# Tag Info

## New answers tagged brownian-motion

1

The Papageorgiou paper is presumably referring specifically to quasi-random sequences used in path generation. Researchers had noticed that, in high dimensions, QR sequences tend to have good space coverage for the first couple of dimensions: but terrible coverage for the latter dimensions: (Plots here are points 101-200 from a 32-dimensional QR ...

3

A key property of Brownian motion is independent increments. So if $x-1 > y$, then $$\mathbb{E}[\Delta W_x \Delta W_y] = 0$$ because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. $x-1 \leq y < x$, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ ...

1

The equation can easily be derived from the characteristic function of the geometric Brownian motion. As stated in the footnote, the authors use $$\frac{dS_t}{S_t} = \sigma dW_t$$ as the underlying model. The change in stock price $X_T = S_T - S_t$ is therefore normally distributed with mean 0 and variance $\sigma^2 (T-t)$. The characteristic function of ...

1

I only glanced at it for a second. But, looks like the underlying process is arithmetic Brownian motion, and they're computing conditional expectation of it's expectation at time t. So, basically, use the fact that $$\exp(\sigma W_t - \frac{1}{2}\sigma^2 t)$$ is a martingale, and then with the remaining independent increment, calculate directly using the ...

3

I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at $$S_t = S_0\exp(\mu t + \sigma B_t).$$ Assume that $S_0$ is known and fixed and look at by Ito's formula $$d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt.$$ Then with some abuse of notation: $$... 4 The convexity of the exponential function of the stochastic variable W makes its expectation greater than the exponentiation of the expectation of W. This is an example of Jensen's inequality, E[e^{\sigma W}]> e^{\sigma E[W]}=1. \sigma can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ... 2$$S_t = S_0\exp((r-\frac{\sigma^2}{2})t+\sigma W_t) is not yet a martingale for it is not dirftless. From a probabilistic point of vew the "drift adjustment" comes into play so that the expected value of $S_t$ will be $e^{rt}$ rathern than $e^{(r+0.5\sigma^2)t}$. For the expected value of a log-normaly distributed variable with mean $\mu$ and vol ...

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