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0

Note that \begin{align*} dJ_c(s) = dB_s -cJ_c(s) ds. \end{align*} Moreover, $d\langle J_c, J_c\rangle_t = dt$. Then, by Ito's lemma, \begin{align*} \int_0^1J_c(s) dJ_c(s) &= \int_0^1 d\left(J_c^2(s) -s \right)\\ &= J_c^2(1) - 1. \end{align*} Since \begin{align*} J_c(1) &= \int_0^1 e^{-c(1-s)}dB_s \end{align*} is normal with zero mean and the ...

2

We assume that \begin{align*} d\left(\! \begin{array}{c} S^1(t)\\ S^2(t) \end{array} \!\right) =\textrm{diag}\left(S^1(t), S^2(t)\right)\bigg[\left(\! \begin{array}{c} r\\ r \end{array} \!\right)dt + \left(\! \begin{array}{cc} \sigma_{1,1} &\sigma_{1,2}\\ \sigma_{2,1} &\sigma_{2,2} \end{array} \!\right)d \left(\! \begin{array}{c} W_t^1\\ W_t^2 \end{...

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This is not a complete answer, just a correction for the case of Poisson process. Clearly, given $N(t) = n$, $$K(t) = \sum_{k=1}^n (t-\tau_k),$$ where $\tau_k$ is the $k$th jump time. It is well known that, given $N(t) = n$, the jump times $\tau_1,\dots,\tau_n$ are distributed as order statistics $U_{(1)},\dots, U_{(n)}$ of an iid $U[0,t]$ sample $U_1,\... -1 Finally, I found it wrong. The correct way is dB_x = mvnrnd(zeros(K, 1), Omega_xx/n, n); % n-by-m matrix B_x = cumsum(dB_x,1); % n-by-m matrix B_x = [zeros(1,K); B_x]; J_c = exp(C*[1:n]/n)'.*cumsum(exp(-C*[1:n]/n).*[zeros(K,1) dB_x(1:n-1,:)'])'; dJ_c=[J_c(1,:); diff(J_c)]; % m-by-n matrix int_J = J_c'*dJ_c; 0 Define$X_t = -\int_0^t\beta_s\,dW_s$and$Y_t = -\frac{1}{2}\int_0^t\beta_s^2\,ds$. Then$Z_t = e^{X_t+Y_t}$. Even though you did not mention it I am guessing that there is a condition on$(\beta_s)_{s\geq 0}$that makes$X$a local martingale. Since Brownian motion has continuous paths,$X$is continuous as well, which implies$Z$is continuous as well. ... 3 Check out Girsanov's paper from 1962 (here or here) for the proof of why his counterexample works; he gives the counterexample: $$dX_t = \frac{|X_t|^{\alpha}}{1+|X_t|^{\alpha}}dB_t, \quad 0\le \alpha < \frac{1}{2}$$ where$B_t$is Brownian motion/standard Wiener process. 1 By the stationarity of the increments, this is equivalent to $$\min_{s \in [0,t]} \mathbb{P}(|X_s| \leq \epsilon)>0$$ for all$t>0$. Without any additional assumptions this does, in general, not hold true. Consider for example$X_t := t$(i.e. a deterministic drift process), then $$\mathbb{P}(|X_t| \leq \epsilon) = 0$$ for all$t>\epsilon$. ... 1 You'll want to have a look at "A Signed Measure on Path Space Related to Wiener Measure" by K. Hochberg http://www.jstor.org/stable/2243147 Hocberg constructs a Markovian signed measure (of unbounded variation) on the space of continuous paths, associated with the operator$Lu:={\partial^4u\over\partial x^4}$. The transition density$p$is the fundamental ... 0 Since$k>1$, your inequality won't give you anything, at least for positive$C$. By telescoping product of your inequality,$\frac{S_{t_2}}{S_{t_1}}\le k^\infty\frac{C_{t_2}}{C_{t_1}}=\infty,\,\forall t_1<t_2$. 0 You're right, this is no hard, although is quite tricky. The only thing you need is self-similarity. First note that$x^qa^{-q} = q\cdot\sup_{\lambda>0}(\lambda x - p^{-1}\lambda^p a^p)$. Then $$\sup_{t\ge 0} \left(\frac{X_t}{1+t^{p/2}}\right)^q = q\cdot\sup_{t\ge 0}\sup_{\lambda>0}\left(\lambda X_t - p^{-1}\lambda^p (1+t^{p/2})^p\right)\\ = \sup_{\... 2 As it was hinted in the comments, the result is not true unless you assume uniform convergence somewhere. Here is a simple example. Let X_n = X have non-degenerate Bernoulli distribution and Y_{un} = X \mathbf{1}_{u\ge n}. Then P(Y_{un}\neq X_n)\to 0, u\to\infty, and e^{itY_{un}} \to 1, n\to\infty, in all senses. However, e^{itX_n} = e^{itX}\... 3 See https://en.wikipedia.org/wiki/Renewal_theory#The_elementary_renewal_theorem The elementary renewal theorem 0 For all random variables U,V and constants a,b: Expectation is Linear :$$\mathsf E(aU+bV) ~=~ a\,\mathsf E(U)+b\,\mathsf E(V)$$Variance is not linear, but we have the rule that :$$\mathsf {Var}(aU+bV) ~=~ a^2\,\mathsf {Var}(U)+b^2\,\mathsf {Var}(V)-2ab\,\mathsf {Cov}(U,V)$$However, when U,V are uncorrelated :$$\mathsf {Var}(aU+bV)~=~ a^2\,\... -2 A shorter answer: The trick is remove the denominator$(1-t)$, by defining$X_t = Y_t(1-t)$, then $$dX_t = (1-t)dY_t - dY_t = -dY_t + dW_t \implies dY_t = \frac{1}{1-t} dW_t,$$ hence $$X_t = \int_0^t \frac{1}{1-u}dW_u.$$ 1 The notation seems to change somewhat between arxiv versions and journal version. My guess is that the "1/2" is a typo. I agree that example should have been more clearly explained. The idea is to compare to an optimal randomized stationary algorithm, which is one that makes decisions without using queue backlog. It would have been nice for the paper to ... 0 Example: An excursion of a Brownian motion$(B_t: t\ge 0)$is a piece$(B_t: a\le t\le b)$of its path such that$B_a=B_b=0$but$B_t\not=0$for all$t\in(a,b)$. Let us use the left endpoint of the time interval of a Brownian excursion as a label for the excursion. The collection of these left endpoints is a point process on$(0,\infty)$comprising ... 2 It seems that when we talk about Poisson random measure, we often assume that the characteristic measure$\nu$is$\sigma$-finite. Then the random measure$\tilde{N}(t,dx)$is also$\sigma$-finite. Of course,$\tilde{N}(t,A)$is not a martingale if$\nu(A)=\infty$. To define the stochastic integral $$\int^T_0 f(t,x)\tilde{N}(dt,A),$$we first consider the ... 1 If$X_i$s are independent and you have already their standard deviations, say$\sigma_i$, for$X_i$, which seem to be the same, then the variance of each$X_i$, for$i=1,\cdots, n$is$\sigma^2_i$, and so the variance of$X$is equal to$\sum_{i=1}^n\sigma^2_i=n\times \sigma^2_i$. As a result, the standard deviation of$X$is$\sqrt{n\times \sigma^2_i}=\...

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You have to prove the convergence of the integral $$\int_{-\infty}^{+\infty}\left|e^{i\xi \sqrt tu}\right|e^{-u^2/2}\mathrm du$$ for each complex number $\xi$. Since we can write $\xi=a+ib$ where $a$ and $b$ are real numbers, it suffices to prove the convergence of the integral $$\int_{-\infty}^{+\infty}e^{b\sqrt tu}e^{-u^2/2}\mathrm du$$ for each real ...

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This answer comes from the lecture notes by David Nualart, contained in ch. 1 of "Stochastic Equations for Complex Systems: Theoretical and Computational Topics", Springer, 2015. The Malliavan derivative is defined in the following way. Let $W$ be defined as the canonical process on the probability space $(\Omega, \mathcal{F}, P)$ where $\Omega = C([0, T])$ ...

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The derivation is rather lengthy, and is found in many textbooks. Here's an sketch: $$S(\omega )=\sum_{k=-\infty}^\infty R_k \, e^{-i\omega k } \tag{1}$$ relates the spectral density of an stationary signal with its second order statistics. We assume $R(k)$ is fixed for $|k|\le m$, and we want to find $R(k)$ for $|k|> m$ so that the entropy $H$ is ...

0

The equation you are solving is not the equation that you want to solve. The FPE for the one dimensional autonomous Ito SDE $$dX=b(X) dt + \sigma(X) dW$$ is $$\frac{\partial P}{\partial t}=-\frac{\partial}{\partial x}(b(x) P(x,t))+\frac{\partial^2}{\partial x^2} \left ( \frac{\sigma^2}{2} P(x,t) \right ).$$ In your case, taking $D=1$, $b(x)=-x,\sigma(x)=\... 0 It has been a while that I posted this question. I unfortunately did not quite understand the proof on a second visit to the same question. Below is my proof. Is this right? The proof uses the fact that the indexing set is$\mathbb{R}$. We will use also the fact that the infimum of a set is the same as the infimum of its closure and the fact that$X$is ... 2 The key point is the following lemma which has nothing to do with probability theory, but is just elementary "deterministic" calculus. Lemma: Let$f: [0,\infty) \to \mathbb{R}$be a càdlàg function and fix$M>0$and$0 \leq T_0<T_1<\infty$. Then the following statements are equivalent: There exists$t \in (T_0,T_1]$such that$|\Delta f(...

1

Note that $\sup_t E[|X_{t\wedge n}; |X_{t\wedge n}|\ge M]=\sup_{t\in [0,n]} E[|X_{t}|; |X_{t}|\ge M]<\infty$ (left-limits existing implies bounded on compact intervals [need to verify; know it's true if left and right limits exist]). Your reasoning doesn't work. If each sample path $[0,n] \ni t \mapsto X_t(\omega)$, this does not imply $$\sup_{t \in [0,... 2 Another way of looking at it is : Any email sent in the interval has an independent probability p=\dfrac{2\lambda_A}{2\lambda_A+\lambda_B} of being one sent by Alice, because Alice sends emails at a constant average rate per unit-interval of \lambda_A over two unit-intervals, and Bob sends them at rate per unit-interval of \lambda_B over just the ... 0 Each voter who turns up in {10} hours has a probability of \frac{4}{10} of being in the first 4 hours and independently a probability of voting for the first candidate of \frac12, so a probability of \frac15 of both. So, given 1000 voters in total, you have a binomial distribution A_4 \sim Bin(1000,\frac15), i.e.$$P(A_4=k\mid N_{10}=1000)={...

1

Suppose $f:\mathbb R \to [0,\infty)$ is even, decreasing on $[0,\infty),$ with $\int_{\mathbb R} f <\infty.$ Let $g$ be any translate of $f$ and let $c_i\in [(i-1)h,ih]$ for each $i\in \mathbb Z.$ Then $$\tag 1 \int_{\mathbb R} g - 3f(0)h \le \sum_{i\in \mathbb Z} g(c_i)h \le \int_{\mathbb R} g + 3f(0)h.$$ This result answers your questions 1. and 2. in ...

3

We can compute the covariance between $W_t$ and $I_t$ as follows: \begin{align*} covar(W_t, I_t) &=E\left(W_t \int_0^t W_s ds \right)\\ &= E\left( \int_0^t W_t W_s ds \right)\\ &= \int_0^t E\left(W_t W_s \right) ds\\ &= \int_0^t \min(t, s)\, ds\\ &=\frac{t^2}{2}. \end{align*} Alternaticely, note that \begin{align*} d(tW_t) = W_t dt + t ...

0

Hint: If $c_nY_n(\omega)\to0$ and $\frac{c_n}{c_{n+m}}\to c\in\mathbb R$ then $$c_nY_{n+m}(\omega)=\frac{c_n}{c_{n+m}}c_{n+m}Y_{n+m}(\omega)\to c\times0=0$$

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As suggested by you, I put my comment as an answer. I suggest you have a look at lemma 3 (the iff part of the assertion) here . Best regards.

1

xivaxy, I don't think that's correct. Suppose $X_n = X_{n-1}$ with probability .5, and $X_n = 0$ with probability .5. We set $X_0 \ne 0$. Let $N$ be the first $n$ such that $X_n = 0$. $N$ is a stopping time because $1_{\{N=n\}}$ is a function of $(X_1,...,X_n)$ only. Then we have $\mathbb{E}[X_n 1_{\{N \ge n\}}] = \mathbb{E}[X_{n}]$ because $X_n \ne 0$ ...

3

The goal is to find a formula for the $n$-fold composition of $f$ with itself. This is difficult to do directly from the definition, so the author uses a clever trick. The idea is that $f$ is a fractional linear transformation, and for every fractional linear transformation there is a corresponding matrix. Explicitly, if $T(z)=\frac{az+b}{cz+d}$ then the ...

0

Suppose $\{X_{n,k}\}$ are i.i.d. with mean $\lambda^{-1}<\infty$ and $\{\gamma_n\}$ are i.i.d. with $\mathsf{Geo}(p)$ distribution. Then the sequence $$Y_n := \sum_{j=1}^{\gamma_n} X_{n,j}$$ is i.i.d. and thus defines a renewal sequence with mean $$\mathbb E[Y_1] = \mathbb E[X_{1,1}]\mathbb E[\gamma_1]=(\lambda p)^{-1}$$ (by Wald's identity), or ...

1

Let $$X_1(t)=B_1(t)$$$$X_2(t)=\rho B_1(t)+\sqrt{1-\rho^2}B_2(t)$$ where $B_1(t)$ and $B_2(t)$ are independent. we have $$\operatorname{cov}\left( {{X}_{t}}^{1}-X_{s}^{1}\,,\,{{X}_{2}}^{t} \right)=\operatorname{cov}\left( {{B}_{1}}(t)-{{B}_{1}}(s),\,\rho {{B}_{1}}(t)+\sqrt{1-{{\rho }^{2}}}{{B}_{2}}(t) \right)={{\rho }}t-\rho \min \{t,s\}$$ then $${{\rho }^{*}... 0 I would work with more equalities rather than inequalities. Beside that for the second and third constraint we have to sum up the amounts to be sent. \sum_{j=1}^R t_{ij}=x_i \ \ \forall \ i \in \{1,\ldots, P \} \sum_{i=1}^P t_{ij}=d_j \ \ \forall \ j \in \{1,\ldots, R \} -1 [Drawing out a simple case, e.g. two circles of length 2 might help visualising the problem (their product becomes a "cross").] So as OP suggests, for irreducible Z we need \forall i, i', j, j' \exists n : P_{(i,j)(i',j')}^{(n)}>0, i.e. there is some n for which the n-step transition probability from (i,j) to (i',j') is nonzero (for all ... 1 For 5.: The key is this: A previsible martingale is constant in time. In more detail, if X admits a second decomposition X_n=X_0+M'_n+A'_n then by subtracting you find that M_n-M'_n(=A'_n-A_n) is both a martingale and previsible, hence constant in time a.s., whence M_n-M'_n=M_0-M'_0=0-0=0 for all n, a.s. 1 This is discrete time, so right continuity is not an issue. It is simply the fact that if P[D_n]=1 for all n then P\left[\cap_n D_n\right]=1 (and conversely). 1 Looks not that complicated, if I understood the statements correctly. For simplicity let me set x=0, y_0=0. i\geq 0\,\, \sum_{i\geq 0} p(h,0,\xi _i)h\leq \frac{h^\frac{1}{2}}{2\pi}+ \sum_{i\geq 0} q(h,0,\xi_i)h\leq\frac{h^\frac{1}{2}}{2\pi}+\int _{y\geq0}p(h,0,y)dy\leq \frac{h^\frac{1}{2}}{2\pi}+\frac{1}{2}\,\,\, where\,\, q(h,0,\xi_i)=inf\{p(h,0,y)|y\... 2 We may assume y_k=kh (k\in{\mathbb Z}) and 0\leq x\leq h. Then it is easy to see that$$|x-x_k|\geq\bigl(|k|-1\bigr)h\qquad(k\ne0)\ .$$It follows that$$p(h,x,x_k)={1\over 2\pi\sqrt{h}}e^{-(x-x_k)^2/(2h)}\leq{1\over 2\pi\sqrt{h}}e^{-(|k|-1)^2h/2}\qquad(k\ne0)\ .$$Summing over k\in{\mathbb Z} and taking proper care of k=0 we therefore obtain$$\...

1

(I've revised my answer to account for the authors' use of the Dirac delta function, which I had misinterpreted as a Kronecker delta function. This is sometimes called a delta-correlated random process.) The authors appear to apply the following simulation procedure (for heuristics, see here): If $\{X(z)\}_{z\in\mathbb{R}}$ is a continuous Gaussian ...

1

It follows from the continuity of the sample paths and the definition of $\tau_a$ that $|M_{t \wedge \tau_a}| \leq a$ for all $t \geq 0$. This means that $M^{\tau_a}$ is, indeed, a bounded process. Moreover, the optional stopping theorem implies that it is a local martingale. By the definition of the quadratic variation, we know that $(N_s^2 - \langle N \... 3 Unfortunately, I have never really found an explanation for basic stochastic calculus which is both mathematically accessible and technically correct. I will give the usual mathematical treatment, but I concede that it is not particularly accessible.$W$is a process on$[0,\infty)$with the following basic properties:$W(0)=0W$is a Gaussian process.$...

1

Although Brownian paths are not differentiable point wise, we may interpret their time derivative in a distributional sense to get a generalized stochastic process called white noise. We denote it by $$\eta (t,\omega )=\overset{\centerdot }{\mathop{B}}\,(t,\omega )$$ We also use the notation $$d\eta=dB_t$$ The term white noise arises from the spectral ...

0

I hope this is right! If anyone happens along later, please let me know if I'm on the right track. References: Da Prato, et. al., "Stochastic Equations in Infinite Dimensions" Carmona, et. al., "Interest Rate Models: an Infinite Dimensional Stochastic Analysis Perspective" The above expression can be rewritten (assuming $G_t(x)$ admits a density) as $$\... 0 https://arxiv.org/abs/0804.4689 Introduction to Potential Theory via Applications Christian Kuehn (Submitted on 29 Apr 2008) We introduce the basic concepts related to subharmonic functions and potentials, mainly for the case of the complex plane and prove the Riesz decomposition theorem. Beyond the elementary facts of the theory we deviate slightly from ... 1 Your argument is that s(t-T) and s(T) have identical distributions due to periodicity and symmetry. This is okay. However that then gives you:$$\begin{align}\mathsf P(X(t)\leq x) ~=&~ \mathsf P(s(t-T)\leq x) \\[1ex] ~=&~ \mathsf P(s(T)\leq x)\\[1ex] ~=&~ \int_{-1}^x f_{s(T)}(z)\operatorname d z~\mathbf 1_{x\in(-1;1)}+\mathbf ...

2

Since the process is Gaussian and one-dimensional, its probability density function (PDF) must be $$f_X(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\tag{1}$$ where $\mu$ is the mean, and $\sigma^2$ is the variance. From the problem statement we know that $\mu=0$, so we only need to find $\sigma^2$. From the definition of the auto-...

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HINT Markov process has future independent of its past. Formally if $\mathcal{F}_{s \le t}$ is the sigma-algebra generated by $(X_s)_{s \le t}$ then $$\mathbb{P}\left[X_{t+\epsilon} \in A| \mathcal{F}_{s \le t}\right] = \mathbb{P}\left[X_{t+\epsilon} \in A| \mathcal{F}_t \right]$$ for any event $A$. Can you apply the definitions of the Wiener process to ...

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