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1.2.4) Limiting PageRank values [12 points]. The directedl graph below depicts the hyperlinks among six web pages {A_ B.C.D.E. F} along with proposed PageRank centr...

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1.2.4) Limiting PageRank values [12 points]. The directedl graph below depicts the hyperlinks among six web pages {A_ B.C.D.E. F} along with proposed PageRank centrality values for each Web Wag. In ts exercise assumn teleportation evaporation-Tain component4) {2 points} Are these correct (limniting) equilibriu vales lor the basic PageRank update rule? poinlsh PageRank values correspond t the stationary tlistribution Markov chain (MC), which dleseribes the secpuence ol page visits ol random weh s

1.2.4) Limiting PageRank values [12 points]. The directedl graph below depicts the hyperlinks among six web pages {A_ B.C.D.E. F} along with proposed PageRank centrality values for each Web Wag. In ts exercise assumn teleportation evaporation-Tain component 4) {2 points} Are these correct (limniting) equilibriu vales lor the basic PageRank update rule? poinlsh PageRank values correspond t the stationary tlistribution Markov chain (MC), which dleseribes the secpuence ol page visits ol random weh sucfer. Such MC has trausition prohability matrix (Damt ) - [0, 1]6*6 . whcre Domt dliag(dqut dgut is the out-degree matrix, Andl {0,1}6*6 is the graph $ axljacency matrix Determine the PageRank of the wel pages hy computing the stationaly distribution of thc associated MC. iC.1 solve the following set of cqpuations (YOn Can Use aWY programming language) T =P 1Tw =1. Is this consistent with Four answer to a)? c) /8 points square matrix is called Markov (or right stochastic} if i} all the elements of JOI- negative: and ii) each IOW SUMS "p 10 Show that the transition probability matrix P := (Dout ) MC' is Markov . Prove that for aHy eigenvale Markov matrix |Xl < 4 (Hint: check Gershgorin circle theorem) _ points} Suppose that matrix P is also irreducible Aud "periodic (see Perron-Frobenius theoremn) . Show that lit il t for all x e R" whcrc constant that depends on IF P is the transition prohability matrix of an irreduciblc andd aperiodic MC, what the rclationship betwecn And the stationary dlistribution Motivatecl by you ASWCT Call VOU specily simple iterative procedure to obtain the PageRank vector? Implement such pIO cexlure for OUI network of interest and illustrate the speed of convergence with plot.



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The authors of the article "The Fate of Priority Areas for Conservation in Protected Areas: A Fine-Scale Markov Chain Approach" (Envir. Mgmnt., $2011 : 263-278$ ) postulated the following model for landscape changes in the forest regions of Italy. Each "pixel" on a map is classified as
forested $(F)$ or non-forested $(N F) .$ For any specific pixel, $X_{n}$ represents its status $n$ years after 2000 $($ so $X_{1}$ corresponds to $2001, X_{2}$ to $2002,$ and so on \right$) .$ Their analysis showed that a pixel has a 90$\%$ chance of being forested next year if it is forested this year and an 11$\%$ chance of being
forested next year if it non-forested this year; moreover, data in the twenty-first century are
consistent with the assumptions of a Markov chain.
$$
\begin{array}{l}{\text { (a) Construct the one-step transition matrix for this chain, with states } 1=F \text { and } 2=N F \text { . }} \\ {\text { (b) If a map pixel was forested in the year } 2000, \text { what is the probability it was still forested in }} \\ {2002 ? 2013 ?}\end{array}
$$
$$
\begin{array}{l}{\text { (c) If a map pixel was non-forested in the year } 2000, \text { what is the probability it was still }} \\ {\text { non-forested in } 2002 ? 2013 ?} \\ {\text { (d) The article's authors use this model to project forested status for several Italian regions in }} \\ {\text { the years } 2050 \text { and } 2100 . \text { Comment on the assumptions required for these projections to be }} \\ {\text { valid. }}\end{array}
$$

In this problem, we're told that X seven is the state oven area, whether it's forced it or non forested, and that is in years after the year 2000. So we have a state space that is either forced ID or non forested. For part, A were asked to construct the one step transition matrix for this chain by denoting ST F as one and state N f. As to in the question, we're told that probability of going from forested state to a forested state in the next step is equal to 90% 0.90 And we can also denote this small p sub 11 which is the single step transition probability we're going from state one to state one. Based on that, we can also say that the transition probability from state one to state to is 0.1. Because there is only two possibilities. You're either going to force its date or you are going to a non forced its state. We're also told that if it is forced it this year, the probability of it being none if it is not forested this year, the probability of it being forested next year is 11% or 0.11 which means that if it is not forced it this year, the probability of it being not forced it in the next year is equal to 0.89 which is the complement of 11%. Now the transition matrix, the one step transition matrix, is simply these entries. So if we substitute in the actual values for these probabilities, we get falling for Pete, Part B were asked if the area was forced it in the year 2000. What is the probability that it was forested in 2002 and in 2013? So if the starting point is 2000 the year 2000 and two implies and is equal to two, which means that there would be two transitions from the year 2000 until the year 2000 and two. So what we're looking for is the two step probability from forested to forested now defined the two step transition probabilities. We can multiply the one step transition probabilities matrix by itself, using matrix multiplication. You can do this by hand. For a small matrices like this one. Breaking is software, but it comes out to these entries zero point 8 to 1, 0.179 0.1969 and 0.8031 And so now the two step probability of transitioning from forced it deforested is found by looking at entry one one in the Matrix, which is this entry here. So the two step transition probability going from forested to forested is equal to zero point 8 to 1. And so that is the probability of having a forested region in 2002 if you started off as forested. And so the second part is what is the probability of still being forced ID in the year 2013? So if we go from 2000 to 2013 that means that and is equal to 13 so we can find the 13 step transition probabilities matrix by raising the Matrix to the power off 13 and using software that comes out to the following. And so the probability of going in 13 steps from forested to force did is once again the entry 11 in this matrix. So it zero point 546 and then for part C. We're asked if the pixel ah was non forced it in the year 2000. What is the probability that it is non forced it in the years 2002 and 2013? So we've already made the two step and 13 step transition probability matrices so we can use the entries from these matrices that should be one right there. So the probability of going in two steps from non forced it to non forced it is 0.831 that 0.8031 and the probability of going in 13 steps from non forced into non forced it is equal to 0.500 And to be more precise, that should actually be 0.501 So these were the answer is to part C and then last for party were asked if this kind of model was used to project to forecast the forest status in the years 2050 and 2100. What assumptions would be required for these projections to be valid. So one assumption would be that with the single step, transition probabilities still hold that far into the future. you could imagine that. I mean, that's five and 10 decades away of. We may treat forests differently in that period of time. These forests are already in a protected area, so it might be the case that we're protecting forests even more in the future, or perhaps even less so. That is a question of whether at time homogeneous Markov chain would be appropriate for modeling that far into the future.

First one. The test for a unit root in series you rate unemployment rate using the usual dickey fuller test with a constant yeah. And the augmented dickey fuller with two legs of change of unemployment rate. I find that seven both times we are unable to reject the now hypothesis that unemployment rate series is a unit fruit. The legs are not significant. However, the significance of the legs matters. So the outcome of the unit root test, we will repeat what we have done in part one two series vacancy rate and report the result in part two. I guess similar result. So the rate is a unit root. Well part one and two. I use package the R. Package A. T. S. A. And the function is a D. F. Dot test. R. Three. We assuming that unemployment rate and vacation re rate are both integrated of level one. We test for co integration using the angle grandeur test with no legs. So the step the steps are as follow. We first regress, you read on the rate then we yet the residual and we run the key fuller has on the residual to see whether the residuals our unit root. I find that you're right and we rate Arco integrated at the 5% level. Yeah Heart Forest. I get the leads and lacks estimator of the change in vacancy rate and I did note that uh CB rates up minus one. This is for the lack and plus one is for the lead. This is a regression result. So the usual centered errors are in green and in round brackets, the robots that Iran's are in blue and in square brackets you can see that the main estimate on vacancy rate is highly significant. This one is not correct. So the centered errol the usual one for the estimate of the first lack of change in vacancy rate is 164 In all cases except for the estimate of the lead of C. V. Right. The robust standard Iran's are larger than the usual standard errors. This is usually the case it happens but rare that the robot standard errors are smaller than the usual standard errors. The r square of this regression is 0.77 So for the rate, because the robot standard error is larger than the usual standard error. So we will get a wider confidence interval if we use a robot standard error and for confidence interval you will run this function in our count in and you impose the name of the regression. It was spits all the 95% confidence intervals for all explanatory variables. The default version is the 95% interval. But because the standard barrel of this estimate is are very close, two versions are very close to each other so the confidence intervals should be roughly equal. Yeah. Last part. What you could say about real business of the claim that you rate and the rate are co integrated. Yeah. When I run the test and good grandeur, the results are not consistent across alternative types of process. In one case I can reject the notion that the residuals are united and for all the cases I cannot reject. So I conclude that the claim that you rate and be rate our co integrated is not robust.

In this problem, we're told that the number of visits that a person makes to an auto repair shop in a given your follows a personal distribution where the personal print parameter is based on the number of visits that they made in the previous year. So, for example, if they make if they made zero visits in the previous year, then the number of visits in the next year is a possible distribution with this parameter. And so for part, they were asked to construct the one step transition matrix for the number of repair shop visits by a randomly selected customer in the observed year. Now recall that postal distribution for X and parameter mu is given by this equation, and so the probability that except n plus one equals X Given except Ben was I visit. Is it possible with this probability distribution? So we could say that the one step probability of going from zero visits in the previous year two zero visits in the next year is equal to the probability of zero given this rate and so that rate amuse sub zero is this value and so that's equal to you, you to the miners 1.93 nine approximately times 1.939 to explain it. Zero over zero factorial, which comes out to approximately 0.144 And we continue with those count calculations for all single step transitions. Because there are five possible states, we know that the transition matrix is going to be five by five. So you would save a lot of time by calculating these probabilities using software. So that's what I did. And here's the one step transition matrix that resulted. And so this answer is part A. So for Part B were asked that if a customer makes two visits this year, what is the probability that customer makes two visits in the next year and two visits in the year after that? So it helps to index the is just Teoh. Make sure we make less mistakes. So we're looking for the one step probability that we go from two visits to two visits, and then we again go from two visits to two visits. So this represents two visits in the next year, and this represents two visits in the year after that, and they're both one step probabilities one step transitions because the first time we're just going from this year into the next year. So what? We know we went to two visits this year. What is the probability of two visits in the next year and then in the next year, Once we've gone to visits, it's just one more step to the next year. So it's going from two visits next year, 22 visits the year after. So that's one transition. So the one step transition probability from two visits to you to visit is 0.27 So this comes out to you. Zero point 27 squared and that equals 0.73 for Part C were asked if a customer makes no visits last year, what is the probability? They make a total of exactly two visits in the next two years? So we're looking for all of the ways to have a total of two visits over two transitions, assuming the current state is zero visit. So one way is to go ahead and make two visits in the next year, and then we would need to also make zero visits in the next year that is in the year after the next year. So next year we make two visits and the year after we make zero visits for a total of two visits and another way is to do one visit in each of the next years. So that's the problem. One step probability of going from 0 to 1 times the single step probability of going from 1 to 1. So here again we will have a total of two visits and the third way is to have zero visits in the next year, but have two visits in the year after that. And so this comes out to 0.27 times 0.148 plus zero point 279 time, 0.333 the 0.144 time, 0.27 and that comes out to about 0.172


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