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The initial formula can be thought of as Taylor expansion out to second order. The rules can be understood in the framework of Riemann sums. An integral $dB$ in sum form looks like: $$\sum f_i (B_{i+1}-B_i).$$ An integral $dt$ in sum form looks like $$\sum f_i (t_{i+1}-t_i).$$ An integral $(dB)^2$ in sum form looks like $$\sum f_i (B_{i+1}-B_i)^2.$$ An ...

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In general, $X \in \mathcal{V}$ does not imply $\mathbb{E}(|X_t|)<\infty$. Just consider the process $$X_t(\omega) := \begin{cases} 0, & t < 1, \\ \frac{1}{\omega}, & t \geq 1 \end{cases}, \qquad \omega \in \Omega := (0,1).$$ on the measurable space $((0,1),\mathcal{B}((0,1)))$ endowed with the Lebesgue measure. Obviously, $X \in ... 3 I'm not too familiar with Brownian motions, but I think the problem is basically the same as in the case of individual random variables as in muaddib's answer: "Take the process$X$where$X_0$,$X_1$are iid positive random variables and$X_2$is a normal random variable with variance$X_0+X_1$and mean zero." Here$X_2$could be said to have "random ... 2 Start by conditioning on the first step: \begin{eqnarray*} \rho &=& P(Y_1=1)P(S_n\geq 1\mid Y_1=1) + P(Y_1=-k)P(S_n\geq 1\mid Y_1=-k) \\ &=& p\cdot 1 + (1-p)P(S_n \geq k+1) \\ && \qquad\text{since, if$Y_1=-k,\;$subsequently reaching$S_n=1,\;$for some$n,\;$is} \\ && \qquad\text{equivalent to reaching$S_m=k+1,\;$... 2 Since$T$is a hitting time and$T(\omega)>S(\omega)$we can write$T(\omega)=S(\omega)+T(\theta_S\omega)$. Expressing the Brownian motion explicitly as a function of two variables, i.e.,$B_T(\omega)=B(\omega,T(\omega))$we have $$B_T(\theta_S\omega)=B(\theta_S\omega,T(\theta_S\omega)) =B(\omega, ... 2 If you have an Ito process$$dX_t=\mu_tdt+\sigma_tdB_t$$then Ito's lemma states that$$df(X_t)=f'(X_t)dX_t+\frac{1}{2}f''(X_t)\sigma_t^2dt$$Here \mu_t=0.4S_t and \sigma_t=0.5S_t. We also have f'(x)=\frac{1}{x} and f''(x)=-\frac{1}{x^2}. So d(\log S_t)=\frac{dS_t}{S_t}-\frac{1}{2}\frac{(0.5S_t)^2}{S_t^2}dt=0.4dt+0.5dB_t-0.125dt 2 A simplified way of thinking about this is to interpret the equation as saying what approximately happens to the value in the left hand side when time advances a little. If the present time is t and a very short interval of time \Delta t passes, then the value of f(t,X(t)) will change approximately by the amount$$ \dfrac{\partial ... 2 There is no problem integrating a brownian motion pathwise with respect to a continuous (say) integrand. This is because a brownian motion has continuous almost everywhere sample paths. In the Ito integral the brownian motion is used as an integrand, i.e. it has the form$\int_{[0,T]} f(s,\omega) dW(s,\omega)$. One can show that if every continuous ... 1 My bet would be that the definition is in fact through the quadratic covariation:$\langle w_t,\bar w_t\rangle = \rho t$for all$t$. Then, as a result from Ito lemma you get: $$\mathrm dw_t\bar w_t = w_t\mathrm d\bar w_t + \bar w_t\mathrm dw_t + \mathrm d\langle w_t,\bar w_t\rangle$$ which gives you that$\Bbb Ew_t\bar w_t = \rho t$. I am not sure ... 1 Ito's formula tells you that $$f(t,X_t)=f(0,X_0)+\int_0^t \frac{\partial f}{\partial s}(s,X_s) ds + \int_0^t \frac{\partial f}{\partial x}(s,X_s) dX_s + \int_0^t \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(s,X_s) (dX_s)^2$$ where$(dX_s)^2$is a notational shorthand for the quadratic variation of$X_s$. If$X_t$is a 1D diffusion process satisfying ... 1 Partial answer: We will derive the unconditional variance. By the Ito isometry, we get that$ \mathbb{E}(\int_{0}^{r}\exp\{(r-s)\phi+\lambda(W_r-W_s)\}dB_s)^2 = \mathbb{E}\int_{0}^{r}\exp\{(r-s)\phi+\lambda(W_r-W_s)\}^2ds $Noting that$\exp\{r\phi + \lambda W_r\}$is a constant, we move that outside of the integral. ... 1 Rewriting the first formula of the proof gives $$M_t = \underbrace{\int_0^t q \, \text{sgn}(M_s) |M_s|^{q-1} \, dM_s}_{=:X_t} + \underbrace{\frac{1}{2} q (q-1) \int_0^t |M_s|^{q-2} d\langle M \rangle_s}_{=:A_t}. \tag{1}$$ Since$(M_t)_{t \geq 0}$is a local martingale with continuous sample paths, we know that$(X_t)_{t \geq 0}$is a local martingale with ... 1 Following up on @saz's comment: In expanding$Y:=\log X$using Ito's formula you express$Y$in terms of$\mu$,$\sigma$, and$W$. This then leads to an explicit formula for (a putative solution)$X=e^Y$in terms of those same three ingredients. One of your jobs is then to check that this formula does indeed provide a solution. The first thing to notice is ... 1 Here are a few things to look at. Elton Hsu has a brief intro here https://www.math.kyoto-u.ac.jp/probability/sympo/PSS03abstract.pdf He also has a old paper published by the AMS http://www.math.northwestern.edu/~ehsu/Brownian%20Motion%20and%20Riemannian%20Geometry.pdf Both of these have a nice list of references. As far as books in print I would ... 1 It is not sufficient to consider component wise independence, or even independence between the components. You need exactly the independence that is written in the definition of independent increments: The two-dimensional random variables$F_{t_i} - F_{t_{i - 1}}$must be independent. 1 That the stochastic integral $$\int_0^t \sigma(s,X_s) \, dW_s$$ is not defined as an pathwise integral does not mean that $$\int_0^t \sigma(s,X_s) \, dW_s$$ has no "a.s." meaning. Note that for each$t$$$M_t := \int_0^t \sigma(s,X_s) \, dW_s$$ is a random variable; so in particular it does make sense to state that$\$X_t = X_0 + \int_0^t b(s,X_s) \, ds + ...

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Not directly your question, but Wales, Energy landscapes (2003, 681p) might be interesting. That leads to "basin-hopping", a practical stochastic algorithm for minimizing e.g. configurations of molecules. scipy basinhopping says The acceptance test used here is the Metropolis criterion of standard Monte Carlo algorithms, although there are many other ...

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