Difference between revisions of "Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 30/Solution 1"

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(Created page with "Let us examine what happens to the individual states <math display="block">\bold{e_1} = \begin{bmatrix}1\\0\\0\\0\end{bmatrix}\quad \bold{e_2} =\begin{bmatrix}0\\1\\0\\0\end{b...")
 
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Let us examine what happens to the individual states <math display="block">\bold{e_1} = \begin{bmatrix}1\\0\\0\\0\end{bmatrix}\quad \bold{e_2} =\begin{bmatrix}0\\1\\0\\0\end{bmatrix}\quad \bold{e_3}=\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\quad \bold{e_4}=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}.</math> Examining <math display="inline">\bold{e_2}</math>, we can see the 2nd column of <math display="inline">P</math> is <math display="block">\begin{bmatrix}0\\1\\0\\0\end{bmatrix}</math> so the second state is stationary (<math display="inline">P^n\bold{e_2}=\bold{e_2}</math> for all <math display="inline">n</math>). Likewise <math display="inline">P^n\bold{e_3}=\bold{e_3}</math>.
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Let us examine what happens to the individual states <math display="block">\bold{e_1} = \begin{bmatrix}1\\0\\0\\0\end{bmatrix},\quad \bold{e_2} =\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\quad \bold{e_3}=\begin{bmatrix}0\\0\\1\\0\end{bmatrix},\quad \bold{e_4}=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}.</math> Examining <math display="inline">\bold{e_2}</math>, we can see the 2nd column of <math display="inline">P</math> is <math display="block">\begin{bmatrix}0\\1\\0\\0\end{bmatrix}</math> so the second state is stationary (<math display="inline">P^n\bold{e_2}=\bold{e_2}</math> for all <math display="inline">n</math>). Likewise, <math display="inline">P^n\bold{e_3}=\bold{e_3}</math>.
  
Examining <math display="inline">\bold{e_1}</math>, we can see that every iteration this state has probability <math display="inline">1/2</math> of becoming third state and <math display="inline">1/2</math> of remaining the same (by the first column of <math display="inline">P</math>). Since <math display="inline">\bold{e_3}</math> is stationary this means that <math display="block">P^n\bold{e_1}=\frac{1}{2^n}\bold{e_1}+(1-\frac{1}{2^n})\bold{e_3}</math> So as <math display="inline">n</math> goes to infinity, the probability of <math display="inline">\bold{e_1}</math> becoming the third state goes to one.
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Examining <math display="inline">\bold{e_1}</math>, we can see that every iteration this state has probability <math display="inline">1/2</math> of becoming third state and <math display="inline">1/2</math> of remaining the same (by the first column of <math display="inline">P</math>). Since <math display="inline">\bold{e_3}</math> is stationary, this means that <math display="block">P^n\bold{e_1}=\frac{1}{2^n}\bold{e_1}+\left(1-\frac{1}{2^n}\right)\bold{e_3}</math>, so as <math display="inline">n</math> goes to infinity, the probability of <math display="inline">\bold{e_1}</math> becoming the third state goes to one.
  
 
Similarly, as the number of iterations increases, the probability of <math display="inline">\bold{e_4}</math> becoming the second state goes to one.
 
Similarly, as the number of iterations increases, the probability of <math display="inline">\bold{e_4}</math> becoming the second state goes to one.
  
Putting this all together, we have that the probability of ending up in the second state as <math display="inline">n\rightarrow \infty</math> is the sum of the probabilities of starting in the second and fourth states, ie. <math display="inline">\frac{7}{11}</math>, while by the same reasoning the probability of ending up in the third state is <math display="inline">\frac{4}{11}</math> (the remaining states have probability zero). This means our answer is <math display="block">\lim_{n\rightarrow\infty}P^n\bold{x_0}=\color{blue}{\begin{bmatrix}0\\7/11\\4/11\\0\end{bmatrix}}</math>
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Putting this all together, we have that the probability of ending up in the second state as <math display="inline">n\rightarrow \infty</math> is the sum of the probabilities of starting in the second and fourth states, i.e., <math display="inline">\frac{7}{11}</math>, while by the same reasoning the probability of ending up in the third state is <math display="inline">\frac{4}{11}</math> (the remaining states have probability zero). This means our answer is <math display="block">\lim_{n\rightarrow\infty}P^n\bold{x_0}=\color{blue}{\begin{bmatrix}0\\7/11\\4/11\\0\end{bmatrix}}.</math>

Latest revision as of 18:34, 8 March 2018

Let us examine what happens to the individual states Examining , we can see the 2nd column of is so the second state is stationary ( for all ). Likewise, .

Examining , we can see that every iteration this state has probability of becoming third state and of remaining the same (by the first column of ). Since is stationary, this means that , so as goes to infinity, the probability of becoming the third state goes to one.

Similarly, as the number of iterations increases, the probability of becoming the second state goes to one.

Putting this all together, we have that the probability of ending up in the second state as is the sum of the probabilities of starting in the second and fourth states, i.e., , while by the same reasoning the probability of ending up in the third state is (the remaining states have probability zero). This means our answer is