# Tag Info

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I think Matt Wolf had the best answer by far, but the only point I would add is that the normal distribution can actually be a bit of a dangerous assumption at times, I actually believe this is the reason that more emphasis has been placed on risk management (especially recently) as opposed to pricing models. The main reason for GBM is that it creates ...

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Number one, the central limit theorem means a lot of things that may not be normal end up looking normal when lots of little 'experiments' or impacts are added up. Number 2, when dealing with finance you need a model that seems plausible. An arithmetic Brownian motion could go negative, but stock prices can't. On the other hand, it seems quite plausible ...

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They won't be the same. If you run a discrete simulation you will get the actual (or an instance of an actual path) price process for the future value of the stock using the real probability measure. If you do the same thing using the closed form solution, the path will look very similar but will drift downwards. Why are they different? To see it ...

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Were we doing physics and we said there was an arithmetic Brownian motion we could indeed have a drift rate other than $\mu=r$ and it would make sense. Suppose, for example, that a fluid is moving at velocity $v$ and we have a random walk of particle in it. This would be reality. The whole point of using SDEs in finance is to identify what ought to be ...

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I think, that the answer by Taylor Martin contains the explanation, but should be much shorter, so I'll put it here separately. Essentially, Brownian motion is a measure on the space of continuos functions (trajectories), say on an interval on the real line . How does one describe this measure in probabilistic terms? One sets the probability space ...

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Exhibiting a counter-example is straight-forward enough. For example, let $B_{t}(\omega)$ be a Brownian motion and $\mathcal{T}(\omega)$ a stopping time on $(\Omega,\mathbb{P})$ with a continuous distribution. Then with ...

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If you say that in probability theory an event happens almost surely it happens with probability one. From Wikipedia: The difference between an event being almost sure and sure is the same as the subtle difference between something happening with probability 1 and happening always. If an event is sure, then it will always happen, and no ...

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$W_{t_i}$ and $W_{t_i}$ are not independent (in fact $W_{t_i}=W_{t_i}$), e.g. $$E[W_t^2]=t^2\neq E[W_t]\cdot E[W_t]=0$$ So you cant separate the expectation in your equation.

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You might want to give us the exact statement of the author. Let the Wiener process $W_{s}$ be a r.v. from $\left(\mathcal{F}_{s},\Omega\right)\to\left(\mathcal{B}\left(\mathbb{R}\right),\mathbb{R}\right)$. The Borel-$\sigma$-algebra $\mathcal{B}\left(\mathbb{R}\right)$ contains all intervals of the form $\left[x,y\right]$ for $x\neq y\in\mathbb{R}$, ...

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The above question was a typo due to the author -- the expression should be evaluated as $$E(t|\mathcal{F}_{s}^{W}) = t$$ due to the reasoning in the question. Sorry for the noise.

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I use straightforward approach: Generate "returns"; Make cumulative sum of returns from Step 1; Take any Nth (N should be "big enough") point for series obtained on Step 2. That would be "closes"; Then take max and min between "closes" = highs and lows. In R: n <- 10000 # quantity of "ticks" inside 1 day m <- 200 # number of days rets <- ...

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I am not sure if I understood your question correctly but I will try to answer it anyway. If you have a standard normal random vector $z \sim N(\mathbb{0},I_n)$ (where $z,0 \in \mathbb{R}^{n\times1}$ and $I_n \in \mathbb{R}^{n\times n}$ is the identity matrix) and you want to transform it into a multivariate normal $x \sim N(\mu,\Sigma)$ you do it the ...

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The error is in the application of Girsanov theorem. We have multivariate Black-Sholes market, however I apply one-dimensional Girsanov theorem. I should apply multi-dimensional Girsanov theorem. Then there would be now such equations, except the case for $\rho=1$. The alike task is formulated here ...

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