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Your solution to (b) looks fine. You can solve (a) in a similar fashion, but it may be easier to first note that $X_n$ is a function of $S_0, \dots, S_n$ (write $Y_k = S_k - S_{k-1}$). Then to show it is a martingale, it is equivalent to show that $\mathbb{E}[X_{n+1} - X_n \mid S_0, \dots, S_n] = 0$, which saves you from having to work with the sum. Each ...

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Hint: By flipping the sign of $X_j$, all $j=1,2,\dots,n$, we have $\mathbb{P}(S_n=s\mid S_n^2=s^2)=\mathbb{P}(S_n=-s\mid S_n=s^2)$ for all $s$, so ...

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Let $X_t$ denote the number of parrots at time $t$ and let $Y_t$ denote the number of crows at time $t$. \begin{aligned}P\left(X_{t}=k\mid X_{t}+Y_{t}=n\right) & =\frac{P\left(X_{t}=k\wedge Y_{t}=n-k\right)}{P\left(X_{t}+Y_{t}=n\right)}\\ & =\frac{P\left(X_{t}=k\right)P\left(Y_{t}=n-k\right)}{P\left(X_{t}+Y_{t}=n\right)}\\ & =\frac{e^{-\... 2 Let \{t_1, t_2, t_3, ...\} be arrival times of a Poisson process of rate \lambda>0. Let X(t) be a real-valued random process that possibly depends (causally) on the Poisson process. Assume that \overline{X} is a real number such that: \lim_{T\rightarrow\infty} \frac{1}{T}\int_0^T X(t)dt = \overline{X} \quad (\mbox{ with prob 1}) $$The PASTA ... 1 According to "A first course in Stochastic Processes" by Samuel Karlin and Howard M. Taylor, this process is called "Multi-Type Branching Process". The extinction criterion for it is: \lim\limits_{n \to \infty} P(v_n = 0) = 1 iff the absolute value of all eigenvalues of EA does not exceed 1. 1 No you can't. Let Y,Z be independent copies of the two state symmetric Markov chain starting with the invariant distribution. Let X=1_{Y=Z}-1_{Y\neq Z}. It is easy to check$$ \mathbb{E}[X\mid Y]=\mathbb{E}[X\mid Z]=0, $$but of course X\neq 0. 1 Independence is not necessary for the weak law of large numbers. For a sequence of correlated random variables X_i, if the covariance r(j) = Cov(X_k,X_{k+j}) goes to zero as j \to \infty, then the weak law of large numbers still holds. To see this, suppose X_1, X_2, \dots are a sequence of correlated random variables such that r(j)=Cov(X_k,X_{k+j}... 1 We use the strong Markov property to remove the condition on \mathcal{F}^W_{\tau_a} and say that X(t-\tau_a) is just another Brownian motion, hence$$ \mathbb{P}(X(t-\tau_a)<0\mid\mathcal{F}^W_{\tau_a})=\frac12 $$and we conclude$$ \mathbb{E}\left[\chi_{\sup_{s\in[0,t]} W(s)\geq a}\mathbb{P}(X(t-\tau_a)<0\mid\mathcal{F}^W_{\tau_a})\right]= \mathbb{...

1

If you take $f(x) = x$, you see that $W_t$ is a continuous local martingale. Then, taking $f(x) = x^2$ gives us that $W_t^2 - W_0^2 - t$ is also a continuous local martingale so that $W_t$ has quadratic variation at time $t$ equal to $t$. Now apply Levy's characterisation of Brownian motion to conclude.

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No. You'll have to make a decision about those missing entries in the covariance matrix, and by extension the sort of function you want out. Choose them all zero and you'll get a function that is continuous nowhere. Choose something like the 'Orstein-Uhlenbeck exponential kernel' (referenced in the comments in your link) and you'll get something much more ...

3

$W(1)$ is a random variable and you cannot just suppose that it is equal to $a$. In fact, you know its distribution. If $W$ is a standard Brownian motion then $W(t) \sim \mathcal{N}(0,t)$ for every $t$. In particular, $W(1) \sim \mathcal{N}(0,1)$ which is why $\mathbb{E}[W(1)] = 0$.

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Allow me to rewrite the PDE in question in my own notation. We are seeking to solve: $$\frac{\partial }{\partial t} P(x,t) = -\frac{\partial}{\partial x}\left[\left(\sin(k x) + F \right) P(x,t)\right] + \frac{\partial ^2}{\partial x^2} \left[D \cdot P(x,t)\right]$$ where $D$, $F$ and $k$ are constants. By using separation of ...

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The hypothesis can be written as $Ee^{-tX}=e^{-\sqrt {2t} b}$ for all $t>0$. Differentiating w.r.t. $t$ (from the right) and setting $t=0$ we get $EX=\infty$.

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No! You need much more than just $W_t-W_{k-1}\sim N(0,t-k+1)$. You want to show that $W_{t+k}-W_k$ is another Wiener process, i.e., you want to show it is almost surely $=0$ at $t=0$, independent Gaussian increments, and continuous paths with probability 1 These follows from $W_t$ being a Wiener process (why?).

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Feller obtains in his book An Introduction to Probability Theory. Vol II (p.342) the following result (... unfortunately he does not give a detailed proof): Let $(B_t^x)_{t \geq 0}$ be a Brownian motion started at $x \in \mathbb{R}^d$ (i.e. $B_t^x = x+B_t$ where $(B_t)_{t \geq 0}$ is a standard Brownian motion). For fixed $a>0$ define a stopping time $\... 2 If you do not know the graph, then you need to think about something general. One piece of useful information that you do have is the degree of each vertex. Intuitively, it sound plausible that the invariant distribution is related to the degree of each vertex. If you draw some simple graphs, then you can see that this is indeed the case and that the ... 0 Finally, I think that I proved that$Y$is adapted. We define$\tau_k$as the time of the$k$-th jump ($\tau_0 := 0$). This is equivalent to the recursive definition $$\tau_{k+1} := \inf \{ t \in (\tau_k;T] | X_t \ne X_{\tau_k} \}.$$ To prove that$\tau_{k+1}is a stopping time, note that $$\{ \tau_{k+1} > t \} = \{ \tau_k \ge t \} \cup ( \{ \tau_k &... 1 They are not that complicatedly nested. You just need the following result: Let T be a stopping time, r be a \mathcal F_T-measurable positive random variable and$$ S=\inf\{t>T\mid d(B_t,B_T)=r\}. Then S is a stopping time. Proof: \begin{align*} \{S\le t\}&=\{T<t\}\cap\left(\bigcup_{u\in]T,t]}\{d(B_u,B_T)=r\}\right)\\ &=\{T<t\... 0 This statement is true. For any q \in \mathbb{Q}, there exists a null set N_q such that\lim_{n \to \infty}\sum_{s \in \pi_n}(B_{s'\wedge t}(\omega)-B_{s\wedge t}(\omega))=t$$for every \omega \in \Omega \setminus N_q. Define$$\Omega'=\bigcap_{q \in \mathbb{Q}}\Omega\setminus N_q.$$We have$$P(\Omega')=1.$$Let \omega \in \Omega'. Then for ... 5 Write \| \mu - \nu \|_{TV} := \frac{1}{2}\sum_{i \in S} |\mu(i) - \nu(i)| for the total-variation distance between two measures \mu and \nu on the countable state space S. We know the following facts: Fact. Let X, Y be RVs with marginal distributions \mu and \nu under \mathbf{P}, respectively. Then$$ \| \mu - \nu \|_{TV} \leq \mathbf{P}(... 4 Approach I (via characteristic functions): The identity $$E(1_A g(B_{t_1},\ldots,B_{t_n})) = P(A) E(g(B_{t_1},\ldots,B_{t_n})) \tag{1}$$ can be easily extended to complex-valued continuous functions (just writeg= \text{Re} g + i \, \text{Im g}$and apply$(1)$separately to the real and imaginary part of$g). Choosing $$g(x_1,\ldots,x_n) := \exp \left( i ... 1 Please feel free to amend the solution to improve it or correct some mistakes or you can even propose a new one of your own. Solution to Exercise 1.19 : 1 . Almost sure continuity results from the composition of the continuous mapping <x,.> with the a.s. continuous trajectories of X_t(\omega). So we only have to prove that increments over the ... 0 Take m=2. Let \mathcal F_t be the natural filtration and consider a variable X such that$$F_{X,B_{1,1},B_{2,1}}(x,y,z) = 2F(x)F(y)F(z) \text{ if } xyz>0, 0 \text{ otherwise}$$where F stands for the standard normal cdf, and this definition is extended by "independence" (i.e. you build the rest of your Brownian motions by drawing mutually ... 1 The fundamental matrix of an ergodic Markov chain is$$Z = \left(I-P+W\right)^{-1}$$where P is the transition matrix and W is a matrix where each row is the fixed probability vector w = (w_i) (the limiting distribution). You can calculate the mean first passage times m_{ij} (expected number of steps to reach state j when starting from the state ... 0 You can't draw that conclusion from the model. Suppose there are 4 men and 4 women each divided into two groups of 2. Pair the groups and have each mixed sex group of 4 shake hands all around their group. Then everyone shakes twice so the average is 2 but hlaf the people are clearly disease free. Of course that distribution of handshakes is not ... 2 This can be expressed as: 1. go from state 1 to 2, and stay there for at least N-1 units of time. ie we have: p_{12}p_{22}^{N-1} 2 You formula is not correct. The answer should be P_{12} P_{22}^{N-1}, the first step is move from state 1 to state 2, then the chain has to stay in state 2 for the next N-1 steps. 0 After doing some digging around, I think I have found the answer. I am posting it here so that it might help somebody who may have the same question. As written in An Introduction to Computational Stochastic PDEs (page 314) and Oksendal's Stochastic Differential Equations: An Introduction with Applications, 6e (page 65), a White Noise process in continuous ... 0 Your notation (x,y)=\{(2,0),(0,2),(1,1)\} is fine, but P(E_3|E_1E_2) = \frac{1}{3} is not true. Let the two aces be A\clubsuit and A\spadesuit. Case 1: (x,y)=(2,0). There is only one way to have 2 aces in E_3 (both A\clubsuit and A\spadesuit) and {24\choose 11} ways to have non-aces in E_3, while overall there are {26\choose 13} ... 0 Another way to compute P(E_3\mid E_1,E_2) (from your first attempt) is to consider the third and fourth piles as an array of 26 locations where cards can be put, one card in each location. (Top of the third pile, card next to the top of the third pile, second card from the top of the third pile, etc.) There are two aces that can be put in two of these ... 1 The probability for selecting 1 from 2 aces and 12 from the remaining 24 cards when selecting 13 from 26 cards is:$$\begin{align}\mathsf P(E_3\mid E_1,E_2)&=\left.\dbinom{2}1\dbinom{24}{12}\middle/\dbinom{26}{13}\right.\\[1ex] &=\dfrac{13}{25}\end{align}$Let$(x,y)= (\text{number of ace in the third pile}, \text{number of ace in the fourth ...

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[1] $E[Y_kY_i]=E[E[Y_kY_i|Y_i]]=E[Y_iE[Y_k|Y_i]]=E[Y^2_i]$ changing $k \rightarrow j$ and subtructing we have the first equality. [2] $E[(Y_k-Y_j)^2|F_i]=E[Y_k^2|F_i]+E[Y_j^2|F_i]-2E[Y_kY_j|F_i]$ but now: $E[Y_kY_j|F_i]=E[E[Y_kY_j|F_j]|F_i]$ $=E[Y_jE[Y_k|F_j]|F_i]=E[Y^2_j|F_i]$ This way we have the second equality. Use has been made as suggested ...

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The answer above is circular to me (since the proof of martingale convergence that I know goes through the upcrossing inequality), so I will try to give a different answer (I will be paraphrasing Durrett PTE version 5 Theorem 4.2.10 and 4.2.11, so feel free to read those up; I think they are very accessible). The intuition is the following: since a ...

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