7

I'll expand on Mark's and SRKX's answers which are both correct but brief. To be clear the words long and short have been generalized in finance. They used to mean that you owned a stock or had sold a stock short. Now they are often used to say you make money when a value goes up (long) or make money when some value goes down (short). In this case ...


6

the vega of a call is always positive. The holder of a call option is therefore long volatility whatever the spot price.


4

Mark Joshi's answer is absolutely right, but just to elaborate a bit: The Vega of an option is the sensitivity of its value with respect to volatility $\nu = \frac{\partial V}{\partial \sigma}$. For calls, it makes sense that the Vega is always positive, not matter the level of the underlying. If you take the Black-Sholes model, you can find the ...


4

For NYSE EuroNext data I think this is the page to start at. This seems to be what you are after here There are sample files provided here showing hte format of their many historical data sets.


4

The Black-Scholes price of this option is approximately $14.8$. When I run a Monte Carlo simulation with $10000$ paths and "exact" time stepping, I get results very close to this value. You are simulating the terminal asset price with the first-order Euler approximation over multiple time steps: $$S(t+\Delta t)= S(t) + rS(t)\Delta t + \sigma S(t)\sqrt{\...


4

(1) No, the stochastic differential equation for Heston model does not have an explicit solution. What does exist is an explicit formula for the Fourier transform of a call option price. See e.g. http://www.zeliade.com/whitepapers/zwp-0004.pdf for a decent survey. (2) Yes, implied vol always exists. You can check that the Black-Scholes price of an option ...


4

Note that, \begin{align*} \frac{\partial{C}}{\partial{\sigma}} &=\frac{S_0}{\sqrt{2\pi}}{e^\frac{-d_+^2}{2}}(\frac{-1}{\sigma})(d_-)-\frac{Ke^{-rt}}{\sqrt{2\pi}}e^{\frac{-d_-^2}{2}}(\frac{-1}{\sigma})(d_+)\\ &=\frac{1}{\sqrt{2\pi}}e^{\frac{-d_+^2}{2}}\left[-\frac{S_0 d_-}{\sigma} + \frac{Ke^{-rt}d_+}{\sigma} e^{\frac{d_+^2}{2} - \frac{d_-^2}{2}} \...


3

Consider two options with maturity $T$ that only differ in their exercise styles, one being European (holder can only exercise at $T $), the other American (holder exercises when it's best for him/her). These options need not necessarily be vanilla options. Let us further denote by $I (S_t) $ the intrinsic value of these contigent claims at time $t $, i.e. ...


3

Let $C\left(S,t\right) $ represent the price of the call option when the underlying price is S at time t. Now if S changes by h instantaneously, the call price becomes $C\left(S+h, t\right) $. So the change in the call option price is: $C\left(S+h, t\right) - C\left(S,t\right) $ Which you can approximate via first order Talyor series: $C\left(S+h, t\right)...


3

If $\sigma=0$, the stock price is deterministic and grows at rate $r$. In one year, it is thus worth $100\cdot e^{0.05}\approx 105.13$. The strike is $K=100$. Your payoff is thus $5.13$. Discounting at rate $r$, you get as today’s fair option price $5.13\cdot e^{-0.05}\approx4.88$. Note that there is no randomness and the stock price is perfectly predictable....


2

From the format of your question, I imagine it comes from some exercises set. If so, I would be curious to see it, because it looks really weird to me. Calculate the price the investor is willing to pay for the option assuming they want to make 3% expected return over the period. That doesn't make sense. Indeed, what you are trying to price is a ...


2

You can derive these formulae by tweaking the black scholes derivation. If you are using PDE method, you will use different boundary conditions. If you are using integration over the risk neutral probability , you will use a different payoff function but the same risk neutral density. Alternatively , you can observe that these payoffs are combinations ...


2

A convex function is when the line between two points on the graph always lies above the graph. And this does hold for the put, its also sometimes called a sublinear function. Also see http://en.wikipedia.org/wiki/Convex_function So the author is correct in saying that $(K-s)^+$ is convex.


2

I think you are right. Now when I check papers I've used for my thesis I don't see almost any with empirical data section. Maybe this one will be helpful: Roswell E. Mathis, III, Gerald O., Bierwag Pricing Eurodollar Futures Options with Ho and Lee and Black, Derman, and Toy Models: An Empirical Comparison


2

I think after spending some time on the topic with research papers, I can come up with a satisfying answer. I will list them item by item so (I hope) it would be more clear. Starting from the most obvious. Theoretically, adjusting your model to the market price has no benefit except to calibrate your greeks (prominently Delta and Gamma). It literally says "...


2

If the market prices the option at USD5 and your model says the price is USD4.9. Assuming all parameters between model and market, then the difference comes from the vol you have and the implied. If you sell the option at USD5 and you delta hedge "continuously", then you will scalp the difference between realized and implied vol. You will realize a pnl that ...


2

I will answer in two parts: 1): Assuming this is meant in the Black Scholes sense (one dimensional diffusion), here is an alternative explanation motivated by the convexity of the option price. In simple terms, if you plot the Back Scholes Call price as a function of S, you get this convex curve (blue curve): Now if you draw the tangents at two different ...


2

As @AlexC shows this is true in the Black Scholes model. However as the late Mark Joshi comments in the link, the result is not true in all models. For example , in the stochastic vol model , suppose implied vols fall if the underlying stock moves away from at the money in either direction. Perhaps the losses on implied vol exceed the gamma profits from the ...


1

Here is the reasoning I use. (Assuming familiar facts from the standard BSM setup). The Delta of a Call is positive, as S goes up the Delta goes up towards 1. (Why? The call has more moneyness, more probability of exercise, so the option behaves more like the stock). The Delta of a Put is negative, as S goes up the Delta goes towards zero (less chance of ...


1

Have a look at Tick Data. They sell trade (and optionally quote) data for a lot of different instruments including the EUA futures. Their pricing is 125 USD for a one year tick history and you need to order for at least 250 USD at a time. I ordered some Eurex DAX future tick history from them. The order process was flawless and their pricing was ...


1

You are mostly right, I don't really get what you don't understand. The answer in the book is quite clear, but let me put it that way : Selling a put and buying a call on the same underlying $S$ with same maturity and same stike $K$ is always equivalent to a long position in a forward contract on $S$ with delivery price $K$. The easiest way to see that is ...


1

first, there is a formula for the continuously monitored case. second, if you use log coordinates the Euler discretization is exact so this should be done. third, the convergence for discretely monitored to continuously is actually very slow so you will need a lot of steps. fourth, it's actually better to draw the hitting time to the barrier rather than ...


1

One solution is to calculate the annual dividend yield implied by that. $Div_{yield}=\delta=1.5/40$ and then replace the $r$ on $d_+$ by $r-\delta$. A cleaner way would be to compute it using a binomial tree.


1

The value of an cash-or-nothing option is just the discounted expected payoff of the option. So the value of such a call should be $e^{-r (T - t)} N \mathbb{P} \left\{ S_T > K \right\}$, where $\mathbb{P} \left\{ S_T > K \right\} = \mathcal{N} \left( d_2 \right)$, and $N$ is the cash agreed to be paid. The asset-or-nothing is a bit more complicated ...


1

You can also include variance reduction techniques in you monte carlo simulations, such as control variates or antithetic variates. Both aim at reducing the variability of your simulated option price and are very popular for monte carlo simulations. http://en.wikipedia.org/wiki/Antithetic_variates http://en.wikipedia.org/wiki/Control_variates Both are ...


1

If I understand correctly you have calculated our investors expected payoff using his probabilities to 11.177USD. He wants a three percent return so the value he assigns is 11.177/1.03 = 10.85USD. Simple as that. You can then have another argument a la Black and Scholes to show that you can replicate the payoff to another cost. If that cost is lower, your ...


1

Found.. Doesn't contain the aggregations but it's easy to scrape the detailed codes from here: http://www.cso.ie/px/u/nacecoder/index.asp


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