# Tag Info

3

To provide a straight forward answer: It is not a good model. It never was, it never will be. Until we all do not come up with a better model that provides better modeling accuracy while it is equally intuitive and makes similarly simplifying assumptions the BS model with its geometric brownian motion component is here to stay. It actually does not matter ...

2

Basically, Black-Scholes is an "industry standard" formula. It is widely used by practitioners and usually augmented with extra specifications or intuition. It has a closed form solution, which is rare in option pricing models. It is also relative to simple to understand. Otherwise, you usually need to rely on Monte Carlo simulation or some other way. And ...

3

Brownian motion - because it is simple, and results in intuitive closed form solutions, and it's not a terrible description of asset prices, especially when employed in high-frequency event time. Geometric - because the returns compound, and equities cannot go below zero due to the fact that they are limited liability corporations There are many, many ...

4

If at first you don't have a model at all, then geometric Brownian motion is not bad. As others before me said: log-returns are normally distributed in this model. This is debatable and there are times and markets where this is not true. There is more than enough research about this. But why is a model based on Brownian motion not that bad? The reason is ...

0

The normal distribution is very powerful distribution: By the central limit theorem, the mean of any large sample always converges to the normal distribution Considering the most simplistic Binomial Tree model, where price goes only up or down each period, it can be shown that the distribution of returns of this tree converges to Normal for infinetesimal ...

0

So Black-Scholes came around and it made pricing options mathematically viable in a (seemingly) objective sense. The assumptions underlying the model are flawed, but they work reasonably well in a lot of market environments. The problem lies in when normal market conditions start to change character. When things start to get wacky, like when a giant selloff ...

1

First of all, a filtration $( \mathscr{F}_t )_{t \geq 0 }$ is a "set" of sigma algebras indexed usually by time t that are increasing. That is, for every $t>0$, $\mathscr{F}_t$ is a sigma algebra and $\mathscr{F}_t \subseteq \mathscr{F}_T$ for all $0\leq t \leq T$. The canonical example, is the filtration generated by a process, say Brownian Motion $W$: ...

2

You have $$\widetilde{W}_t=W_t+\int\Theta(u)du$$ which is in general not a Brownian motion, because it has a drift component. But 5.3.1 states $$M_t=M_0+\int \Gamma(u)dW_u\tag{5.3.1}$$ , which holds only for a Brownian motion $W$ (and $M_t$ martingale). So one cannot trivially replace $W_t$ and $W_t+\int\Theta(u)du=\widetilde{W}_t$ in 5.3.2 aswell by ...

1

The GBM is a continuous model, so using large integer time steps naturally leaves large discretization error (which vanishes when you increase the number of steps). Use small time step 0.001: paths(j + 1,i) = paths(j,i) * exp((mu - vol^2/2)*0.001 + vol * 0.001^0.5*shocks_ant(j,i)); Then the mean is almost exactly 100 as expected.

0

appearantly your sampling variance is too large. I reimplemented your example in R. What I first saw is, that the mean got worse if I took more time steps (you take $300$). Your volatility is $0.3$ which is $30\%$ per year and you sample $300$ years. What you should do is the following: define a variable nbr_steps_peryear choose the number of years then ...

2

You should look at confidence interval. Normally, your confidence interval size is proportional to the standard deviation, looking something like: with probability $p$ your value will be in the interval: $$[\bar{S} - k*StdDev, \bar{S} + k*StdDev]$$ Then, getting back to your simulation, we can say that your time step is very big (1 year) and you simulate ...

3

You have typo "vol^2", but it should be "vol". Its $$\sqrt{\sigma^2T}=\sigma\sqrt{T}$$

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