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For a Brownian motion, if you wait $dt$, the variance will grow linearly with (proportionally to) $dt$. For a fractional Brownian motion, it will grow with a power law of $dt$, in fact in $dt^{H}$, where $H$ is the Hurst exponent. See wikipedia for more details. It means the fBM will somehow keep memory of the past. When $H$ is lower than 1/2, it will mean ...

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The more phenomenological definitions in his books are probably more helpful. Whether one uses the fractal dimension, Hurst coefficient, or exponential coefficient alpha, there is a value that corresponds to pure Brownian motion, a regime relative to this value that corresponds to persistence of motion, and the opposite regime that corresponds to ...

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Let's consider the following example: the process is initialized randomly with $\pm1$ and then stays there forever. Seems stationary to me, but it would never cross its mean.

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There is a shortcut around the Forward Equation when you are looking for the stationary distribution. Let me write $$dX = \mu(X)dt +\sigma(X)dW$$ for $$\mu(x)=b(1-x)-ax\ \text{ and }\ \sigma^2(x)=x(1-x)$$ The Forward Equation indeed states that the stationary distribution $p(x)$ will be satisfied for $\partial p/\partial t = 0$, therefore $$... 2 The portfolio is self-financing. You simply forgot a term in b and a -t term in V: \begin{eqnarray} V_t &=& a_t S_t + b_t \beta_t = (2B_t ) (10+ B_t) + (- t - B_t^2 - 20B_t)1 \\ &=& 20B_t + 2B_t^2 - t - B_t^2 - 20B_t \\ &=& B_t^2 - t \end{eqnarray} Applying Ito's lemma \begin{eqnarray} dV_t &=& (2B_t dB_t + ... 2 I think to gain intution you have to understand that the same agents that value the stocks will value the options. And agents compensate for volatility by demanding higher expected returns. Therefore you should ask: Why are stocks priced as they are in the first place? In your example, the stock with higher volatility has much lower expected return. This ... 1 This is a special case of the question of why$$ \int_0^T f(t) dW_t $$is normally distributed for a continuous function f(t). This Ito integral can be approximated by a sum$$ \sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) . The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point ...

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For simplicity, We assume that $\alpha$ is a positive constant. You need to show that, for any $t>0$, \begin{align*} \int_0^t e^{\alpha u} dW_u \end{align*} is normally distributed. Consider the process $\{X_t, t \geq 0\}$, where \begin{align*} X_t = \frac{1}{\sqrt{\frac{1}{t}\int_0^t e^{2\alpha u} du}}\int_0^t e^{\alpha u} dW_u, \end{align*} for ...

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As per your comments, this is the Kunita Watanabe decomposition. See the post at http://math.stackexchange.com/questions/413103/kunita-watanabe-decomposition and the presentation http://www.eurandom.nl/events/workshops/2011/ISI_MRM/Presentation/Vanmaele.pdf

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