Find a semigroup, stationary distribution and calculate a probability
Let ${{X_t | tge 0}}$ be a Markov Process on a state space $S={{1,2,3}}$ with a generator
$$ G=begin{bmatrix}
-1 & 0 & 1 \
3 & -4 & 1 \
2 & 0 & -2
end{bmatrix}$$
(a) Find a semigroup ${{ P_t | t ge 0}}$ and find a stationary distribution
(b) Calculate $P(X_{3t}=1 | X_{5t}=3, X_{0}=2, X_{4t}=3, X_{t}=2)$
(a) $G=B;A;B^{-1}$. I've got a formula that $P_t=sum_{n=0}^{infty} frac{t^n}{n!}A^n=B(sum_{n=0}^{infty} frac{t^n}{n!}A^n)B^{-1}$, so I'm diagonalizing my matrix G, substitute to my formula and that's all.
According to stationary distribution we have that $pi G=0$, so $(pi_1,pi_2,pi_3)G=(0,0,0)$, and from that I've found that $(pi_1,pi_2,pi_3)=(2/3,0,1/3)$, because $sum pi_i=1$.
(b)I suppose that I need some formulas to calculate this probability but I can't recollect any.
Please, correct me where I'm wrong and help me with a subpoint (b).
probability-theory stochastic-processes markov-process
asked Dec 26 '18 at 11:39
MacAbraMacAbra
17519
add a comment |
Let ${{X_t | tge 0}}$ be a Markov Process on a state space $S={{1,2,3}}$ with a generator
$$ G=begin{bmatrix}
-1 & 0 & 1 \
3 & -4 & 1 \
2 & 0 & -2
end{bmatrix}$$
(a) Find a semigroup ${{ P_t | t ge 0}}$ and find a stationary distribution
(b) Calculate $P(X_{3t}=1 | X_{5t}=3, X_{0}=2, X_{4t}=3, X_{t}=2)$
(a) $G=B;A;B^{-1}$. I've got a formula that $P_t=sum_{n=0}^{infty} frac{t^n}{n!}A^n=B(sum_{n=0}^{infty} frac{t^n}{n!}A^n)B^{-1}$, so I'm diagonalizing my matrix G, substitute to my formula and that's all.
According to stationary distribution we have that $pi G=0$, so $(pi_1,pi_2,pi_3)G=(0,0,0)$, and from that I've found that $(pi_1,pi_2,pi_3)=(2/3,0,1/3)$, because $sum pi_i=1$.
(b)I suppose that I need some formulas to calculate this probability but I can't recollect any.
Please, correct me where I'm wrong and help me with a subpoint (b).
probability-theory stochastic-processes markov-process
asked Dec 26 '18 at 11:39
MacAbraMacAbra
17519
add a comment |
Let ${{X_t | tge 0}}$ be a Markov Process on a state space $S={{1,2,3}}$ with a generator
$$ G=begin{bmatrix}
-1 & 0 & 1 \
3 & -4 & 1 \
2 & 0 & -2
end{bmatrix}$$
(a) Find a semigroup ${{ P_t | t ge 0}}$ and find a stationary distribution
(b) Calculate $P(X_{3t}=1 | X_{5t}=3, X_{0}=2, X_{4t}=3, X_{t}=2)$
(a) $G=B;A;B^{-1}$. I've got a formula that $P_t=sum_{n=0}^{infty} frac{t^n}{n!}A^n=B(sum_{n=0}^{infty} frac{t^n}{n!}A^n)B^{-1}$, so I'm diagonalizing my matrix G, substitute to my formula and that's all.
According to stationary distribution we have that $pi G=0$, so $(pi_1,pi_2,pi_3)G=(0,0,0)$, and from that I've found that $(pi_1,pi_2,pi_3)=(2/3,0,1/3)$, because $sum pi_i=1$.
(b)I suppose that I need some formulas to calculate this probability but I can't recollect any.
Please, correct me where I'm wrong and help me with a subpoint (b).
probability-theory stochastic-processes markov-process
asked Dec 26 '18 at 11:39
MacAbraMacAbra
17519
Let ${{X_t | tge 0}}$ be a Markov Process on a state space $S={{1,2,3}}$ with a generator
$$ G=begin{bmatrix}
-1 & 0 & 1 \
3 & -4 & 1 \
2 & 0 & -2
end{bmatrix}$$
(a) Find a semigroup ${{ P_t | t ge 0}}$ and find a stationary distribution
(b) Calculate $P(X_{3t}=1 | X_{5t}=3, X_{0}=2, X_{4t}=3, X_{t}=2)$
(a) $G=B;A;B^{-1}$. I've got a formula that $P_t=sum_{n=0}^{infty} frac{t^n}{n!}A^n=B(sum_{n=0}^{infty} frac{t^n}{n!}A^n)B^{-1}$, so I'm diagonalizing my matrix G, substitute to my formula and that's all.
According to stationary distribution we have that $pi G=0$, so $(pi_1,pi_2,pi_3)G=(0,0,0)$, and from that I've found that $(pi_1,pi_2,pi_3)=(2/3,0,1/3)$, because $sum pi_i=1$.
(b)I suppose that I need some formulas to calculate this probability but I can't recollect any.
Please, correct me where I'm wrong and help me with a subpoint (b).
probability-theory stochastic-processes markov-process
probability-theory stochastic-processes markov-process
asked Dec 26 '18 at 11:39
MacAbraMacAbra
17519
asked Dec 26 '18 at 11:39
MacAbraMacAbra
17519
edited 2 days ago
MacAbra
asked Dec 26 '18 at 11:39
MacAbraMacAbra
17519
asked Dec 26 '18 at 11:39
MacAbraMacAbra
17519
asked Dec 26 '18 at 11:39
MacAbraMacAbra
17519
17519
add a comment |
add a comment |
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