Search results
When the Markov process is discrete-valued (i.e discrete state space), it is called a Markov chain. Markov chain can further be divided into Continuous –parameter (or time) Markov chain and discrete – parameter (or time) Markov chain for continuous and discrete parameters respectively. Random process and stochastic process are synonymous.
Apr 29, 2020 · 1. Interesting problem. Here are some thoughts. The usual Markov criterion is that each item depends only on the one before it. That is, its probability distribution is the same regardless of the prior elements: Your problem is slightly different. You have deleted some elements from the sequence, and you want to prove that the next element ...
May 2, 2015 · If pA was 1 then the Markov chain would never get out of state A. Then, of course, 1 = P(heads) =πA. If pA <1 then the solutions above are valid and we may compute the probability of the heads in general. Watch this: P(heads) = (pB +pC)pA + (1 −pA)pB + (1 −pA)pC −2pA +pB +pC + 2 = πA !
Feb 17, 2020 · Y3 is already 0 and Y4 cannot be two - so it's not possible. For second case : when X2 =0 which means both Y1 and Y2 are zero. X3 =1 means Y3 has to be 1 as Y2 is zero. Now, for calculating X4 =2 , Y4 has to be 1 because Y3 is also 1. In short X4 depends on what we have in X2 but to be a markov chain it should not depend on X2 right? $\endgroup$
Oct 2, 2016 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Feb 3, 2018 · Let h(k) h (k) be the expected number of steps / years in this example until we reach the state 0 0 when you are in state 2 2. So we have that. h(2) = 0 h (2) = 0. because when you are in state 2 2, you need 0 0 steps / years to reach 2 2. Then for k = 1 k = 1.
Feb 14, 2019 · My intuition and solution: It is not a Markov chain. P(Yn = 1 | Yn − 1 = 1, Yn − 2 = 0) = 1 4 P(Yn = 1 | Yn − 1 = 1) = 1 2 ∗ p(1), where p (1) is a probability that we are at the state 1, given we are either at the state 1 or 2. Surely probability that we are at the state 1 isn't equal to half, which is obvious if we look at the ...
Jun 15, 2012 · Can anyone give an example of a Markov Chain and how to calculate the expected number of steps to reach a particular state? Or the probability of reaching a particular state after T transitions? I ask because they seem like powerful concepts to know but I am having a hard time finding good information online that is easy to understand.
May 2, 2017 · This depends on $f$. In fact, $Y_n=f(X_n)$ is a Markov chain in $\mathcal Y$ for every Markov chain $(X_n)$ in $\mathcal X$ if and only if $f$ is either injective or ...
It seems to me that this is not a Markov chain because in order to predict the value of Xn+1 X n + 1 we need the value of Yn Y n which we can only get by comparing Xn−1 X n − 1 and Xn X n. It just occurred to me that I might try making a table of all differing values x_n+1 can assume given x_n at a certain value.
