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The rules $$(dt)^2 = 0 \qquad dW_t \, dt = 0 \qquad (dW_t)^2 = dt \tag{1}$$ are heuristic rules to simplify calculations when applying Itô's formula. Mind that this is the only application; do not use them anywhere else. In Itô's formula expressions of the form $$\int_0^T f(X_t,Y_t) dX_t \, dY_t$$ pop up. Using $(1)$, we get $$dX_t \, dY_t = (\mu_t dt ... 3 No, it is not correct; the identity$$\mathbb{E}(W(k/n) \cdot W(t))=0$$(which you used in your calculation) does not hold true. Hint: A Brownian motion has stationary increments, i.e. W_t-W_{k/n}\stackrel{d}{=} W_{t-k/n}. So,$$\mathbb{E} \left( \left[ W \left( \frac{k}{n} \right)-W(t) \right]^2 \right) = \mathbb{E}\left( \left[ W \left( t-\frac{k}{n} ...

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If $U(t)=e^{-\mu t}W(\frac{\sigma^2e^{2\mu t}}{2\mu})$, then \begin{eqnarray*}\mbox{Cov}[U(t),U(t+s)]&=&\mbox{Cov}\left[e^{-\mu t}W\left(\frac{\sigma^2e^{2\mu t}}{2\mu}\right),e^{-\mu (t+s)}W\left(\frac{\sigma^2e^{2\mu (t+s)}}{2\mu}\right)\right]\\[10pt] &=& e^{-\mu t} e^{-\mu (t+s)} \mbox{Cov}\left[W\left(\frac{\sigma^2e^{2\mu ...

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\begin{align} \Pr(Y=0) = \Phi\left( \frac{50-\mu} \sigma \right) = {} & \int_{-\infty}^{50} \varphi\left(\frac{z-\mu}\sigma\right) \, \frac{dz} \sigma = \int_{-\infty}^{(50-\mu)/\sigma} \varphi(z)\,dz \\[12pt] & \text{ where } \varphi(z) = \frac 1 {\sqrt{2\pi}} e^{-z^2/2}. \end{align} For $y>50$, $$\Pr(50 \le Y \le y) = \int_{50}^y \varphi\left( ... 2 Stochastic calculus is a huge area in physics, engineering, and pure math. What is a really huge topic in research right now are SPDEs. Stochastic partial differential equations. Take your favorite PDE and add some noise to it. Now you have a SPDE. These equations have numerous mathematical challenges, such as issues of roughness and defining solutions, but ... 2 From what I've been told, Brownian Motion and Stochastic Calculus by Karatzas and Shreve is the gold standard. Continuous Martingales and Brownian Motion by Revuz and Yor is also a great reference. What you have listed as background knowledge is sufficient. 1 The question doesn't ask you to compute E\int_0^\infty X_n^2(t)dt, it asks you to show that it is finite. Do you need to know \sum_{k=0}^{n-1}\frac{k}{n^2} = \frac{n-1}{2n} in order to realize this is a finite number? Nope!, finite sum of finite numbers is finite. Also you should mention something about why X_n is adapted (trivial but since the ... 1 I think you already have it (I say this without knowing what \rho_n is). You have shown that$$ \int_0^t \left( \int_{-\infty}^\infty u^\prime_n( W_s - a ) h(a) da \right)dW_s = \int_{-\infty}^\infty h(a) \left( \int_0^t u^\prime_n( W_s - a ) dW_s \right)da, and that, as n \to \infty, the left hand side converges to \int_0^t \left( ... 1 Freidlin & Wentzell and its community is interested in a set of topics a little bit different from yours (metastability and exit problems). Your kind of cases have been studied, see for example http://repository.ias.ac.in/1132/1/323.pdf and its reference for further information. In a nutshell, the sde you describe, under general conditions for c and ... 1 Let A = \left(\begin{array}{rr} 0 & 1\\ -1 & 0 \end{array} \right), X_t = \left(\begin{array}{r} X_1(t)\\ X_2(t) \end{array} \right), and B_t = \left(\begin{array}{r} B_1(t)\\ B_2(t) \end{array} \right). Then \begin{align*} dX_t = AXdt + \alpha dB_t. \end{align*} Note that \begin{align*} d\left(e^{-At} X_t \right) &= -Ae^{-At} X_t dt + ... 1 By assumption,P[|H\cdot M|^*_t\ge c]\le P[|H\cdot U|^*_t\ge c/2]+P[|H\cdot V|^*_t\ge c/2]\le 18/c(\|U_t\|_1+\|V_t\|_1)$$The M=U-V decomposition as described by Writing a martingale as the difference of two non-negative martingales satisfies \|M_t\|_1=\|U_t\|_1+\|V_t\|_1. 1 Sufficient statistics should be M = \text{card}\{i: Y_i > 0\}, S = \sum_i Y_i and T = \sum_i Y_i^2, with likelihood function$$ \eqalign{L(Y) &= {n \choose M} \Phi\left(\dfrac{50-\mu}\sigma\right)^{n-M} \prod_{i: Y_i > 0} \dfrac{\exp(-(Y_i-\mu)/(2\sigma^2)}{\sqrt{2\pi \sigma^2}}\cr &= {n \choose M} ...

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I believe your argument provides a good interpretation for total (not quadratic) variation, because it is the one associated to the first derivative. For quadratic variation, one should note first that squaring an infinitesimal element will cause it to shrink even more. So if you are adding up smaller things than before and it still results in a great ...

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For reference what you have is an exponential martingale. For A) I would just go about it that way $$E[M_\alpha(t)]= e^{-\alpha^2/t}E[e^{\alpha W_t}]=e^{-\alpha^2/t}\int_Re^{\alpha \sqrt tZ} \phi(z) dz$$ where $\phi(.)$ is the normal pdf, compute the integral, done. B) is a generalization of the preceding steps. Second question: E[M_\alpha(t)^p]= ...

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