# Tag Info

9

The probability of any chair being occupied can be calculated exactly and, for the third chair as asked for in the question, it comes to $513/1943=0.26402\ldots$. More generally, let $\lambda$ be the rate at which customers arrive multiplied by the time taken for a haircut. So, here, we have $\lambda=(10{\rm mins})^{-1}(15{\rm mins})=3/2$. Then, after a long ...

6

The difference between Markov chains and Markov processes is in the index set, chains have a discrete time, processes have (usually) continous. Random variables are much like Guinea pigs, neither a pig, nor from Guinea. Random variables are functions (which are deterministic by definition). They are defined on probability space which most often denotes all ...

5

Queueing theory uses Kendall's notation, as you described. There are three components describing the behavior of a queue: The customers arriving for service, which is usually described by a Poisson process (random arrivals), but sometimes by non-Poisson processes or even deterministic arrivals rates The time required to service each customer, which is ...

5

This is not a solution, but it might be helpful, and it is too long for a comment. Your $N^N$ equations can be simplified, because you can exploit symmetry in the structure of the problem. Say that $L(S)$ is the expected number of rounds for the game to end after reaching state $S$, where $S$ is some string of length $N$. Then: $$\begin{eqnarray} ... 4 I can improve a little on MJD's method. This was based on computing for each state S (the last N values) the expected remaining number of steps L(S) until a final state (last N values equal) is reached. Let me change the notation slightly. Let T_S be the number of steps from state S until a final state is reached, and assume that this final ... 4 This has nothing to do with being a Lévy process or even with randomness. Assume that the function f:t\mapsto f(t) has no positive jump. Let t_y=\inf\{t\gt0\mid f(t)\gt y\}. Assume that t_y is finite and that y\gt f(0). Then f(t_y-h)\leqslant y for every h\gt0, by definition of t_y, hence \limsup\limits_{s\to t,s\lt t}f(s)\leqslant y. ... 4 For a Markov process (X_t)_{t \geq 0} we define the generator A by$$Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} = \lim_{t \downarrow 0} \frac{P_tf(x)-f(x)}{t}$$whenever the limit exists in (C_{\infty},\|\cdot\|_{\infty}). Here P_tf(x) := \mathbb{E}^xf(X_t) denotes the semigroup of (X_t)_{t \geq 0}. By Taylor's formula ... 4 Note that both (Z\circ \theta_t)\, V and  \mathbb{E}_{Y_t}[Z]\, V are product functions whose expectation can be found using your main equation. Define$$\phi(z)=\mathbb{E}_z(Z)=g_0(z)\int K_{s_1}(z,dy_1)g_1(y_1)\cdots\int K_{s_m-s_{m-1}}(y_{m-1},dy_m)g_m(y_m).$$Then we calculate$$ \begin{eqnarray*} ...

4

Poisson process is memoryless so after $n$ arrival, the probability of $A$ be the first arrival (or similarly for $B$) remains same for all $n$. Suppose that $X$ is the first arrival of the Poisson process with parameter $\lambda$ and $Y$ is the first arrival of the Poisson process with parameter $\mu$. From another question we know that $P(X>Y)=\frac ... 4 Suppose that for$t\geq s$, we have some function$g$so that $$E[f(X_{t}) \,\vert\, \mathcal{F}_{s}]=g(X_{s}) \quad \mbox{a.s.}\tag1$$ Conditioning on$X_s$in (1) gives $$E[f(X_{t}) \,\vert\, X_s]=g(X_{s}) \quad \mbox{a.s.}\tag2$$ From this we deduce $$E[f(X_{t}) \,\vert\, \mathcal{F}_{s}]=E[f(X_{t}) \,\vert\, X_s] \quad \mbox{a.s.}\tag3$$ Equation (3) ... 4 Let$f: [0,\infty) \to [0,\infty)$and$s,t \in [0,\infty)$. Then $$\DeclareMathOperator{cov}{cov} \cov(B_{f(s)},B_{f(t)}) = \min \{f(s),f(t)\}$$ since$\cov(B_u,B_v)=\min\{u,v\}$holds for all$u,v \in [0,\infty)$, so in particular for$u:=f(s)$,$v:=f(t)$. Concerning your second question: Theorem: Let$(X_t)_{t \geq 0}$a continuous local ... 3 The transition kernel$K_t$is defined by some measurability conditions and by the fact that, for every measurable Borel set$A$and every (bounded) measurable function$u$, $$\mathrm E(u(X_t):X_{t+1}\in A)=\mathrm E(u(X_t)K_t(X_t,A)).$$ Hence, each$K_t(\cdot,A)$is defined only up to sets of measure zero for the distribution of$X_t$, in the following ... 3 Indeed$\sigma_x$is exponentially distributed under$P_x$. To see this, note that$[\sigma_x\gt s]\subseteq[X_s=x]$hence $$P_x[\sigma_x\gt t+s\mid\sigma_x\gt s]=P_x[\sigma_x\gt t+s\mid X_s=x]=P_x[\sigma_x\gt t].$$ This is specific to the starting distribution$P_x$. For other starting distributions, the distribution of$\sigma_x$is the barycenter of ... 3 This uses two conventions at once: For every integer$t\geqslant0$,$P_t$is the transition matrix of$t$steps of the Markov chain with transition matrix$P$. Hence$P_0$is the identity and$P_{t+1}=P_tP=PP_t$for every$t\geqslant0$. For every transition matrix$P$and every$B$in$\Sigma$,$P(x,B)=\sum\limits_{y\in B}P(x,y)$. In the discrete ... 3 We will use the result if$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=a $and$|a|<1$, then$\lim_{n\to \infty}{a_n}=0$. Let $$a_n = \binom {2n}{n} p^n(1-p)^n \implies \frac{a_{n+1}}{a_n}= {\frac { 2p(1-p)\left( 2\,n+1 \right) }{(n+1)}}.$$ Taking the limit of the last expression $$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}= 4 p(1-p).$$ Now, note ... 3 We can show that${2n \choose n}p^n(1-p)^n \rightarrow 0$as$n \rightarrow \infty$by the ratio test. Let$a_n = {2n \choose n}p^n(1-p)^n$, then $$\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n} = \lim_{n \rightarrow \infty} p(1-p) \frac{(2n+2)(2n+1)} {(n+1)^2} = 4p(1-p)$$ Since$p$is a probability,$4p(1-p) \leqslant 1$, if$p \neq \frac{1}{2}$, we're ... 3 Copying my answer from MO... There is an improper marginalization in the equation that defines your overall strategy: $$P(X_2,\dots,X_n|X_1)=P(X_n|X_{n-1},\dots,X_1)\cdot P(X_{n-1}|X_{n-2},\dots,X_1)\cdot \dots \cdot P(X_2|X_1)?$$ For simplicity, take$n=3$: $$P(X_2,X_3|X_1)=P(X_3|X_2,X_1) \cdot P(X_2|X_1)?$$ Now remember that each$P$above is really a ... 3 We define a sequence of (discrete) stopping times $$\tau_j := \frac{\lfloor 2^j \tau \rfloor+1}{2^j}, \qquad j \in \mathbb{N}.$$ It is not difficult to see that$\tau_j$is indeed a stopping time and$\tau_j \downarrow \tau$as$j \to \infty$. Since the Brownian motion has continuous paths, this implies$B(\tau) = \lim_{j \to \infty} B(\tau_j)$. Let ... 3 Since$X_s$is$\mathcal F_s$-measurable, the tower property shows that $$\mathrm{P}_x\left[\left.X_{t+s}\in A\space\right|X_s\right]=\mathrm E_x\left[Y\left.\space\right|X_s\right],\qquad \text{where}\quad Y=\mathrm{P}_x\left[\left.X_{t+s}\in A\space\right|\mathcal{F}_s\right].$$ Item (3) of Definition 17.3 asserts that$Y$is$\sigma(X_s)$-measurable, ... 3 First of all, it would be nice if you link the definitions of F-K formula and KBE you have in mind. Anyway: KBE is an equation used to study$P_t f(x):=\mathsf E_x[f(X_t)]$in terms of the generator $$\mathcal A :=\lim_{t\downarrow 0}\frac{P_t - P_0}{t}.$$ This is indeed applies to general Markov processes, as its definition depends only such notions ... 3 It has been a while since I've done anything with Markov chains, so I apologize for any poor notation in advance. Notice that$ \{X_n=x_n, X_k=x_k\} $is$\sigma$-measurable from$ \{X_n=x_n, \ldots,\ X_1=x_1 \},$for$k< n$. Really the Markov property for finite state spaces is equivalent to$ P(X_{n+1}=x_{n+1}\,|\, \sigma\{X_n,\, \ldots,\, X_1 \} ...

3

Both answers are wrong... The system of three equations can be readily rewritten as $A=2B=C$ hence the only solution such that $A+B+C=1$ is $(A,B,C)=(\frac25,\frac15,\frac25)$. If $(A,B,C)=(\frac14,\frac12,\frac14)$ (your solution), then $A\ne2B\ne C$. If $(A,B,C)=(\frac15,\frac25,\frac15)$ (their solution), then $A+B+C\ne1$.

3

We have to show that for each $n\geqslant 1$, if $x\in S$ is fixed, then the map $S\in\mathcal{S}\mapsto P^n(x,S)$ is a probability measure, and if $B\in\cal S$ is fixed, the map $x\in S\mapsto P^n(x,B)$ is measurable. We proceed by induction. We assume that $P^{n-1}$ is a transition kernel. If $x\in S$ is fixed, since for all $y$ we have ...

3

The process $(r_c(t))$ based on $P(\xi(t,n)=1)=c^n$ jumps from $n$ to $n+1$ after a geometric time with mean $1/c^n$ hence $(r_c(t))$ hits $n$ after $\Theta(1/c^n)$ steps. Since $E(r_c(t))=\Theta(\log t)$ and $r_c(t)\geqslant r(t)$, $E(r(t))=O(\log t)$. On the other hand, $r(t)\to+\infty$ almost surely. Otherwise, $r(t)$ would stay at some level $n$ forever ...

2

The reason number 2 is wrong is that each number is predicated upon being able to either go to or come from 5 in each state, which does not uniformly affect each of the states. The degree to which each state's steady state varies depends upon the column corresponding to 5 in the transition matrix, which is not the same for each.

2

In general, any directed graphical model can be converted into an undirected graphical model (though the converse is not true). They're just two different ways to express a joint probability distribution. An excellent review of graphical models is given here, and some discussion of converting directed to undirected models is given in section 2.5. The ...

2

Pick some $a$ such that $a\geqslant-g_{ii}$ for every $i$ (this assumes the state space is finite). The matrix $G_a=G+aI$ has only nonnegative entries hence, for every $n\geqslant0$, so has $(G_a)^n$. Now, $\mathrm e^{tG_a}=\sum\limits_{n\geqslant0}(t^n/n!)(G_a)^n$ is a linear combination with nonnegative coefficients of matrices with nonnegative entries, ...

2

The notion of time-homogeneity of a stochastic process $(X_t)$ can refer to the invariance of its distribution, that is, to the fact that the distribution of $X_t$ does not depend on $t$ and more generally to the fact that, for every set $T$ of nonnegative time indices, the distribution of $X_{t+T}=(X_{t+s})_{s\in T}$ does not depend on $t$. Or it can ...

2

The notation can be a bit cumbersome because of the nested integrals, but this solution relies only on very basic properties of integration and is direct (no induction). Consider the difference $a(x_0,F)=\mathsf P'_{x_0}(F)-\mathsf P_{x_0}(F)$. By uniqueness of measure it follows from the definition of $\mathsf P_{x_0}$ that: \begin{align} a(x_0,F) ...

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