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anastassius [24]
3 years ago
11

Let P = 0.50.30.50.7 be the transition matrix for a Markov chain with two states. Let x0 = 0.50.5 be the initial state vector fo

r the population.Find the steady state vector x. (Give the steady state vector as a probability vector.)
Mathematics
1 answer:
pav-90 [236]3 years ago
4 0

Answer:

Probability distribution vector = \left(\begin{array}{c}0.375\\ 0.625 \end{array} \right)

Step-By-Step Explanation

If P=\left(\begin{array}{cc}0.5&0.3\\ 0.5&0.7 \end{array} \right)  is the transition matrix for a Markov chain with two states.  

x_{0}=\left(\begin{array}{c}0.5\\ 0.5 \end{array} \right)  be the initial state vector for the population.

X_{1}=P x_{0}=\left(\begin{array}{cc}0.5&0.3\\ 0.5&0.7 \end{array} \right) \left(\begin{array}{c}0.5\\ 0.5 \end{array} \right) =\left(\begin{array}{c}0.4\\ 0.6 \end{array} \right)  

X_{2}=P^{2} x_{0}=\left(\begin{array}{c}0.38\\ 0.62 \end{array} \right)  

X_{3}=P^{3} x_{0}=\left(\begin{array}{c}0.38\\ 0.62 \end{array} \right)  

X_{30}=P^{30} x_{0}=\left(\begin{array}{c}0.37499\\ 0.625 \end{array} \right)  

In the long run, the probability distribution vector Xm approaches the probability distribution vector \left(\begin{array}{c}0.375\\ 0.625 \end{array} \right) .

This is called the steady-state (or limiting,) distribution vector.

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x = ounces of 14% iodine.

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we know that in a 14% iodine solution, 14% is iodine and the rest is something else, we also know that "x" ounces of the solution have 14% of iodine, how much is that?  (14/100) * x, or 0.14x.

likewise, in the 89 ounces of 47% solution there is (47/100) * 89 of iodine.

and likewise as well, in "y" ounces of 29% iodine there is (29/100) * y, or 0.29y of iodine.

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\bf \begin{array}{lccclll}
&\stackrel{ounces}{amount}&\stackrel{\%~of~iodine}{quantity}&\stackrel{iodine~oz}{amount}\\
&------&------&------\\
\textit{14\% solution}&x&0.14&0.14x\\
\textit{47\% solution}&89&0.47&41.83\\
------&------&------&------\\
mixture&y&0.29&0.29y
\end{array}
\\\\\\
\begin{cases}
x+89=\boxed{y}\\
0.14x+41.83=0.29y\\
--------------\\
0.14x+41.83=0.29\left( \boxed{x+89} \right)
\end{cases}
\\\\\\
0.14x+41.83=0.29x+25.81\implies 16.02=0.15x
\\\\\\
\cfrac{16.02}{0.15}=x\implies 106.8=x
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