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For $T/2 \leq t \leq T$, \begin{align*} E(X\mid \mathcal{F}_t) &= \exp\big(W_{\frac{T}{2}}+\frac{1}{2}T\big) E\big(\exp\big(W_{T}-\frac{1}{2}T\big)\mid \mathcal{F}_t\big)\\ &= \exp\big(W_{\frac{T}{2}}+\frac{1}{2}T\big) \exp\big(W_{t}-\frac{1}{2}t\big)\\ &=\exp\big(W_{\frac{T}{2}}+W_{t} + \frac{1}{2}T-\frac{1}{2}t\big). \end{align*} For $0 \leq t ... 1 If$\sigma=0$there is no randomness: the spot follows a single deterministic path. That is, the measure consists of a point mass at that path. Any equivalent measure can again only give a point mass at that same path, with the same drift. So in this case we must have$\mu = r$to have an equivalent martingale measure. This is arbitrage free, but there ... 3 For a Brownian motion, if you wait$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$is lower than 1/2, it will mean ... 3 The more phenomenological definitions in his books are probably more helpful. Whether one uses the fractal dimension, Hurst coefficient, or exponential coefficient alpha, there is a value that corresponds to pure Brownian motion, a regime relative to this value that corresponds to persistence of motion, and the opposite regime that corresponds to ... 1 I will assume that the convergence is in probability and the partition$\Pi_nis given by \begin{align*} 0=t_0 < t_1 < \cdots < t_n = t. \end{align*} Note that $$\mathbb{E}\big((W_{t_i}-W_{t_{i-1}})^3 \big)=0,$$ and $$\mathbb{E}\big((W_{t_i}-W_{t_{i-1}})^6 \big)=15(t_i-t_{i-1})^3.$$ Let\|\Pi_n\| = \max_{i=1}^n|t_i-t_{i-1}|$. Then, for any small ... 2 Note that$Xis a continuous martingale. Moreover, the quadratic variation is given by \begin{align*} \langle X_t, \, X_t\rangle = \int_0^t |\sigma_u|^2 du = c^2 t. \end{align*} That is, \begin{align*} \langle X_t/c, \, X_t/c\rangle = t. \end{align*} From Levy's characterization,X/c$is by law a Brownian motion, which we denote by$\beta$. Then, by law, ... 0 You shall distinguish between the random variables and their distributions. That is, when we talk about Brownian motion, we often define it in terms of finite-dimensional distributions (that is, conditional distribution is normal with the current value as the mean and the variance of$1$). By Kolmogorov's Extension Theorem (KET), there exists a unique ... 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 + ... 0 Your choice of$a_t$and$b_t$is feasible. For a self-financing portfolio, the units invested should be static within an infinitesimal time interval, that is, no extra investing or withdrawing during this period. In other words, the portfolio value changes only through its underlyings. For a further discussion, See the article ... 4 Since$W_{2t}-W_{t}$is independent of$W_t$and has the same law as$W_{2t-t}=W_t$we only have to compute $$P(X(X+Y)<0)$$ where$(X,Y)$follows a bivariate normal distribution (with zero correlation). From there you can split the probability in two cases : either$X<0$and$X+Y>0\$ or the opposite. The two events have the same probability since ...

1

Your decomposition is correct. I will show here the computation for one term: \begin{align*} P(W_t < 0, W_{2t} >0) &= E(W_t < 0, W_{2t}-W_t > -W_t)\\ &= E\Big(E\big(\mathbb{1}_{\{W_t < 0\}}\mathbb{1}_{\{W_{2t}-W_t > -W_t\}}\mid W_t\big)\Big)\\ &= E\Big(\mathbb{1}_{\{W_t < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ ...

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as I mentioned here, this paper provides some theoretical insight (and a way to approximate the true value). The authors end up with an approximative series for the density. It is implemented in the function maxdd of the R-package fBasics. There are convenient functions dmaxdd, pmaxdd and rmaxdd. Calculating the Expected Drawdown should be easy. (to be ...

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