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Finding conditional and joint probabilities in a Markov Chain.

Probability distribution of markov chainProbabilities in markov chainMarkov chain and conditional entropyMarkov Chain Exercise.Markov Chain Example.Computing with initial distribution and transition matrix of a finite Markov chainQuestion about transition probabilities for Markov ChainGiven Markov chain transition matrix, calculate $mathbbP(X_1=3|X_2=1).$Getting started with Markov Chain.What would be the value of $P$ in this example?

0

$begingroup$

Let $X_0, X_1, ...$ be a Markov Chain with transition matrix
beginbmatrix
0 &1/2 & 1/2\
1&0&0\
1/3&1/3&1/3
endbmatrix

and initial distribution $α = (1/2, 0, 1/2)$.

Find the following:

(a) $P(X_2 = 1 | X_1 = 3)$

(b) $P(X_1 = 3, X_2 = 1)$

(c) $P(X_1 = 3 | X_2 = 1)$

(d) $P(X_9 = 1 | X_1 = 3, X_4 = 1, X_7 = 2)$

My solution:

(a) $P(X_2=1|X_1=3) = frac13$.

---

(b)

$P(X_2=1, X_1=3) Rightarrow P(X_2=1|X_1=3) cdot P(X_1=3)$

Now, $
alpha P = beginbmatrix frac12 & 0 & frac12 endbmatrixcdot
beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix
Rightarrow beginbmatrix frac16 & frac512 & frac512 endbmatrix
$

So, $
P(X_2=1, X_1=3) = frac13 cdot frac512 = frac536
$

---

(c)

$P(X_1=3|X_2=1) = fracP(X_2=1P(X_2=1)$

Now, $P(X_2=1|X_1=3) = frac13$

$alpha P = beginbmatrix frac16 & frac512 & frac512 endbmatrix$. So, $P(X_1=3) = frac512$.

$alpha P^2 = beginbmatrix frac16 & frac512 & frac512 endbmatrixcdot
beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix = beginbmatrix frac59 & frac29 & frac29 endbmatrix$. So, $P(X_2=1) = frac59$.

Hence, $P(X_1=3|X_2=1) = fracfrac13 cdot frac512frac59 = frac14$.

---

(d)

$P^2 = beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix cdot beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix =
beginbmatrix
frac23 & frac16 & frac16 \
0 & frac12 & frac12 \
frac49 & frac518 & frac518 \
endbmatrix$.

So, $P(X_9=1|X_1=3,X_4=1,X_7=2) = P(X_9=1|X_7=2) = 0$

---

Is my calculation correct?

random-variables markov-chains

share|cite|improve this question

edited Mar 15 at 21:21

user366312

asked Mar 15 at 2:18

user366312user366312

663318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Might be time to come up with titles that are more descriptive of the specific questions you’re asking.
  $endgroup$
  – amd
  Mar 15 at 6:09
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $X_0, X_1, ...$ be a Markov Chain with transition matrix
beginbmatrix
0 &1/2 & 1/2\
1&0&0\
1/3&1/3&1/3
endbmatrix

and initial distribution $α = (1/2, 0, 1/2)$.

Find the following:

(a) $P(X_2 = 1 | X_1 = 3)$

(b) $P(X_1 = 3, X_2 = 1)$

(c) $P(X_1 = 3 | X_2 = 1)$

(d) $P(X_9 = 1 | X_1 = 3, X_4 = 1, X_7 = 2)$

My solution:

(a) $P(X_2=1|X_1=3) = frac13$.

---

(b)

$P(X_2=1, X_1=3) Rightarrow P(X_2=1|X_1=3) cdot P(X_1=3)$

Now, $
alpha P = beginbmatrix frac12 & 0 & frac12 endbmatrixcdot
beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix
Rightarrow beginbmatrix frac16 & frac512 & frac512 endbmatrix
$

So, $
P(X_2=1, X_1=3) = frac13 cdot frac512 = frac536
$

---

(c)

$P(X_1=3|X_2=1) = fracP(X_2=1P(X_2=1)$

Now, $P(X_2=1|X_1=3) = frac13$

$alpha P = beginbmatrix frac16 & frac512 & frac512 endbmatrix$. So, $P(X_1=3) = frac512$.

$alpha P^2 = beginbmatrix frac16 & frac512 & frac512 endbmatrixcdot
beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix = beginbmatrix frac59 & frac29 & frac29 endbmatrix$. So, $P(X_2=1) = frac59$.

Hence, $P(X_1=3|X_2=1) = fracfrac13 cdot frac512frac59 = frac14$.

---

(d)

$P^2 = beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix cdot beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix =
beginbmatrix
frac23 & frac16 & frac16 \
0 & frac12 & frac12 \
frac49 & frac518 & frac518 \
endbmatrix$.

So, $P(X_9=1|X_1=3,X_4=1,X_7=2) = P(X_9=1|X_7=2) = 0$

---

Is my calculation correct?

random-variables markov-chains

share|cite|improve this question

edited Mar 15 at 21:21

user366312

asked Mar 15 at 2:18

user366312user366312

663318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Might be time to come up with titles that are more descriptive of the specific questions you’re asking.
  $endgroup$
  – amd
  Mar 15 at 6:09
  

  
  
  
  

add a comment | 

$begingroup$

Let $X_0, X_1, ...$ be a Markov Chain with transition matrix
beginbmatrix
0 &1/2 & 1/2\
1&0&0\
1/3&1/3&1/3
endbmatrix

and initial distribution $α = (1/2, 0, 1/2)$.

Find the following:

(a) $P(X_2 = 1 | X_1 = 3)$

(b) $P(X_1 = 3, X_2 = 1)$

(c) $P(X_1 = 3 | X_2 = 1)$

(d) $P(X_9 = 1 | X_1 = 3, X_4 = 1, X_7 = 2)$

My solution:

(a) $P(X_2=1|X_1=3) = frac13$.

---

(b)

$P(X_2=1, X_1=3) Rightarrow P(X_2=1|X_1=3) cdot P(X_1=3)$

Now, $
alpha P = beginbmatrix frac12 & 0 & frac12 endbmatrixcdot
beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix
Rightarrow beginbmatrix frac16 & frac512 & frac512 endbmatrix
$

So, $
P(X_2=1, X_1=3) = frac13 cdot frac512 = frac536
$

---

(c)

$P(X_1=3|X_2=1) = fracP(X_2=1P(X_2=1)$

Now, $P(X_2=1|X_1=3) = frac13$

$alpha P = beginbmatrix frac16 & frac512 & frac512 endbmatrix$. So, $P(X_1=3) = frac512$.

$alpha P^2 = beginbmatrix frac16 & frac512 & frac512 endbmatrixcdot
beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix = beginbmatrix frac59 & frac29 & frac29 endbmatrix$. So, $P(X_2=1) = frac59$.

Hence, $P(X_1=3|X_2=1) = fracfrac13 cdot frac512frac59 = frac14$.

---

(d)

$P^2 = beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix cdot beginbmatrix
0 & frac12 & frac12 \
1 & 0 & 0 \
frac13 & frac13 & frac13 \
endbmatrix =
beginbmatrix
frac23 & frac16 & frac16 \
0 & frac12 & frac12 \
frac49 & frac518 & frac518 \
endbmatrix$.

So, $P(X_9=1|X_1=3,X_4=1,X_7=2) = P(X_9=1|X_7=2) = 0$

---

Is my calculation correct?

random-variables markov-chains

share|cite|improve this question

edited Mar 15 at 21:21

user366312

asked Mar 15 at 2:18

user366312user366312

663318

$endgroup$

share|cite|improve this question

edited Mar 15 at 21:21

user366312

asked Mar 15 at 2:18

user366312user366312

663318

share|cite|improve this question

edited Mar 15 at 21:21

user366312

asked Mar 15 at 2:18

user366312user366312

663318

asked Mar 15 at 2:18

user366312user366312

663318

asked Mar 15 at 2:18

user366312user366312

663318