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Point) Find the vector of stable probabilities for the Markov chain whose transition matrix is0.3 0.7 0.2 0.8 0.2 0.8W =...

Question

Point) Find the vector of stable probabilities for the Markov chain whose transition matrix is0.3 0.7 0.2 0.8 0.2 0.8W =

point) Find the vector of stable probabilities for the Markov chain whose transition matrix is 0.3 0.7 0.2 0.8 0.2 0.8 W =



Answers

Verify that $P$ is a regular stochastic matrix, and find the steady-state vector for the associated Markov chain. $$P=\left[\begin{array}{ll} 0.2 & 0.6 \\ 0.8 & 0.4 \end{array}\right]$$

So for the first part of this problem we need to check that our matrix is regular, sarcastic so we check that it is a stochastic matrix. First we see that all the entries are between zero and one and we look at the sums of the columns and we see the some of the columns are also equal to 11 half plus one, 4004th is +11 half plus one half is one and one third plus two thirds is one. Now. Next we need to check that we have a regular stochastic matrix. Now to check that it's regular. Um we see that we do have some zero entries so we need to then go ahead and square are matrix to see if we get some non zero entries. So let's go ahead and rewrite our matrix really quick. So now we're gonna go ahead and do this matrix multiplication. So we'll take the first row of the first matrix have the first column of the second. We get one half times one half inches. 1/4 plus one half times 1/4 which is 1/8. So that means our top entry will be three. Hope I could write the three 3/8. Here we go. 38 Now for our next row down we do one half times 1/4 which is 1/8 plus one half and 1/4 which is 1/8 plus one third times 1/4 which is 1/12. And then if we add those all together, we get one third and we're going to continue doing this matrix multiplication. And as we do so, we end up with the following matrix and then we get this matrix and I corrected that a typo right here on it said two thirds. It should have been one. All right. Now we look at this new matrix and we see that each entry is non zero. So that means this matrix the original matrix must have been regular. So we had a regular stochastic matrix. All right. So now to find the steady state distribution, we want to look at the matrix i minus P. Where this is R P matrix. So let's go ahead and write down I minus P. And then we want to find the null space of this matrix. So to find the no space we're just going to do our standard row reduction. So we'll keep our first tried the same one half negative one half zero. And then our second World, We're going to take the first road plus two times the second row. So that's going to give us a zero here. That's going to give us a one half here and that's going to give us a negative two thirds here and then the last year we're gonna do the same. We're gonna do the first road plus two. Yeah, two times a second row. There's times the third row that's going to give us a zero. That's going to give us a negative one half and that's gonna give us a two thirds. All right. And then we continue our row reduction in continuing. We'll use we'll do the first row plus the second row to cannes to replace the first row. And that gives us a one half a zero and a negative two thirds. We'll keep the second row the same and then some are lost two rows together and we'll get 000 So what does this tell us? Tell us? This tells us that the vector of our no space is a vector given by the following equations. We have one half times X one where X one is the first entry of our city state, vector is equal to two thirds times X three. Similarly, we have one half times X two is equal to two thirds times X three. And that's just from solving the system of equations that we were given from this matrix. Now to solve for X one and X two, we multiply both sides by two. So if we multiply both sides of each equation by two. These tools that we have right here are going to turn into force and we get the following system of equations. So now using this system of equations, we can write our steady state vector. All we want to do is write a vector where our dependent variable has a one in its entry. So our dependent variable is X three. To open a one there and then our independent variables. We will put the coefficient in front of our dependent variable for each of the respective equation. So X one has a four thirds and next to has a four thirds. All right. So this factor that we have is close to our steady state vector, but not quite. To get our steady state vector. We need a probability distribution, which means we need to scale this vectors to set all of its entries sum to one. So the way that we do that is we write the sum of our vector. So, I mean we take four thirds plus four thirds plus one and find the summation. But that gives us eight thirds. Sorry, there should be a one there. This 4 30 plus four thirds. That's going to be eight thirds plus one is going to be 11 3rd. All right. And once we do that, we then multiply this vector, our study state vector by the reciprocal of the summation. So I'm going to multiply this vector by 3/11. And I was playing it by its reciprocal. And what does that give us? Well that's going to give us 4/11. It's going to give us 4/11 and then it's gonna give us 3 11. It's we see that this is just a scale or multiple. So um this vector is still in the null space of I minus P. But it's scaled such that when we summit centuries for 11 plus 4, 11 plus 3 11 we get one. So this must be our steady state vector of our matrix.


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