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2

I think you can simplify things a lot by just using two states, foot (state $1$) and bicycle (state $2$). Then you have the transition matrix: $$\begin{pmatrix} 0.2 & 0.8 \\ 0.3 & 0.7 \end{pmatrix}$$ In fact, the way you've done it is a bit confusing because falling off a bicycle is not a state in the same sense that going to school by foot/bicycle ...


2

Let $X_n$ be the number of white balls in poll $1$. Then $X_n\in\{0,1,2,3\}$. The probability transition matrix is $$ \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1/9 & 4/9 & 4/9 & 0 \\ 0 & 4/9 & 4/9 & 1/9 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} $$ How is this derived? Take the ...


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The equality $$\mathbb{E}\left[ {\frac{d} {{dt}}\mathbb{P}\left( {A|{X_t}} \right)|B} \right] = \sum\limits_{x \in S} {\mathbb{P}\left( {{X_t} = x|B} \right)\frac{d} {{dt}}\mathbb{P}\left( {A|{X_t} = x} \right)} $$ is wrong since $$\mathbb{P}\left( {A|{X_t}} \right) = \sum\limits_{x \in S} {\mathbb{P}\left( {A|{X_t} = x} \right){1_{\left\{ {{X_t} = x} ...


1

I do not know which state does each row and column of the Markov Chain represents. However, you already have the transition matrix, which is irreducible and aperiodic, so you just need to solve the equation $\pi P = \pi$ where $\pi$ is a row vector, corresponding to each row of the matrix, then $\lim_{n\rightarrow\infty}P(X_n=0)=\pi_{TTT}$ You should ...


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EDIT: found an error, fixed. Hm, seems that I lied to you. We have 2 principal eigenvalues here, 1 and -1. Here's the output of python (check that I'm not mistaken in the matrix): >>> import numpy >>> numpy.linalg.eig(numpy.array([[0, 1.0/6, 1.0/3, 1.0/2, 0, 0, 0, 0], [1.0/6, 0, 0, 0, 1.0/3, 1.0/2, 0, 0], [1.0/3, 0, 0, 0, 1.0/6, 0, ...


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First work out the state space. It's built into the setup that the state space contains $\{ 0,1,2,3 \}$, and in fact it doesn't contain anything else, as you can check by checking that if $X_n \in \{ 0,1,2,3 \}$ then no matter what $Y_n$ is, you will have $X_{n+1} \in \{ 0,1,2,3 \}$. From there just build the transition matrix case by case. (Here I use the ...


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The state space is $\{0,1,2,3\}$. When at state $0$, the next state has the distribution $\mathcal{Y}$ which is the distribution of $Y_n$. When at state $1$, the next state also has the distribution $\mathcal{Y}$, while at state $2$, it has the distribution $\mathcal{Y}+1$. Finally, at state $3$, it has distribution $3-\mathcal{Y}$. All in all, we obtain the ...


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The steady state is the solution of a system of linear equations. As such, it can be obtained by dividing suitable determinants, which themselves are polynomial expressions in the given entries. As it is always the same deteminant that occurs in the denominator, your observation follows.


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First, I wouldn't modelise the markov chain like that , I would consider four states 1 = "player A play", 2 = "player B play", 3 = "player A had won" and 4 = "player B had won" The transition matrix would be $$M = \begin{pmatrix} 9/52 & 10/13 & 3/52 & 0 \\ 10/13 & 9/52 & 0 & 3/52 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 ...


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There can be two interpretations of your question. Under the first, you are interested in the maximal population existing at any given time, i.e., $\max_i Y_i$, where $Y_i$ is the population at time $i$. Under the second, you are interested in what is knows as total progeny, which is $\sum_i Y_i$. Lindvall showed that for a critical branching process, ...


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Based on your attempt, you seem to want to know the probability that you arrive to state $3$ from either state $2$ or state $1$. Since this is an absorbing chain, you will need to break out the transient portion $M'$ from the absorbing portion: $$M':=\left(\begin{array}{c} 0.5/.8 & 0.3/.8\\ 0.1/.5 & 0.4/.5 \end{array}\right) = ...


1

Actually, your friend is right, the chain is not ergodic. By definition, a Markov chain with transition matrix $P$ is ergodic if there exists $n \geq 1$ such that $$P^n > 0. \tag{1}$$ A direct calculation shows that $P^n$ either equals $P$ or $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ for any $n \geq 1$. This means that $(1)$ doesn't ...


1

I'm going to assume that "being tied" is considered losing, thus having that the team loses 40% of its games if the second team scores the first goal. Then we have $H_1$={the team scored the first goal} and $H_2$={the opponent scored the first goal}. $A$={the team won}. $$ P(A)=\sum_{i\in I}P(A|H_i)P(H_i)=0.6*0.3+0.4*0.7 $$


1

Assuming one of the two teams always scores a goal, the probability of winning is the probability of each of the two mutually exclusive and exhaustive possibilities (i.e., scoring first or scoring second) times the probability of winning each way. In other words, $$\begin{align}\mathsf P(\textrm{winning}) & = \mathsf P(\textrm{first}) \cdot \mathsf ...


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It depends on $W$. For example if $W=f(Y)$, then you are right, however if W is independent of $Y$, then conditioning on $W$ can increase the conditional mutual information $I(X;Z|YW)$! For example if $W$ is $mod 2$ sum of $X,Z$ which is independent of $X,Z$ then $I(X;Z|Y)=0, I(X;Z|YW)=1$ which violates the desired Markov chain.


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Yes, your proof is correct. Alternative argumentation: For $$P^n := \underbrace{P \cdots P}_{\text{$n$ times}} = (p_{ij}^{(n)})_{i,j \in E}, \qquad n \in \mathbb{N},$$ we have $P^{k+l} = P^k \cdot P^l$. Therefore, $$p^{(k+l)}_{ii} = \sum_{j \in E} p_{ij}^{(k)} p_{ji}^{(l)} \geq p_{i j_0}^{(k)} p_{j_0 i}^{(l)}$$ for any fixed $j_0 \in E$.



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