a-find-the-steady-state-vector-for-the-markov-chain-in-exercise-3-b-what-is-the-probability-that-aft

Question

a. Find the steady-state vector for the Markov chain in Exercise $3 .$
b. What is the probability that after many days a specific student is ill? Does it matter if that person is ill today?

Alvar Garcia-Fernandez · Accepted Answer

Hello, Numerator. Okay, we&#x27;re doing number 13, but it&#x27;s based on number three. So number three were given this stochastic matrix. Um, as you see described right there. We played around with what happens if we start with 20% L on Monday. What happens on Tuesday? Wednesday for the ills. And what if we start with 100%? Uh, well, on Monday, What happens to the Wells on Tuesday? Wednesday. Okay, but now, question 22 I started question 13. Is this? Let&#x27;s see, I&#x27;m just gonna copy this and some pages to 63 actually. Okay. They&#x27;re just separate Male, but a little bit. It&#x27;s his question 13. They want you to find the stochastic. I&#x27;m sorry, the City state vector. So what we&#x27;re gonna do is gonna apply, Okay, just a minute. I&#x27;m recording. Okay, Stop. Okay. So regardless of what matrix you start off with, let&#x27;s say we start with 50% Well, in 50% ill, right? It&#x27;s a probabilistic myth. Probabilistic factor. So I stand up 200%. Okay, there. Go. Let&#x27;s get rid of this. Let&#x27;s see what happens if we apply em several times. That&#x27;s one way to get the study state factor if it exists. So how about four and in range 1 to 10? That&#x27;s what we get we&#x27;re gonna do is just to play em again and again and again so m v 061 m of x one x two M o M times x two x three Well, that&#x27;s just powers of AM times like zero, isn&#x27;t it? So I&#x27;m gonna say, Let&#x27;s show m to the n times zero and again again. The question was also doesn&#x27;t matter what x zero starts at. Do you still get the same result will change up zero see what happens. It shouldn&#x27;t, right, So I&#x27;m gonna say em to the N Times x zero. Let&#x27;s make some more fancy Was print out the power that we&#x27;re talking about the X one that we&#x27;re talking about. So let&#x27;s put a quote there, comma a comma there In a quote I answer, we get so this question 13 but with the output for a setting up the stochastic matrix, there&#x27;s output. Be the analyzing the hills and see analyzing the wells. Okay, so we start with 50 50 and eso we get x one is 70. 30 x two is 80 20 and so forth. Now is there a limit? While it looks like it might be going a 0.9 point one, So what&#x27;s the long term effect in the hills? It was. The question looks like 10%. Let&#x27;s see if that&#x27;s true. Let&#x27;s throw in some bigger exponents. Just make sure, um, how about from 10 to 100? Let&#x27;s suffering from all outlets, print out everything. Alrighty. So let&#x27;s see 0.89 point one. But eyes getting really, really, really close to 10.9 point one. I think so. 10% ales in long term. Let&#x27;s go crazy. How about 100 to 1000? Let&#x27;s print out every 100 every 100th one. Alrighty. So to the 100th power, we get basically the same thing to the 5/100 power. Same thing with some crazy Randall Verity end there, but especially point line, 90% well and 10% ill in the long term. That doesn&#x27;t matter if we change zero. Let&#x27;s see, let&#x27;s make eggs. Zero. Let&#x27;s start with 100% well, so we get the same thing and question 13 so that that that started 100% well and zero. Il still seems to be going to 00.9 point once a 0.10% ill in the long term. And how about if we start off with How about we start off with 0%? Well, on 100% ill? Does that matter? Let&#x27;s see. Question 13. No, I don&#x27;t think so. Alright, Looks good. So that&#x27;s your answer. Okay. The raid. Hope that was helpful. Look what your homework. There&#x27;s question 13 right there. That&#x27;s it. You wanted to look at the question three. I have a separate video. You can look at that. Here&#x27;s the up for question 13. Right here. So the final result is 10% ill of long term. Okay. Good luck with, um, you guys hope that was helpful.

Michael Jacobsen · Answer

in this video, we have a stochastic matrix p, which is provided here, and our first goal is to find the steady state vector of this matrix P. So to that end, what we need to do for the first step is to solve the equation that p times a vector X results in a vector X. This can be rearranged into the matrix equation p minus the identity matrix Times X equals E zero vector. Let&#x27;s take this system here and go to an augmented matrix will be of size to buy three and from the Matrix p p minus. I tells us that we&#x27;re subtracting one from the main diagonal. So here and here, that means will have a negative 0.3. Then the rest of the row is copy 0.6 and zero. Since we&#x27;re augmenting with the zero vector, then copy points three into the second row and now take 0.4 minus one for a negative 10.6 and copy a zero. So this is the occupant in matrix for that matrix equation. And notice immediately that the second row is equal to the first row times negative one. That means in real reduction we can immediately wipe out Row two, then in Row one. If we divide by negative, 10.3 will obtain one negative to zero, and this matrix is already in reduce row echelon form. This implies that X one is going to be equal to two times x two x two is a free variable saw right X to his ex too. And now the solution for X from P X equals X is going to be if we pick a one for X two, we&#x27;ll have a two x two in the first entry, which will be a two. This is not the steady state vector, but it&#x27;s very close. We just need to multiply it by the right constant. If we take the some of the entries of X two plus one, resulting in three, then we can conclude that Q will be equal to 1/3 of the Vector X that we&#x27;ve just found. So this is going to be equal to 2/3 and 1/3. Let&#x27;s pause for a moment and described what the steady state vector Q is indicating for us if we focus on the first entry 2/3 which is consistent with transitioning to news. We find the following that after many transitions, 2/3 of the viewers or approximately 2/3 we&#x27;ll be listening to the news. So this is how to find the steady set victor Q as well as a bit of its interpretation.

Robin Corrigan · Answer

So in this question, we refer back to you exercise one, which was in the previous chapter, which was the situation where we had to negotiators who could either be cooperating or be competitive. And for we were we determined in exercise one that the the single step transition probabilities were these four probabilities here. And so, for part A of this exercise were asked to construct the one step transition matrix for this Markov chain where X aban is the strategy at the end stage of negotiation. So it is straight forward to construct that, given these transition probabilities. So p sub 11 is simply entry 11 In this matrix, we have 0.6 p. Sub 12 is entry 12 So 0.4 goes here and then we have 0.3 here in 0.7. So this is the one step transition matrix. Next in part B, we rest were essentially asked, What is the to step probability that we will go from state one to state one. Remember, we had two states we had cooperative and competitive. So if we live with this one in this too, we&#x27;re asking what is the probability we call from cooperative to competitive in two steps. So I should have said competitive. It&#x27;s the probability the two step probability have going from one to you, too. And so the way to calculate this is defined the two step transition matrix and then look at entry 12 in that matrix. So using software have calculated this, and this matrix is as follows 0.48 0.52 0.39 and 0.61 So entry 12 is a probability of 0.52 So we can say that this probability is 0.52 So now, for part C, the scenario has been modified a little bit so that there are three states where the third state is the end of negotiation and the single step transition matrix. Yeah, it looks like this, and you can see here that once you get into this third state, you just keep self looping. That is the end of negotiation. So there&#x27;s no way out of there. That is the end of the of the chain. And you can also see that whether you are cooperating or being competitive. You have probabilities of 0.2 and 0.3, respectively, of transitioning into the final state or the end of negotiation state. And now we&#x27;re asked if, if what? The current state is cooperative, what is the probability that we end negotiations within three steps? So that is the three step probability of going from cooperative, which is indexed by one to ending negotiations indexed by three. And so to do this, we find the three step transition matrix, which is the single step transition matrix to the power three. And we look at the entry 13 and that will be our probability of making a three step transition from 1 to 3. So I&#x27;ve gone in, calculated P cubed and looked at the entry 13 and that happened to be zero point 5 to 4. So we can say that this probability is 0.5 to 4. And finally, for party, it&#x27;s essentially the same as part C, except that we&#x27;re starting in a competitive state. So if we&#x27;re in a competitive state, what is the probability that negotiations end within three steps? So we&#x27;re looking for three step probability of going from 2 to 3. So this time we look at the three step transition matrix, except that we look at the value that&#x27;s in entry to three, and that value is 0.606

Amy Jiang · Answer

Okay, So have a P matrix of points wide 0.25 What you find to five here? Fine. Why put too fine. You point your fight. So let&#x27;s say this Executive One two on three. Okay, Now we&#x27;re asked. We were given our state victor, our initial vector, which is one no one we want to put out. We get quite five 0.25 important to flight. Okay, now let&#x27;s multiply. P times are in a director, Our little dish x one and it&#x27;s multiply P Temper X one back there. That gives us what? Why, What if I point you flying kind here and put fire rest of fortune five everywhere. Okay, It was an excellent actor. That was 0.5 point 25 0.25 And that just gives me play. 375 0.3125 0.3 once in flight through. So the probability of getting are true. The food three is 30.312

Robin Corrigan · Answer

In this exercise, we refer back to example of 6.13 which had to do with the cell phone contracts in China and for which we found the one step transition matrix to be equal to this matrix right here. And we can index this as China Telecom, Johnny Unicom and China Mobile and for part A were asked to determine these steady state probabilities. Now, to do that, we can create a matrix, which is like this, and this is the transpose of the one step transition matrix minus the identity matrix. And we want to find the role reduced echelon form for this matrix. It can be done manually using Gulshan Jordan elimination, or you can use software to do it. So for this part of the Matrix, that&#x27;s the transpose of this matrix minus the identity matrix. So it&#x27;s this calculation here. So then this matrix becomes and this is the Nature X that results. So now the trick is to you. The last step is to find the row reduced echelon form for this matrix and using computer software, it comes out to the following and in this form, these number here numbers here correspond to the steady state probabilities for China Telecom, China Unicom and China Mobile, so we can call it steady. State probabilities are 0.368 0.215 and 0.417 So that&#x27;s part A for Part B were asked in the long run. In the long run, what proportion of Chinese cell phone users I will have contracts with China Mobile. So that is the last number. So the probability the steady state probability for China Mobile is 0.417 That also means that in the long around 0.417 uh, the proportion of contracts that are with China Mobile will be 0.417 And finally, for part C, we&#x27;re told that a certain user is currently with China Telecom and were asked to determine the average number of contract changes that that customer would make before signing with China Telecom again. So this is the average number of transitions that take place before returning to China Telecom, and this is given by one over price of one which is this value, the proportion or the probability in the long run of being with China Telecom and this comes out to you 2.717

Charles Carter · Answer

This is problems using, um, this joint relative frequency in the to a table is basing location versus whether or not you&#x27;re staying in your home state. So we have yes and no to staying in your home state. And then we have nothing. We have three different states. So Nebraska, North Carolina, and other Nebraska, North Carolina and other states. So it&#x27;s just filling in as needed. Some for part A. He wants us to calculate that was probability that, uh, plans to stay in his home stay if they&#x27;re from Alaska. So we&#x27;re looking for a probability of yes, given Nebraska. Okay, so we need our probability of getting of Nebraska. Yes, which is 0.0 for four. Okay. And then the probability of not leaving is I&#x27;m not looking for anything. 0.0 point for Okay. Said it. Find that probability. We&#x27;re going to 0.0 for four, divided by the sum to get our total. Nebraska 0.44 plus 0.4. Okay, So put that calculator when 04 ford, about about 40.0 for four. Last point for there&#x27;s a 9.9% or the probability is 0.99 Have them staying. Give another in Nebraska. OK, Part B says most probably that randomly selected student who does not plan to stay at home and we&#x27;re gonna look at all that knows, um, and North Carolina. So we need the North Korean and other for no. So North Carolina know is 0.193 and other for No 0.256 case. We&#x27;re gonna find the probability of North Carolina and given no yes, not capital. And I need to make that a lower case in for now. Case of over the North Carolina, 0.193 then divided by the sum of all of our nos. 0.4 plus 0.193 plus 0.256 Okay, let&#x27;s put that in the calculator. Was there a 0.193 But about 0.4 waas, 0.193 plus 0.256 It&#x27;s the probability is gonna be around 23% or 0.227 Okay, Now, part C that over here it wants us to know whether planning to stay in their home state and living in Nebraska are independent events. So basically, we see that compare the probability of Nebraska, North Carolina and other. So we need those data points to so 0.51 from North Carolina. And for other, we&#x27;re gonna have 0.56 Case was found The probability of saying yes given North Carolina, that&#x27;s going to 0.5 wine divided by the sum of those two. And so that probably is staying at your from North Carolina is gonna be around 21% 0.209 to which is also is already way different than Nebraska&#x27;s WAAS. The probability of saying yes if you&#x27;re from other stay again. We&#x27;re just going to the probability of 0.56 divided by the sum for other. So I got 0.56 to about about 0.56 plus 0.256 So that probably is gonna agree around 18% or 0.179 So all these probabilities are different from each other. Um, the problem north China and other closer and probably of Nebraska is is definitely off. So I would make the claim that being Nebraska and claiming yes are gonna be dependent for They&#x27;re not independent events. This is called not independent. And that&#x27;s because that its probability of 0.99 is is a lot different from the 18% in the 21%. Okay, thank you very much.

Problem

Refer to Exercise $4 .$ In the long run, how like…

Problem 13 Easy Difficulty

a. Find the steady-state vector for the Markov chain in Exercise $3 .$
b. What is the probability that after many days a specific student is ill? Does it matter if that person is ill today?

Answer

(a) $x = \left[ \begin{array} { l } { \frac { 9 } { 4 } } \\ { \frac { 1 } { 10 } } \end{array} \right]$
(b) no, it does not matter that person ill today. As we can see above
probability for a specific student is ill is $\frac { 1 } { 10 }$ .

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(a) $x = \left[ \begin{array} { l } { \frac { 9 } { 4 } } \\ { \frac { 1 } { 10 } } \end{array} \right]$(b) no, it does not matter that person ill today. As we can see aboveprobability for a specific student is ill is $\frac { 1 } { 10 }$ .

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

(a) $x = \left[ \begin{array} { l } { \frac { 9 } { 4 } } \\ { \frac { 1 } { 10 } } \end{array} \right]$
(b) no, it does not matter that person ill today. As we can see above
probability for a specific student is ill is $\frac { 1 } { 10 }$ .

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications