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Assume that the random walk is defined as $$ X_n = \sum_{k = 1} ^ n \xi_k $$where $(\xi_a)_{a\ge 1}$ are iid and such as $P(\xi_1 = \pm 1) = 1/2$. $$1_{T_0 = m,X_0=0} = 1_{T_0 = m,X_0=0, X_1 = 1}+ 1_{T_0 = m,X_0=0, X_1 = -1} $$ Now using the Markov property, and as $T_0$ depends in a deterministic way on $(X_a)_{a\ge 1}$, you get $$ T_0 |(X_0=a, X_1 = ...


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Let $Y_t$ be a Poisson process with rate $\mu$. Since $\mu_n = \mu$, we can take the deaths as following $Y_t$ until everyone has died. So given you start with $N$ people, and $n \ge 0$, the probability that $X(t) = n$ is the probability that there are exactly $N-n$ occurrences in the $Y_t$ process in the interval $[0,t]$. The number of such occurrences ...


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Since $N(t)$ is Poisson, which is discrete (its pdf is $\frac{e^{-\lambda t}(\lambda t)^k}{k!}$), then you need summation and not integral: $$\mbox{E}(e^{-\alpha t} S(t)) = \sum_{k=0}^{\infty} e^{-\alpha t}s e^{(\alpha-\lambda \sigma)t} (\sigma + 1)^{k} \frac{e^{-\lambda t}(\lambda t)^k}{k!} =\\= se^{-\lambda\sigma t}\sum_{k=0}^{\infty} \frac{e^{-\lambda ...


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Note that if $\langle 1 - \lambda, \vec u \rangle$ is an eigenpair for $N$, then: \begin{align*} N\vec u &= (1 - \lambda)\vec u \\ &= \vec u - \lambda \vec u \\ &= I_2\vec u - M \vec u &\text{since $\langle \lambda, \vec u \rangle$ is an eigenpair for $M$}\\ &= (I_2 - M) \vec u \end{align*} This suggests that $N$ should be the $2 \times ...


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Assume discrete time $t \in \{0, 1, 2, \ldots\}$. Let $Z(t)$ be the value of your Markov chain, so $Z(t) \in \{s_1, s_2\}$. Let $R(t)$ be the observation on slot $t$ (assume $R(t) \in \{a,b\}$). Assume that $Pr[R(t)=a|Z(t)=s_1] = 0.25$, $Pr[R(t)=a|Z(t)=s_2]=0.5$. Let $H(t)$ be the history of observations up to time $t$: $$ H(t) = [R(0), R(1), \ldots, ...


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I believe this approach will work for your problem: Modify the state transition matrix as follows. Since both states $0$ and $6$ are absorbing states, and you start every new run at state $1$, identify states $0$ and $6$ as state $0$, and have state $0$ go to state $1$ with probability $1$. That is, the new matrix will look like this: $$ A = \left[ ...


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With respect to the game, this is exactly the St. Petersburg paradox. reference: http://en.wikipedia.org/wiki/St._Petersburg_paradox The game has infinite expected value but is worth only a finite amount in practice if that makes sense?


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Hence, since 0 < s < t, we know the distribution of arrivals in $[0,t]$ is Uniform. Since each occurrence is uniform and independent of all others, The formula you stated is correct. Basically, view the events $A_i$ = {arrival i occurs in the first s minutes of $[0,t]$} for $1 ≤ i ≤ 8$ $A_i$ occurs with probability $\frac st$ by uniformity. Each ...


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Starting with the definition, $\newcommand{\prob}{\mathbb{P}}$ $$ \prob(X_n = x_n | X_{n-1} = x_{n-1},…,X_0 = x_0 ) = \prob(X_n = x_n | X_{n-1} = x_{n-1}) $$ Then let $j_1...j_N$ be a labelling of $\{0,1,2,3,...,t_n\}\setminus \{t_1,t_2,...,t_n\}$. You just need to sum over the values you don't specify in the conditioning: \begin{align} &\prob(X_{t_n ...


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Regular Markov chain means that all the entries of the transition matrix $P$ are positive. Let Markov chain have $n$ states. If $\pi _i$ is the equilibrium probability for state $i$, you can say: $$ \pi _i = \sum _{j=1}^n\pi _j P_{ji} $$ Assume by contradition that $\pi _i=0$. Since all the entries of $P$ are positive, from equation above you can conclude ...


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It is easier to calculate the expectation directly than it is to write down the distribution. This is usually the case. Since you haven't said that much about what you've tried yet, let me start with a hint. Try to calculate the expected time to hit $\{ 0,n \}$ starting from $x$ with $0<x<n$. Then send $n \to \infty$. You can set up a system of ...


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In continuous time, in general you need the jump matrix in addition to the generator matrix. The jump matrix will tell you the probability of each path from $i$ to $j$, while the generator will then tell you the distribution of the holding times for each step of a specified path. Assuming none of rows of the generator are zero, the jump matrix $\Pi$ has ...



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