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Consider the Markov chain in Example 3.10.2 with initial

probability vector \(v = \left( {\frac{1}{2},\frac{1}{2}} \right)\) Where \(p = \left[ {\begin{array}{*{20}{c}}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}\end{array}} \right]\)

a.Find the probability vector specifying the probabilities

of the states at timen=2.

b.Find the two-step transition matrix

Short Answer

Expert verified
  1. The probabilities of the states at a time\(n = 2\)is

\(\left[ {\begin{array}{{}{}}{\frac{1}{6}}&{\frac{4}{9}}\\{\frac{1}{9}}&{\frac{1}{6}}\end{array}} \right]\)

  1. The two-step transition probability matrix is

\(\left[ {\begin{array}{{}{}}{\frac{1}{3}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{1}{3}}\end{array}} \right]\)

Step by step solution

01

Given Information

The initial probability vector is \(v = \left( {\frac{1}{2},\frac{1}{2}} \right)\) , and the probability matrix is \(p = \left[ {\begin{array}{{}{}}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}\end{array}} \right]\) .

02

State the vector and matrix

For this case, we first calculate the two-step transition probability matrix and then calculate for states at the time \(n = 2\) .

03

Find the probability vector of the states at the time \(n = 2\) 

  1. So firstly, calculate the two-step transition probability matrix from the given probability matrix. Then the two-step transition probability matrix is given by

\(\begin{aligned}{}{p^2} &= p \times p\\ &= \left[ {\begin{aligned}{{}{}}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}\end{aligned}} \right] \times \left[ {\begin{aligned}{{}{}}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{\frac{1}{3}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{1}{3}}\end{aligned}} \right]\end{aligned}\)

Therefore, to calculate the states at the time, \(n = 2\) we have to multiply the two-step transition probability matrix with the initial vector. Then the state is defined as \(\begin{aligned}{}v \times {p^2} &= \left[ {\frac{1}{2},\frac{1}{2}} \right] \times \left[ {\begin{aligned}{{}{}}{\frac{1}{3}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{1}{3}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{\frac{1}{6}}&{\frac{4}{9}}\\{\frac{1}{9}}&{\frac{1}{6}}\end{aligned}} \right]\end{aligned}\)

04

Compute the two-step transition probability matrix

For the two-step transition probability matrix, the matrix will be defined as

\(\begin{array}{p^{\left( 2 \right)}} = {p^{\left( 1 \right)}} \times {p^{\left( 1 \right)}}\\ = \left( {\begin{array}{}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}\end{array}} \right) \times \left( {\begin{array}{}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}\end{array}} \right)\\ = \left( {\begin{array}{}{\frac{1}{3}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{1}{3}}\end{array}} \right)\end{array}\)

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Most popular questions from this chapter

Suppose that the p.d.f. of X is as follows:

\(\begin{aligned}f\left( x \right) &= e{}^{ - x},x > 0\\ &= 0,x \le 0\end{aligned}\)

Determine the p.d.f. of \({\bf{Y = }}{{\bf{X}}^{\frac{{\bf{1}}}{{\bf{2}}}}}\)

Question:A certain drugstore has three public telephone booths. Fori=0, 1, 2, 3, let\({{\bf{p}}_{\bf{i}}}\)denote the probability that exactlyitelephone booths will be occupied on any Monday evening at 8:00 p.m.; and suppose that\({{\bf{p}}_{\bf{0}}}\)=0.1,\({{\bf{p}}_{\bf{1}}}\)=0.2,\({{\bf{p}}_{\bf{2}}}\)=0.4, and\({{\bf{p}}_{\bf{3}}}\)=0.3. LetXandYdenote the number of booths that will be occupied at 8:00 p.m. on two independent Monday evenings. Determine:

(a) the joint p.f. ofXandY;

(b) Pr(X=Y);

(c) Pr(X > Y ).

For the conditions of Exercise 1, find the p.d.f. of the

average \(\frac{{\left( {{{\bf{X}}_{\bf{1}}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}} \right)}}{{\bf{2}}}\)

Question:Suppose thatXandYhave a continuous joint distribution

for which the joint p.d.f. is defined as follows:

\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{\bf{3}}}{{\bf{2}}}{{\bf{y}}^{\bf{2}}}\;{\bf{for}}\;{\bf{0}} \le {\bf{x}} \le {\bf{2}}\;{\bf{and}}\;{\bf{0}} \le {\bf{y}} \le {\bf{1}}\\{\bf{0}}\;\,{\bf{otherwise}}\end{array} \right.\)

a. Determine the marginal p.d.f.โ€™s ofXandY.

b. AreXandYindependent?

c. Are the event{X<1}and the event\(\left\{ {{\bf{Y}} \ge \frac{{\bf{1}}}{{\bf{2}}}} \right\}\)independent?

An insurance agent sells a policy that has a \(100 deductible

and a \)5000 cap. When the policyholder files a claim, the policyholder must pay the first \(100. After the first \)100, the insurance company pays therest of the claim up to a maximum payment of $5000. Any

excess must be paid by the policyholder. Suppose that thedollar amount X of a claim has a continuous distribution

with p.d.f. \({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\frac{{\bf{1}}}{{{{\left( {{\bf{1 + x}}} \right)}^{\bf{2}}}}}\) for x>0 and 0 otherwise.

LetY be the insurance company's amount to payon the claim.

a. Write Y as a function of X, i.e., \({\bf{Y = r}}\left( {\bf{X}} \right).\)

b. Find the c.d.f. of Y.

c. Explain why Y has neither a continuous nor a discretedistribution.

See all solutions

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