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2

This answer is first adapted from comments in: does convergence in $L^p$ imply convergence almost everywhere. Consider a probability space $[0, 1]$ with uniform measure. Define the random variable $X_{2^i + k}$ by $$X_{2^i + k} = 2^{i/2}\chi_{\left[\frac{k}{2^i}, \frac{k+1/2}{2^i}\right]} - 2^{i/2}\chi_{\left[\frac{k + 1/2}{2^i}, \frac{k+1}{2^i}\right]}$$ ...

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No, independence does not follow merely from this observations. This fact gives you only pairwise independence and this is not sufficient to assure independence of events. (See https://notesofastatisticswatcher.wordpress.com/2012/01/02/pairwise-independence-does-not-imply-mutual-independence/) We might think that \begin{align} \Delta_1 &\perp X_s \\ ...

3

$\nu \circ f^{-1}$ denotes the push-forward measure (or image measure) of $f$ with respect to $\nu$, i.e. $$\nu_1(B) = (\nu \circ f^{-1})(B) = \int 1_B(f(x)) \, \nu(dx)$$ for any Borel set $B$. This implies $$\int g(x) \, \nu_1(dx) = \int g(f(x)) \, \nu(dx)$$ for any $g \in L^1(\nu_1)$.

1

Some related summary comments on ergodic Markov chains. (There are several fleas hopping around triangles elsewhere on this site, each according to the same transition matrix, but with varying questions about the associated process; search 'Markov flea vertex'.) With help from @Did, essentially by manipulating difference equations, I believe you have ...

2

To help hone your intuition, it might help to think about the following simple example, which shows that the Blumenthal law is using much more than just the continuity of Brownian motion. Let $Z$ be a random variable with $P(Z=1) = P(Z=-1) = 1/2$, i.e. a coin flip. Set $X_t = tZ$, so that the process $X_t$ just moves with constant speed $\pm 1$ depending ...

3

Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it. If we define $$f_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$ then $f_n \uparrow 1_{(0,\infty)}$. Obviously, $f_n$ is not differentiable at $x=0$ at $x=1/n$, but for the moment we forget about this (that's the non-rigorous part) and apply ...

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Since Poisson process is a process with independent increments, question 2) is equivalent to question: What is the probability that time between 1st and 100th arrival exceeds 10 days? (As Did correctly said in his comment.) Solving this equivalent question is quite simple: Since the number of arrivals in interval (0; 10] follows a Poisson distribution, ...

1

Since $P$ is symmetric, we know from the spectral theorem that $P$ has real eigenvalues and is diagonalizable. Since $P$ is a doubly stochastic matrix, the stationary distribution is uniform, i.e. $$\pi=\left(\frac16,\frac16,\frac16,\frac16,\frac16,\frac16\right).$$ Since $P$ has a stationary distribution, we know that $1$ is an eigenvalue. In particular, ...

3

Set $$M_t := f(X_t) - \int_0^t Af(X_s) \, ds$$ and write $P_t f(x) := \mathbb{E}^x f(X_t)$ for the semigroup associated with $(X_t)_{t \geq 0}$. Using the definition of the generator, it is not difficult to see that $$\frac{d}{dt} (\mathbb{E}^x f(X_t)) = \frac{d}{dt} P_t f(x) = P_t Af(x).$$ By the fundamental theorem of calculus, this implies P_t ... 0 The major application is in Communication Systems,Filtering Theory,Finance,Stochastic electrodynamics .where there is randomness you can apply it. 1 The closed communicating class might be the whole set. For the second part, you will need an infinite state-space. 1 That is mistaken. Let's look at the probability distribution of X_2 given X_0=6. \begin{align} & \Pr(X_2 = 6\mid X_0=6) \\[10pt] = {} & \Pr\Big( \underbrace{(X_1 = 1\ \&\ X_2 = 6)\text{ or }(X_1=2\ \&\ X_2=6)\text{ or }\cdots}_{\text{five disjuncts}}\mid X_0=6\Big) \\[10pt] = {} & \underbrace{\Pr\Big(X_1 = 1\ \&\ X_2 = 6 \mid ... 1 In some sense, there is an x on the right-hand side because the equality\mathbb{E}_x(Y \circ \theta_s \mid \mathcal{F}_s^+)(\omega) = \mathbb{E}_{B_s(\omega)}(Y)$$holds only \mathbb{P}_x-almost surely. For a function f: \mathbb{R} \to \mathbb{R} the Markov property reads$$\mathbb{E}_x (f(B_{t+s}) \mid \mathcal{F}_s^+) = \mathbb{E}_{B_s} f(B_{t}), ...

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I did some further research and found out the following: The statement $p(x_{t-1}|u_t) = p(x_{t-1})$ has nothing to do with the Markov assumption. It rather states that control is randomly chosen and not a function of the state. The assumption of random controls is essential for the Bayes filter: Without this assumption, the Bayes filter algorithm does ...

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Transition probability matrix (P) in this case will be $$P=\begin{bmatrix}0 & 0.5 & 0.5\\ 0.5 & 0 & 0.5\\ 0.5 & 0.5 & 0\end{bmatrix}.$$ Though $p_{ii}=0$ for $i=0,1,2$, it can be observed that $p_{ii}^n\ne0$ for $n\ge 2$ (also intuitive). Rest is a straight forward calculation depending on $n$.

1

We know the distribution of the vector $V:=(X(7)-X(5), X(5)-X(1), X(1))$, since the increments are independent. Defining the function $f\colon (x,y,z)\mapsto (x+y+z)(y+z)z$, the wanted expectation is $\mathbb E\left[f(V)\right]$.

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I think the complete answer is the following. $|Z_n|=|S_n^2-n\sigma^2|\leq|S_n|^2+n\sigma^2$, so $E[|Z_n|]\leq E[|S_n|^2+n\sigma^2]=E[|S_n|^2]+n\sigma^2<\infty$ because $E[|S_n|^2]=Var(S_n)-E[S_n]^2$ $=Var(Y_1,...,Y_n)-E[Y_1,...,Y_n]^2$ $=\sum_{i=1}^nVar(Y_i)-(\prod_{i=1}^nE[Y_i])^2$ (by independence of $Y_1,...,Y_n$) ...

0

I perused the book you mention and found no definition of what is the Markov assumption (the author does have a word on the markov assumption but its more a critique on the hypothesis then a definition I believe the definition is somewhere else in the lines of conditional independence). Usually markov assumption is translated as $P(x_{t+1} \mid x_0,\ldots, ... 2 No. The disjoint events$\{B_n\}$are not independent. They are mutually exclusive. Hint: Begin with the Law of Total Probability:$\mathsf P(A) = \sum\limits_{n=1}^\infty \mathsf P(A\mid B_n)\;\mathsf P(B_n)$Yes. (But watch your shift key; you have a$P(X\mid y=y)$typo. Additionally you need$X=x$there not$X$by itself). 0 With trial answers to some of the questions in my own comment, suppose we begin the time series at$t = 1$with$X_1 = \mu = 50$and that the SD of the normal distribution is$1/X_{t-1}$. Then the iteration seems to converge quickly to$Norm(\mu = 50,\; \sigma = 1/\mu = 1/50).$Similarly for$\mu=5$. A According to a Kolmogorov-Smirnov goodness-of-fit ... 1 Hint: You know that for$n=0$we get $$A + B = I,$$ and for$n=1$we get $$A + B/3 = P.$$ So you can express$A$and$B$in terms of$I$and$P$... 2 Why not just computing$P^2$, then interpolating? We have: $$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix},\qquad ... 3 Hmm. I don't know how you can directly find A, B, C but I can suggest a way to go about the problem. Now that you have found the eigenvalues \mu_i, solve for the corresponding normalized eigenvectors e_i. Then$$P^n = E^{-1}\Lambda^n E$$where \Lambda is the diagonal matrix of eigenvalues and E = \sum_{i=0}^2 {e_ie_i^T}. Then you can read ... 2 The result is false in this generality. Fix some z\in S and define O\mu=\mu_c+\mu_d(S)\delta_z, where \mu_d and \mu_c=\mu-\mu_d are the discrete and continuous parts of \mu, respectively. Then O satisfies your conditions, but cannot possibly be given by a kernel, if there is a non-discrete probability measure \nu on S. Why not? If ... 2 Define an array as a sequence of random variables on a probability space (\Omega, \mathcal{F},P). Introduce doubly infinite arrays of random variables X_{n,j},~\mathcal{F}_{n,j} for j,n\geq 1, and set sub \sigma -algebras of a \sigma -algebra. Adapting the array to the filtration, then \{X_{n,j}\} is a MDA if the relation ... 0 Let S denote the event when the player wins and F denote the event when the player loses i) P(0,0) = 1 P(k,k+1) = p,P(k,k-1) = q P(6,6) = 1 ii) the probability of loosing all the money at the end of 5 plays is given by the following events$$ FF\\ SFFF\\ FSFF $$that occurs with probability \bigg(\frac{1}{2}\bigg)^2+ ... 1 The only requirement is that the random variables X_t are defined on the same probability space. 1 Stochastic process is a collection of random variables — nothing more and nothing less. That means your collection of iid standard normal random variables is an example of stochastic process that even has a special name — it's called a white noise. 0 Y_{i} having finite variance implies that E[|Y_{i}|^{2}]<\infty, since \text{Var}(Y_{i})=E[|Y_{i}|^{2}]-E[Y_{i}]^{2}. 1 First, N_1(0) + N_2(0) = 0 + 0 = 0. Second, we need to show \{N_1(t) + N_2(t)\} has independent increments, namely, for 0 < t_1 < t_2 < \cdots < t_k, N_1(t_1) + N_2(t_1), (N_1(t_2) + N_2(t_2)) - (N_1(t_1) + N_2(t_1)), \ldots, (N_1(t_k) + N_2(t_k)) - (N_1(t_{k - 1}) + N_2(t_{k - 1})) are independent, which follows from that \{N_1(t)\} ... 3 You need to show independence of increments, i.e. if 0\le a<b<c<d then (N_1(d)+N_2(d)) - (N_1(c)+N_2(c)) is independent of (N_1(b)+N_2(b)) - (N_1(a)+N_2(a)), and similarly for more than two intervals. You can prove that by using independence of increments of each of the two processes separately plus independence of N_1 and N_2. You also ... 2 For sake of compact notation, we let \Delta N_i \mathop{:=} N_i(t+\Delta t)-N_i(t), denote the change in event count over a period \Delta t, and make note of the memoryless property of Poisson distribution's waiting times. Meaning, \mathsf P(\Delta N_i=n) depends only on \Delta t, \lambda_i, and n; it is irrespective of t. Indeed: ... 1 N(t) and N(t+s) are not independent. On the other hand, N(t) and N(t+s)-N(t) are independent. Moreover, N(t)N(t+s) = N(t)[N(t+s)-N(t)] + N(t)N(t). Now use linearity of expectation and independence. Can you finish it from here? 1 By Bayes,$$P(N(s)=k\mid N(t)=n)=P(N(t)=n\mid N(s)=k)\frac{P(N(s)=k)}{P(N(t)=n)}$$Your identity gives:$$=P(N(t-s)=n-k)\frac{P(N(s)=k)}{P(N(t)=n)}$$Sub in Poisson distributions, rest is algebra:$$=e^{-\lambda(t-s)}\frac{(\lambda(t-s))^{n-k}}{(n-k)!}\frac{e^{-\lambda s}(\lambda s)^k n!}{e^{-\lambda t}(\lambda t)^n k!}\\={n \choose ... 3$N(t)$is not equal to$N(t-s)+N(s)$, but $$N(t) = \Big(N(t) - N(s)\Big) + \Big(N(s)\Big)$$ and the two expressions inside the$\Big(\text{big parentheses}\Big)$are independent of each other (whereas$N(t-s)$and$N(s)are not independent of each other). So \begin{align} \Pr(N(s) = k \mid N(t) = n) & = \frac{\Pr(N(s)=k\ \&\ N(t) =n) ... 0 Let\lambda$be the interarrival rate and$\mu$the service rate (so the mean interarrival and service times are$\frac1\lambda$and$\frac1\mu$, respectively). We can model the$M/M/1$queue as a continuous-time Markov chain$\{X(t) : t\geqslant 0\}$, where$X(t)$is the number of entities in the system at time$t$, with state space$\mathbb N\cup\{0\}$... 0 The exponential distribution with mean$1$also has standard deviation$1$. If you collect statistics for$1000$time units, you should get a sample mean of about$1$with a standard error of about$0.032$. When you say your values are not so close to the mean, how far off are they? By the way, you should be aware that if both the inter-arrival and the ... 2 This is Chapter 5, Exercise 16 of Introduction to Probability Models by Sheldon Ross ($10^{\mathrm{th}}$) edition. I am not sure that the solution given is actually correct: Let$X_n\sim\operatorname{Exp}(\mu_n)$. If we suppose jobs$i$and$j$are initially begun, then the expected time until job$kstarts is $$\mathbb E[X_i \wedge X_j] = ... 2 This is not a full solution. I simply try to explain how the order of jobs starting affects the min you are after: if jobs 1 and 2 start first the the probability that job 1 finishes before job 2 is$$p_{12}=\int_0^{\infty}\int_0^{t_2}\mu_1\mu_2e^{-\mu_1t_1-\mu_2t_2}dt_1dt_2=\frac{\mu_1}{\mu_1+\mu_2}therefore job 3 may start with probability ... 1 For the first moment we can write \begin{align} \text{E}(T_{(5)}) =& \text{E}(T_{(1)}) + \\ & \text{E}(T_{(2)} - T_{(1)}) + \\ & \text{E}(T_{(3)} - T_{(2)}) + \\ & \text{E}(T_{(4)} - T_{(3)}) + \\ & \text{E}(T_{(5)} - T_{(4)}) . \end{align} The next observation is that T_{(1)} is a minimum of 100 exponential(\lambda = 1/200) random ... 3 Any càdlàg function f: [0,T] \to \mathbb{R} has at most finitely many jumps with jump size >\epsilon for any (fixed) \epsilon>0, see e.g. this answer. The estimate "\leq [X,X]_t" is not used to conclude that \log V_t is of bounded variation, but to show that the series \sum_{s \leq t} (\log(1+U_s)-U_s) is absolutely convergent. Recall ... 1 Hint: \mathsf E(\max_i X_i \mid X_1<X_2<X_3) = \mathsf E(X_3 \;\mathbf 1_{X_3>X_2>X_1})\big/ \mathsf P(X_3>X_2>X_1) \\ = \dfrac{\int_0^\infty \int_0^{x_3}\int_0^{x_2} x_3f_{X_3}(x_3)f_{X_2}(x_2)f_{X_1}(x_1)\operatorname d x_1 \operatorname d x_2\operatorname d x_3}{\int_0^\infty \int_0^{x_3}\int_0^{x_2} ... 0 The cross quadratic variation process [X_1,X_2] is 0. That is why it is not showing up. You may wonder why [X_1,X_2] is 0.[X_1,X_2]_t = [\int_0^t\mu_1\,ds + \int_0^t\sigma_1\,dB_{1,s},\int_0^t\mu_2\,ds + \int_0^t\sigma_2\,dB_{2,s}]_t $$You need a bunch of facts to show that this is 0. 1-) Bilinearity of quadratic variation 2-) Cross ... 1 I don't believe what you wrote is exactly what you want. Since X_2 starts at time t, it gets a total of time t without failing for free. It seems like you should be looking for \mathbb{P}(X_1 < X_2 + t) = \mathbb{P}(X_1 - X_2 < t). From this page, we know the pdf of X_1 - X_2 is$$ f(x) = \frac{\lambda_1\lambda_2}{\lambda_1 + ... 0 Itô's formula states that $$f(W_t)-f(0) = \int_0^t f'(W_s) \, dW_s + \frac{1}{2} \int_0^t f''(W_s) \,ds$$ for anyf \in C^2$. Applying this for$f(x) := x^3$yields $$X_t = 3 \int_0^t W_s^2 \, dW_s + 3 \int_0^t W_s \,ds.$$ Now the claim follows from the fact that$W_s^2 = X_s^{2/3}$and$W_s = X_s^{1/3}$. 1 I believe you are confusing the mean and mode of your distribution, which is asymetric.$S_t$follows a lognormal distribution with parameters$\mu = -\frac{1}{2} t$and$\sigma^2 = t$. So,$E [S_t] = e^{\mu + \frac{1}{2} \sigma^2} = e^{-\frac{1}{2}t + \frac{1}{2} t} = e^0 = 1$, but the mode of the distribution is$e^{\mu - \sigma^2} = e^{-\frac{3}{2}t}$. ... 1 What you want to calculate is$P(T_2+T_3<T_1)$, where$T_i$'s are independent exponentially distributed random variables. First, let's develop some formulas for continuous independent random variables: $$P(X < Y) = \int_{x<y} f_{X,Y}(x,y)\ dxdy = \int_{-\infty}^\infty\int_{-\infty}^y f_X(x)f_Y(y)\ dxdy = \int_{-\infty}^\infty F_X(y)f_Y(y)\ dy$$ ... 1 Answer for Part II Let$T_1,T_2,T_3$be the service times with probability density of$\frac{1}{3}$. In the problem that you have stated, it is descrete uniform with service times 1,2 or 3. The solution hat you have given is right but on other other hand if it is continuous uniform, and with a pdf of$\frac{1}{3}$, the interval must be 0-3. Given this ... 0 [Returning to part (iii) after a delay of several days due to routine medical matters.] Intuitively, in order for A to be the last to leave she must, in effect, lose two fair and independent coin tosses: so her probability of being the last of the three to leave is$(1/2)(1/2) = 1/4.$Here are the details. First, A is in a fair contest with B to see who ... 1 This is only an answer for a part of the question To answer your second question$2)$: No, in general it's not true that a function bounded by a function of finite variation is itself of finite variation, for example take $$f(x)=\begin{cases} 0, x=0 \\ \sin(\frac{1}{x}), \text{otherwise}\end{cases}$$ which is not of bounded variation although$f(x)\$ is ...

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