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We have $\forall t \geq 0$, \begin{align} M_t &= \int_0^t \int_s^t f(u) dW_u ds \\ &=\int_0^t \left(\int_s^t f(u)dW_u+\int_0^s f(u)dW_u-\int_0^s f(u)dW_u\right) ds \\ &=\int_0^t \left(\int_0^t f(u)dW_u\right)ds-\int_0^t\left(\int_0^s f(u)dW_u\right)ds \tag{By linearity} \\ \end{align} Denote here $I_t= \int_0^tf(u)dW_u$, we have then \begin{...

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Remember that by the definition of the limit supremum, we have $$\limsup_{t\to\infty} f(t) \leq M \iff \left(\forall \epsilon > 0, \exists T : t > T \implies f(t) \leq M + \epsilon\right)$$ In particular, note that we cannot conclude $f(t) \leq M$ for any $t.$ (e.g. note that $\limsup_{n\to\infty} 1 + \frac{1}{n} \leq 1$) In your context, this means ...

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Assuming you are talking about probability measures on the reals, and convolution with respect to $(\mathbb R, +)$, the answer is No. A convolution semigroup need not be continuous anywhere. Let $f$ be any discontinuous solution of Cauchy's functional equation $f(x+y)=f(x)+f(y)$ (these are discontinuous everywhere, and unbounded in every neighborhood) and ...

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For $i\in\{1,\dots,6\}$, let $m_i = E(T_B | Z_t = i)$ be the expected number of steps until reaching $\{3,4,5\}$, starting from state $i$. Trivially, $m_3=m_4=m_5=0$. To obtain a system of linear constraints, apply first-step analysis (conditioning on the first step out of state $i$). For $i=1$, we have \begin{align} m_1 &= \sum_{j\in\{2,6\}} E(T_B | ...

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Taking two steps in the Markov chain can lead to one of two things, with equal probability: $1 \to 2 \to 3$ or $1 \to 6 \to 5$ and we're done. $1 \to 2 \to 1$ or $1 \to 6 \to 1$ and we're back where we started. We took $2$ steps. In the first case, we have $0$ steps left, and in the second case, we have $\mathbb E[T_B]$ more steps left in expectation. ...

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You can actually compute the integral with the formula $d(tY_t) = Y_t \, dt + t \, dY_t$. Fixing $Y_t = B_t^2$, you have by Itô formula that $dY_t = 2B_t dB_t + dt$ and therefore \begin{align} \int_0^1 B_t^2 d t &= \int_0^1 d(tB_t^2) - \int_0^1 td(B_t^2)\\ &= B_1^2 - \int_0^12tB_tdB_t - \int_0^1tdt\\\ &= B_1^2 - \frac12 - \int_0^12tB_tdB_t. \end{...

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You have the right mean and variance for $$\int_0^1 B_t^2 \; dt \; ?$$ But, its distribution is not of any known functions. In fact, it is rather complex. You may find some research results in the link below. https://projecteuclid.org/download/pdf_1/euclid.aop/1020107767

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This question has been asked before in this forum. My answer in Determining distribution of $X_t = \int_0^t W_s^2 \mathrm{d} s$ gives a couple of references to the original solution of Cameron & Martin, and a different approach of M. Kac.

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Well, in these sorts of things there's not really much to do except try to exploit the independence of the increments. So here, $W_{t+2}=W_{t}+W_{t+2}-W_{t}$ and $W_{t+1}=W_t+W_{t+1}-W_t$. Thus, we get $$\mathbb{E}W_t(W_{t+1}+W_{t+2})=2\mathbb{E} W_t^2+\mathbb{E}W_t(W_{t+2}-W_{t})+\mathbb{E}W_t(W_{t+1}-W_t)$$ And by independence and Gaussianity, we thus ...

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The step from finite case to infinite case is not so obvious and requires the very famous Kolmogorov extension theorem.

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As an alternative approach in the spirit of your initial attempt, you can condition on the number $2k$ of steps to reach $\{3,4,5\}$ from $1$: \begin{align} E(T_B) &= \sum_{k=1}^\infty 2k\ P(\text{$2k$ steps}) \\ &= \sum_{k=1}^\infty 2k\ 2^k \left(\frac{1}{2}\right)^{2k} \\ &= \sum_{k=1}^\infty k \left(\frac{1}{2}\right)^{k-1} \\ &= \frac{1}{...

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Hints: $\ S_n\$ is odd when $\ n\$ is odd, and even when $\ n\$ is even. If $\ S_n = m\$ then $\ \frac{m+n}{2}\$ of the first $\ n\$ steps must have been to the right and $\ \frac{n-m}{2}\$ of them must have been to the left. The number of rightward steps among the first $\ n\$ follows a binomial distribution with parameters $\ n\$ and $\frac{1}{2}\... 1 Your sollution is overly complicated. The easy path to finding the probability generating function of the product of independent random variables is to use the definition. $$\mathsf G_X(t)=\mathsf E(t^X)$$ Let$Y$be the random variable resulting in$1$if the coin toss is head and$2$if it is tails. So$Z=YX$and$Y\sim\mathcal U\{1,2\}$. $$\begin{... 1 I actually found a way to solve my problem. The rows of the limiting transition matrix are linear combinations of the stationary distributions, \pi_k, k=1,...,K of the K strongly connected components that correspond to a leaf of the condensation graph (i.e. the absorbing states). The coefficients associated with each \pi_k for each row are given by ... 1 This is a birth-death process, so we may instead use the detailed balance equations$$ \lambda(1-\mu)\pi_{n-1} = (1-\lambda)\mu \pi_n,\quad n\geqslant 1. $$This yields the recurrence$$ \pi_n = \left(\frac{\lambda(1-\mu)}{\mu(1-\lambda)}\right)^n\pi_0,\quad n\geqslant 0. $$Now, assuming that \frac{\lambda(1-\mu)}{\mu(1-\lambda)}<1, we have convergence ... 1 Let S be the set of all the states of the Markov chain. Consider a matrix A and B such that A=P^{r-1}=(a_{ij})_{i,j\in S} and B=P^r=(b_{ij})_{i,j\in S} respectively, where P=(p_{ij})_{i,j\in S} is the transition matrix of the Markov Chain. Then, P^{r+1} = BP = \Biggl($$\sum_{k} b_{ik}p_{kj} \Biggr)_{i,j\in S}$ As per the question, \$...

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