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Notice that $u=(S_n+n)/2$ is $Beta-Binomial(n,1,1)=U[0,n]$. Hence $$Y|S_n\sim Beta(1+\frac{n+S_n}2,1+\frac{n-S_n}2)=Beta(1+u,1+d)$$ with average $\hat y_n=\frac {1+u}{2+u+d}=\frac{1+u}{n+2}$ and hence $$S_{n+1}-S_n\sim B(P((Y|S_n)>X_{n+1}))= B(\hat y_n)$$ where $B(p)$ is Bernoulli on $\{-1,1\}$ with $P(B(p)=1)=p$.

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I don't know of a way to use the reflection principle to prove this. It might be easier to think this way: for any walk to go from $0$ to $k$ it must, in turn, go from $0$ to $1$, then from $1$ to $2$, ..., then from $k-1$ to $k$. Conversely, any sequence of walks from $0$ to $1$ can be joined end-to-end to form a single path from $0$ to $k$. So, ...

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A quick answer can be found using the method of indicator random variables. Let the drink types be labelled from $1$ to $n$. For each $i$ from $1$ to $n$, let random variable $X_i$ be equal to $1$ if at least one person has a drink of Type $i$, and let $X_i=0$ otherwise. Then the number $Y$ of different types drunk is given by $Y=X_1+\cdots+X_n$. Thus ...

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1) Firstly, we can consider the $\ell_1$ normalized eigenvector $\mathbf {v}= \begin{bmatrix} 1/4 & 3/8 & 3/8 \end{bmatrix}^T,$ which corresponds to the dominant eigenvalue $\lambda_1 =1.$ The limit $\lim\limits_{n\to\infty} M^n \cdot \mathbf x$ always exists for any probability vector $\mathbf{x}$ and is equal to $\mathbf v.$ Indeed, one way to ...

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Fix $j_0, i, j_n$. By Markov property, for all $j_{n - 1}, \ldots, j_1 \in S$, where $S$ denotes the state space of this Markov chain, we have \begin{align} P[X_{n + 1} = i \mid X_n = j_n, X_{n - 1} = j_{n - 1}, \ldots, X_0 = j_0] = P[X_{n + 1} = i \mid X_n = j_n]. \end{align} Or equivalently, \begin{align} & P[X_{n + 1} = i, X_n = j_n, X_{n - 1} = j_{n ...

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This is a birth-death Markov chain, so $$\nu_n = \left(\frac p{1-p}\right)^n$$ is an invariant measure for $P$, that is, $\nu P=\nu$. To normalize $\nu$, we compute $$\sum_{n=0}^\infty \nu_n = \sum_{n=0}^\infty\left(\frac p{1-p}\right)^n = \frac{1-p}{1-2p}.$$ Set $$\pi_n = \frac{\nu_n}{\sum_{k=0}^\infty \nu_k} = \left(\frac ... 1 The transition matrix has the "block cycle structure"$$P=\pmatrix{0&A_0&0&0&\cdots&0\cr 0&0&A_1&0&\cdots&0\cr 0&0&0&A_2&\cdots&0\cr \vdots&&&\ddots&&\vdots\cr A_{d-1}&0&0&0&\cdots&0}$$For any dth root of ... 1 Suppose you have a positive value for \pi(0), equal to \pi_0. Then you have the recurrence relation$$\pi(x+1)=\frac{p}{1-p} \pi(x),\pi(0)=\pi_0.$$This recurrence can be solved explicitly: you get \pi(x)=\left ( \frac{p}{1-p} \right )^x \pi_0. Now normalize that. 1 This is not true. Consider, e.g.,$$ A=\pmatrix{1&-1\\-1&1}\quad\text{and}\quad B=\pmatrix{2&0\\0&2}. $$Both A and B satisfy the given conditions but B is reducible. It would be true if you replaced "A_{ij}\leq B_{ij}\leq 0 for i\neq j" by "B_{ij}\leq A_{ij}\leq 0 for i\neq j". This is easy: the only way how to make an ... 1 Assume |A|=k and |B|=n. We want to compute the average size of f(A), with f ranging over all the functions from A to B. For any j\in\{1,\ldots,k\}, there are (look at Stirling numbers of the second kind)$$ {k\brace j}\cdot j!\cdot\binom{n}{j}={k \brace j}(n)_j\tag{1} $$functions such that |f(A)|=j. Our expected value is so:$$ ...

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HINT: The stationary distribution $\pi$ satisfies the relationship $\pi P=\pi$, which is equivalent to showing $\forall j\in S$: $$\pi_j = \sum_{i \in S}\pi_i p_{ij}$$ You can also write out $\pi$ explicitly as a row vector where every entry is $\frac{1}{5}$, and $P$ is a matrix with entries $p_{ij}$ in row $i$, column $j$. Expanding this out and using ...

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The solution of the following system of equations will give the left eigenvector (assuming that there is a corresponding eigenvalue: $1$) $$\begin{pmatrix} P_1& P_2& P_3 \end{pmatrix}= \begin{pmatrix} P_1& P_2& P_3 \end{pmatrix} \begin{pmatrix} .4 & .2 & .4 \\ .6 & 0 & .4 \\ .2 & .5 ... 1 The probability of extinction is the smallest positive root of$$G_O(z)=z Where $O$ denotes the offspring distribution, and $G_O(z)$ its generating function at $z$. It is easily seen that $G_O(0)$ is the probability of extinction in the first generation. Second, if you know about generating functions, then you know that the sum: , where $X$ is ...

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Well, if he doesn't have it, then it must be at his other location, right? So the probability of transititioning from not having it to having it is 1, and the probability of transitioning from not having it to not having it is 0.

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1. For $i\lt j,\; M_j-M_i$ is a function of $X_{i+1},\ldots,X_j$ only. So event $S_{b+1,c}^{'}$ involves r.v.'s $X_{b+1},\ldots,X_c$ only. Also, $M_i$ is a function of $X_1,\ldots,X_i$ only. So event $S_{a,b}^{'}$ involves r.v.'s $X_1,\ldots,X_b$ only. By independence of $X_1,X_2,\ldots,\;$ the events $S_{a,b}$ and $S_{b+1,c}^{'}$ are independent. 2. ...

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Where did you get the statement? It looks wrong to me. A counter-example follows. Consider a chain that looks like an "$\infty$"-symbol and consists of two linked cycles of length 1000 and 1001. Let $i$ be the node that links the two cycles. Whenever the chain is at the node $i$, it goes to the left cycle or the right cycle with probability $1/2$, and then ...

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An exponential variable S, by definition, has a density of the form \lambda \exp(-\lambda x) where \lambda is an arbitrary positive number. This number is so characteristic of the variable that it is called the "parameter". The assignment invites you to demonstrate that \lambda S is another exponential variable, presumably by calculating its density ...

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Allow me a divagation. Say we have a discrete Markov chain $X=\{X_n\}$ with space of possible states $S$ and transition matrix $P=(p_{xy})_{x,y\in S}$. A vector $\pi$ with non-negative entries is said to be a stationary distribution of $X$ if $\sum_{x\in S}\pi(x)=1$ and $\pi P = \pi$, that is if it satisfies $\pi(y) = \sum_{x\in S}p_{xy}\pi(x)$ for every ...

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