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Maybe this can help you to understand the concept of conditional expectation, behind your question. Suppose you have a probability space $(\Omega, \mathcal P (\Omega), \mathbb{P})$, where $\mathcal P (\Omega)$ denotes the set of all possible subsets of $\Omega$ (evidently, a $\sigma$-algebra), and $\mathbb{P}$ is a probability measure (in this case, a ...

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Your computation is valid. Probably the easiest way to show it is to note that, for every $t$, $$E(\mathrm e^{\mathrm iuW_\tau}\mid\tau=t)=E(\mathrm e^{\mathrm iuW_t}\mid\tau=t)=E(\mathrm e^{\mathrm iuW_t}),$$ where the second identity holds thanks to the independence of $W_t$ and $\tau$. One knows that, for every $t$, $E(\mathrm e^{\mathrm iuW_t})=\mathrm ... 3 Durrett's probability book appears to still be free (on author's page). Your subject is embedded in Chapter 3 Central Limit Theorems (Weak Convergence, Characteristic Functions etc., leading to Continuity Theorem 3.3.6). Rick Durrett's Probability: Theory and Examples book Theorem 3.3.6. Levy's continuity theorem: Let$\mu_n$,$1\leq n \leq \infty$be ... 3 First of all, we note that for $$b(x,y) := \begin{pmatrix} k_1 \cdot x \\ k_2 \cdot y \end{pmatrix} \qquad \sigma(x,y) := \begin{pmatrix} \sigma_1 & 0 \\ \sigma_2 \varrho & \sigma_2\sqrt{1-\varrho^2} \end{pmatrix} \qquad B_t := \begin{pmatrix} W_t \\ W_t^1 \end{pmatrix}$$ we have $$\begin{pmatrix} X_t \\ Y_t \end{pmatrix} = \int_0^t b(X_s,Y_s) \, ... 2 The converse is not true. The Cauchy random variable is finite a.s. but it does not have defined its mean. Same with the Pareto distribution for \alpha \leq 1 where the mean is infinite. The property you mention comes from measure theory. If a function f > 0 is not finite a.e. then the integral of f is already infinite over \{f=\infty\}. For ... 2 You start by the conditional expectation with respect to N_t. E[S_t|N_t] = S_0\prod_{i=1}^{N(t)} EX_i because X_i are independant, with the additionnal hypothesis that the process N and the X_is are independant, so$$E[S_t|N_t] = S_0 \mu^{N(t)} $$Let r be the rate of the Poisson process, we have$$E[S_t] = EE[S_t|N_t] = S_0 ... 2 The random variable$X$is measurable with respect to$\sigma$-algebra$\mathfrak F$if$X=\mathbb E(X\mid\mathfrak F)$. One can understand this in a few steps:$\mathbb E(X\mid A)$, where$A$is an event, is the expected value of$X$given that$A$occurs;$\mathbb E(X\mid Y)$, where$Y$is a random variable, is a random variable whose value at ... 2 An obvious lower bound is $$\int_0^{t^{\ast}(\omega) \wedge T} f(t,\omega) \, dt \geq \inf_{t \in [0,t^{\ast}(\omega) \wedge T]} f(t,\omega) \cdot (t^{\ast}(\omega) \wedge T).$$ Without further assumptions on the stopping time$t^{\ast}$or$f$, it will be difficult to get a better lower bound. Since the considered integral is defined pathwise you are ... 2 The predictable$\sigma$-algebra$\mathcal{P}$is defined on the product space$\tilde{\Omega} := [0,\infty) \times \Omega$. In order to prove that a process$X$is measurable with respect to$\mathcal{P}$, it suffices to show that there exists a sequence$(X^n)_n$of$\mathcal{P}$-measurable random variables such that$X^n \to X$pointwise as$n \to ...

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I don't understand the fetish for calculus. There should be a way without it (in fact, personally I prefer avoiding it if necessary): Order the data as follows: $x_1 < x_2 \ldots <x_{n}$. Let's suppose that $x \in [x_i,x_{i+1}]$. Then we have the following the $$\|(x_1, \ldots x_n) - (x,x\ldots x) \|_1 = \sum_{j \leq i}(x-x_j) + \sum_{j>i} ... 1 Using the definition of a Riemann integral, we have that the Riemann sum for a partition 0=t_0<t_1<\dots<t_N=T is$$ S_N = \sum_{i=1}^N W(t_i)(t_i - t_{i-1}), $$Now, we have that W(t_i) = \sum_{k=1}^i[W(t_k) - W(t_{k-1})], so$$ S_N = \sum_{i=1}^N\sum_{k=1}^i[W(t_k) - W(t_{k-1})](t_i - t_{i-1}).  S_N = \sum_{i=1}^N[W(t_i) - W(t_{i-1})](T ...

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An alternate way of showing this result is to use the (stochastic)-integration by part formula : $\int_0^tW_sds= W_t-W_0 - \int_0^tsdWs$ ( no quadratic covariation term has s has 0 quadratic variation) The first term is normal and independent from the stochastic integral which is a Wiener integral. As Wiener integrals are normal random variables, we have ...

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The distribution of $X_0$ should be given by some initial condition, typically it is a constant, but not necessarily and $W_t \sim \mathcal{N}(0,t)$ (i.e. the variance is $t$). Then once both distributions are known, you have to take the integral. An alternative approach, when the solution to the SDE is not easily derivable, is to take expected value of ...

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Let us assume that $X_1$ and $X_2$ are independent. Then $E(X_1X_2)=E(X_1)E(X_2)=\lambda^2$. Now we deal with $X_1^2$. It is a standard fact about the Poisson with parameter $\lambda$ that it has variance $\lambda$. It follows that $E(X_1^2)=\lambda+\lambda^2$, so $X_1^2$ is not an unbiased estimator of $\lambda^2$. Remark: If the "standard fact" cannot ...

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I think you are right. The stochastic Itô integral is a Gaussian random variable with mean zero, since the integrand is an $L_a^2([0,T]\times \Omega)$ process (in fact, deterministic) The mean is 0, (it is a martingale) and the variance (second moment), by Itô's isometry, is $$\sigma^2 e^{-2\kappa t} E\left[\int_0^t e^{-2\kappa s}ds\right] = ... 1 For simplicity of notation suppose that \theta=0; then the solution of the SDE$$dY_t = - k Y_t \, dt + \sigma dW_t$$is given by$$Y_t = e^{-kt} Y_0 +\sigma\int_0^t e^{-k(t-r)} \, dW_r.$$If r(0) \sim \log(\mu,\varrho^2), then we can write r(0)= e^{Y_0} where Y_0 \sim N(\mu,\varrho^2) is independent from (W_t)_{t \geq 0}. In particular, we see ... 1 Hi if you take any process that is not a semi-martingale with null quadratic variations (for example a fractional Brownian motion with Hurst parameter less than 1/2) and then you add some jumps, then it has non null quadratic variations (the square of its jumps) and you have a counterexample. Best regards 1 I am not so sure about your result I would have said that it converges to 0 in probability \forall t>0. First observe that \frac{W_t^2}{2\epsilon}\chi_{(-\epsilon,\epsilon)}(W_t)\le \frac{\epsilon^2}{2\epsilon}=\frac{\epsilon}{2} so the first term goes to 0 almost surely. As for the second term using Itô's isometry we see that :$$E[(\int_0^t ...

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Like I said in the comment I think you actually mean random field. Also I'm not sure how the right hand side of your expectation can still depend on $\vec{r}$ if an average has been taken over $(x,y)$. It would be helpful to actually write out what the averages mean. I think perhaps you mean something like this C(\vec{r}') = \langle ...

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It this case, if for example $m_1\neq m_2, \sigma_1\sigma_2\rho\neq 0$, no, because the drift is not a function of $X_t + Y_t$. In a general situation, when $X,Y$ are two non independant Markov chains, $X+Y$ is not. To understand this with a more simple example, in a discret time setting, take for example $(X_n)$ be a simple random walk on $\mathbb Z$, and ...

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