$, ...


0

The derivation in the book appears wrong. However, the results make sense as the option price at time $t$ should not be impacted by prior dividend payments. It may be out-of topic, I would like to provide some justification of the Musiela-Rutkowski formula. Let $\{H_t \mid t >0\}$, where \begin{align*} H_t = \sum_{0 < T_i \leq t} q_i, \end{align*} ...


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http://finance.bi.no/~bernt/gcc_prog/recipes/recipes/img169.png let $q t$ be big (t goes to infinity where q is the yield) and you will see why . The first part of the BS formula becomes zero. Also in accordance to put call parity, the call must be worth zero if the entire stock price has been paid out in dividends: ...


1

In the Black-Scholes framework, we assume the log returns are normally distributed. This is equal to saying the underlying is log-normally distributed. If you look at Geometric Brownian Motion on wikipedia, you'll see this: The above solution S_t (for any value of t) is a **log-normally distributed** random variable The wikipedia is correct.



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