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Below we compute the probability in four ways: Computation with Markov Chain 0.473981098314993 Computation with generating function 0.473981098314988 Estimation false method 0.536438013618686 Estimation correct method 0.473304632462677 The first two are exact methods and differ only a little (probably some round of ...


4

Due to the sum to one criterion and $w_4=2w_3,$ the $a$ and $b$ also has to be chosen to satisfy this condition: $$1-2a-b=2b$$ I think you miss out this condition, that is knowing $a$ would completely determine $b$. Also, we need each component to be nonnegative. To recover your first solution, you can let $a=\frac12, b=0$. To get your second solution, you ...


3

This Markov Chain is not irreducible and is therefore not ergodic. That is the reason why there is no unique equilibrium distribution. More specifically: nonergodicity entails that the equilibrium distribution depends on the distribution of the initial state $X_0$ of this chain. This chain has two irreducible classes, {1,2} consisting of states 1 and 2, and {...


3

Limiting distribution of an ergodic Markov chain with finitely many states. First thing to check is that this matrix $\mathbf{P}$ is a stochastic matrix with rows summing to $1,$ which is true. Because the transition matrix has all positive elements, it describes an (aperiodic) ergodic Markov chain with a single class of intercommunicating states. [To ensure ...


3

The Transition Probability Matrix corresponds to the transition matrix with actions, since for every state you know the appropriate action, i.e. bring the umbrella when it rains, otherwise leave it at home.


2

One solution is to use sample size estimators and your edge detection confidence scores to map the most “confident” edges. For instance if you frequently observe unique traversal between Node A and Node B and with high confidence, this is likely an edge. However if you observe relatively few traversal between node B to C, and with low confidence, this is ...


2

The 2x2 transition matrix consists of 4 probabilities: $p(s|s)$, $p(d|s)$, $p(s|d)$ and $p(d|d)$, where $s$ and $d$ stand for "same votes" and "different votes" respectively. In case when Alice and Bob had different (previous) votes, then if both switch or both do not switch their choices, they will have different votes again. Otherwise, ...


1

Yule process is the same as the pure birth process, although the two names are apparently used in somewhat different contexts. One speaks of birth&death process when it is described in terms of differential equations and Markovian dynamics, as, e.g., here. The pure birth process then would be usually described by a set of rate equations $$ \dot{P_n}(t) = ...


1

The policy is a function you define to act in an environment, it's not a random variable.


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