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There is a Borel set $E$ in $\mathbb R^2$ such that $F := \{x-y\colon (x,y) \in E\}$ is not a Borel set. Let $A := \{f \in \mathbf{C}\colon (f(1), f(0)) \in E\}$. Then $A \in \mathcal{B}_{\left[0,\infty\right)}$. How about $T(A)$? In fact $$T(A) = \{g \in \mathbf{C}\colon g(0)=0, g(1) \in F\}$$ and is not Borel. added Mar 10 Why is $T(A)$ not Borel? ...

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First of all, we choose $M \in \mathbb{N}$ sufficiently large such that $\sum_{m \geq M} 2^{-m} < \delta/3$. Moreover, we note that \begin{align*} \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \right) &\leq \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \cdot 1_A \right)+ \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \cdot ... 2 We need to check that the random variable X(t) is measurable with respect to \sigma-algebra \mathcal F_t for each t\ge0. If t\in[0,1], the only value that X(t) can take is 0. So we need to find the smallest \sigma-algebra that contains X^{-1}(t)(\{0\})=\Omega. Such \sigma-algebra is \{\emptyset,\Omega\}. If t\in(1,2], the range of ... 2 Well it seems that the correct reference for this result is in the book by Revuz and Yor "Continuous Martingales and Brownian Motion" at the Chapter IX - Stochastic Differential Equation, Section 2 Existence and Uniqueness in the Case of Lipschitz Coefficients precisely treated as an Exercise (2.11) about "Zvonkin's Method". The result is far from trivial. ... 2 You seem to be working towards a proof based purely on Markov theory. However, in this particular case, a simpler and more natural proof can be obtained through stochastic calculus, if you are willing to take a few theorems for granted. We proceed as follows. As you write yourself, let B denote a d-dimensional standard Brownian motion. Let \|\cdot\|_2 ... 2 Let (X_t)_{t \geq 0} be a non-negative solution of the SDEX_t - x = 3t + 2 \int_0^t \sqrt{X_s} \, dB_s \tag{1}$$for x \geq 0. Applying Itô's formula to f(y) = \frac{1}{\sqrt{y}}, we find$$\frac{1}{\sqrt{X_t}} - \frac{1}{\sqrt{x}} = - \int_0^t \frac{1}{X_s} \, dB_s.$$For \tau_{a,b} := \inf\{t \geq 0; X_t \notin (a,b)\}, 0<a<b, this ... 1 Set W_t := A \cdot B_t where A is an orthogonal matrix. Since W_0=A \cdot x we see that (W_t)_{t \geq 0} starts at A \cdot x. As t \mapsto B_t is continuous, we find that t \mapsto A \cdot B_t is continuous. Let 0 \leq s \leq t. Then,$$\begin{align*} \mathbb{E}e^{\imath \, \xi^T \cdot (W_t-W_s)} = \mathbb{E}e^{\imath \, \xi^T (A \cdot ...

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Hint: Let $B$ be the event "$n$ events in the first $T$" and $A$ the event $m$ events in the first $t$." We want $\Pr(A|B)$, which is $\frac{\Pr(A\cap B)}{\Pr(B)}$. To compute $\Pr(A\cap B)$, note that for this event to occur we need $m$ events in the first $t$ hours and $n-m$ additional events in the remaining $T-t$ hours. Thus you need to multiply your ...

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Hint: In order for you answer to be correct you also need to add the following term in the numerator \begin{align*}&\phantom{\,\=}P(m \text{ events in t hours | n events in T hours})= \\ \\& =\frac{P(m \text{ events in t hours, n-m events in T-t hours})}{P(n \text{ events in T hours})}=\\\\& =\frac{P(m \text{ events in t ... 1 The first question is true and comes from the property of Lévy processes sometimes taken as a definition as it characterize those processes (see for example Protter's book) : They are continuous in probability From this your first question point is trivially true. For the second the question the answer is obviously no take a T as the 1st jump time of ... 1 First of all note that\{X_t \neq X_{t-}\} = \{N_t \neq N_{t-}\}$$as the Brownian motion has continuous sample paths. Moreover, since (N_t)_{t \geq 0} is a Lévy process the jumps cannot occur at fixed times, i.e.$$\mathbb{P}(X_{t} \neq X_{t-}) = \mathbb{P}(N_t \neq N_{t-}) = 0.$$Indeed: Write \Delta N_t := N_t-N_{t-}. Then, by Fatou's lemma, ... 1 \mathbb{E}x[n+\tau]y[n] := R_{xy}[\tau] is the expected value of the cross-correlation at lag \tau. This is the average over the signal of the degree to which the signal y "now" can be used to estimate the signal x at "now + \tau". Since these signals are assumed stochastic (but stationary enough for these expectations to be stationary) we cannot ... 1 an example is the transition matrix\bigl(\begin{smallmatrix} 0&1\\ 1&0 \end{smallmatrix} \bigr) with \mu_1 = \mu_2 = \frac 12. If \mu_i p(i,j) = \mu_j p(j,i)  then summing over i puts$$ \mu_i= \sum_j \mu_i p(i,j) = \sum_j \mu_j p(j,i) = (\mu p)_i $$that is: \mu is invariant. 1 The result is also mentioned (and proved) in N. Ikeda, S. Watanabe: Stochastic Differential Equations and Diffusion Processes, Section IV.2 and IV.3. In fact, one can show that the SDE$$dX_t = b(X_t) \, dt+ dB_t, \qquad X_0 = x$$has a (strong) unique solution for any Borel-measurable bounded function b. 1 Consider the following two scenarios: 1) You have two independent Poisson processes N_1(t) and N_2(t) with rates \lambda p and \lambda (1-p). N(t) = N_1(t) + N_2(t) is their sum. This is a Poisson process with rate \lambda. 2) You have a Poisson process N(t) with rate \lambda. Each time an event occurs, a coin is tossed: with ... 1 Hint: We get the independence result if we can show that$$\mathbb{E}\left[\mathrm e^{\mathrm i\left(aN_1(s)+bN_2(t)\right)} \right] = \mathbb{E}\left[\mathrm e^{\mathrm i aN_1(s)} \right]\mathbb{E}\left[\mathrm e^{\mathrm i bN_2(t)} \right],  for all $s\leq t$ and $a,b$. Once the case $s=t$ is done, the case $s<t$ is formal, based on conditioning and ...

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