What is the stationary distribution of the following Markov chain?












0












$begingroup$




Consider a chain with state space ${1,2, cdots }.$ If you are at $1$ go to state $j$ with probability $p_j$ $($$ j=1,2,cdots$ $) ,$ where these are non-negative numbers adding to $1$. If you are in a state $i>1 ,$ then go just one step back$,$ that is$,$ to $i-1.$ Discuss the nature of the states and nature of the stationary distribution$.$





Assuming that all $p_j$'s are positive I found that the Markov chain is irreducible with all it's states recurrent$.$ So there is no transient state and hence inessential state in the above Markov chain$.$ While calculating stationary distribution I found a problem$.$ Here it is $:$



Suppose $pi = (pi_1 , pi_2 , cdots )$ be the stationary distribution of the above Markov chain$.$ Then I found that $pi_2 = (1-p_1) pi_1, pi_3 = (1-p_1-p_2) pi_1, cdots$
Since $sumlimits_{i=1}^{infty} pi_i = 1$ so $pi_1 {1 + (1-p_1) + (1-p_1 -p_2) + (1-p_1 -p_2 - p_3) + cdots }= 1.$ Now how do I find the sum



$$1 + sum_{k=1}^{infty} (1-p_1-p_2 -p_3 - cdots - p_k) ?$$



Or in other words



$$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j ?$$



This gives us $pi_1$ and consequently all the $pi_i$'s$.$ where



$$pi_i = frac {sumlimits_{k=i}^{infty} p_k} {sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j}.$$



for $i=1,2,cdots.$



Please help me in this regard. Thank you very much.






probability stochastic-processes markov-chains







share|cite|improve this question











$endgroup$












  • $begingroup$
    I get $sum_{j=1}^infty{jp_j}$ The value of this will depend on the $p_j$ of course. I think you've solved the problem already.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:02










  • $begingroup$
    Whose value is $sum_{j=1}^{infty} jp_j$?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:12










  • $begingroup$
    $1+(1-p_1)+(1-p_1-p_2)+(1-p_1-p_2-p_3)+dots=sum_{j=1}^{infty} jp_j$
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:17










  • $begingroup$
    How? I got $$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j.$$ You may check. How do you get your expression? Would you please share @saulspatz?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:29










  • $begingroup$
    Yeah I have understood. Anyway same expression. Actually I love multiple sums.😀
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:30


















0












$begingroup$




Consider a chain with state space ${1,2, cdots }.$ If you are at $1$ go to state $j$ with probability $p_j$ $($$ j=1,2,cdots$ $) ,$ where these are non-negative numbers adding to $1$. If you are in a state $i>1 ,$ then go just one step back$,$ that is$,$ to $i-1.$ Discuss the nature of the states and nature of the stationary distribution$.$





Assuming that all $p_j$'s are positive I found that the Markov chain is irreducible with all it's states recurrent$.$ So there is no transient state and hence inessential state in the above Markov chain$.$ While calculating stationary distribution I found a problem$.$ Here it is $:$



Suppose $pi = (pi_1 , pi_2 , cdots )$ be the stationary distribution of the above Markov chain$.$ Then I found that $pi_2 = (1-p_1) pi_1, pi_3 = (1-p_1-p_2) pi_1, cdots$
Since $sumlimits_{i=1}^{infty} pi_i = 1$ so $pi_1 {1 + (1-p_1) + (1-p_1 -p_2) + (1-p_1 -p_2 - p_3) + cdots }= 1.$ Now how do I find the sum



$$1 + sum_{k=1}^{infty} (1-p_1-p_2 -p_3 - cdots - p_k) ?$$



Or in other words



$$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j ?$$



This gives us $pi_1$ and consequently all the $pi_i$'s$.$ where



$$pi_i = frac {sumlimits_{k=i}^{infty} p_k} {sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j}.$$



for $i=1,2,cdots.$



Please help me in this regard. Thank you very much.






probability stochastic-processes markov-chains







share|cite|improve this question











$endgroup$












  • $begingroup$
    I get $sum_{j=1}^infty{jp_j}$ The value of this will depend on the $p_j$ of course. I think you've solved the problem already.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:02










  • $begingroup$
    Whose value is $sum_{j=1}^{infty} jp_j$?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:12










  • $begingroup$
    $1+(1-p_1)+(1-p_1-p_2)+(1-p_1-p_2-p_3)+dots=sum_{j=1}^{infty} jp_j$
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:17










  • $begingroup$
    How? I got $$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j.$$ You may check. How do you get your expression? Would you please share @saulspatz?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:29










  • $begingroup$
    Yeah I have understood. Anyway same expression. Actually I love multiple sums.😀
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:30
















0












0








0





$begingroup$




Consider a chain with state space ${1,2, cdots }.$ If you are at $1$ go to state $j$ with probability $p_j$ $($$ j=1,2,cdots$ $) ,$ where these are non-negative numbers adding to $1$. If you are in a state $i>1 ,$ then go just one step back$,$ that is$,$ to $i-1.$ Discuss the nature of the states and nature of the stationary distribution$.$





Assuming that all $p_j$'s are positive I found that the Markov chain is irreducible with all it's states recurrent$.$ So there is no transient state and hence inessential state in the above Markov chain$.$ While calculating stationary distribution I found a problem$.$ Here it is $:$



Suppose $pi = (pi_1 , pi_2 , cdots )$ be the stationary distribution of the above Markov chain$.$ Then I found that $pi_2 = (1-p_1) pi_1, pi_3 = (1-p_1-p_2) pi_1, cdots$
Since $sumlimits_{i=1}^{infty} pi_i = 1$ so $pi_1 {1 + (1-p_1) + (1-p_1 -p_2) + (1-p_1 -p_2 - p_3) + cdots }= 1.$ Now how do I find the sum



$$1 + sum_{k=1}^{infty} (1-p_1-p_2 -p_3 - cdots - p_k) ?$$



Or in other words



$$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j ?$$



This gives us $pi_1$ and consequently all the $pi_i$'s$.$ where



$$pi_i = frac {sumlimits_{k=i}^{infty} p_k} {sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j}.$$



for $i=1,2,cdots.$



Please help me in this regard. Thank you very much.






probability stochastic-processes markov-chains







share|cite|improve this question











$endgroup$






Consider a chain with state space ${1,2, cdots }.$ If you are at $1$ go to state $j$ with probability $p_j$ $($$ j=1,2,cdots$ $) ,$ where these are non-negative numbers adding to $1$. If you are in a state $i>1 ,$ then go just one step back$,$ that is$,$ to $i-1.$ Discuss the nature of the states and nature of the stationary distribution$.$





Assuming that all $p_j$'s are positive I found that the Markov chain is irreducible with all it's states recurrent$.$ So there is no transient state and hence inessential state in the above Markov chain$.$ While calculating stationary distribution I found a problem$.$ Here it is $:$



Suppose $pi = (pi_1 , pi_2 , cdots )$ be the stationary distribution of the above Markov chain$.$ Then I found that $pi_2 = (1-p_1) pi_1, pi_3 = (1-p_1-p_2) pi_1, cdots$
Since $sumlimits_{i=1}^{infty} pi_i = 1$ so $pi_1 {1 + (1-p_1) + (1-p_1 -p_2) + (1-p_1 -p_2 - p_3) + cdots }= 1.$ Now how do I find the sum



$$1 + sum_{k=1}^{infty} (1-p_1-p_2 -p_3 - cdots - p_k) ?$$



Or in other words



$$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j ?$$



This gives us $pi_1$ and consequently all the $pi_i$'s$.$ where



$$pi_i = frac {sumlimits_{k=i}^{infty} p_k} {sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j}.$$



for $i=1,2,cdots.$



Please help me in this regard. Thank you very much.






probability stochastic-processes markov-chains




probability stochastic-processes markov-chains






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 15:14







Dbchatto67

















asked Dec 5 '18 at 14:34









Dbchatto67Dbchatto67

536116




536116












  • $begingroup$
    I get $sum_{j=1}^infty{jp_j}$ The value of this will depend on the $p_j$ of course. I think you've solved the problem already.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:02










  • $begingroup$
    Whose value is $sum_{j=1}^{infty} jp_j$?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:12










  • $begingroup$
    $1+(1-p_1)+(1-p_1-p_2)+(1-p_1-p_2-p_3)+dots=sum_{j=1}^{infty} jp_j$
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:17










  • $begingroup$
    How? I got $$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j.$$ You may check. How do you get your expression? Would you please share @saulspatz?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:29










  • $begingroup$
    Yeah I have understood. Anyway same expression. Actually I love multiple sums.😀
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:30




















  • $begingroup$
    I get $sum_{j=1}^infty{jp_j}$ The value of this will depend on the $p_j$ of course. I think you've solved the problem already.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:02










  • $begingroup$
    Whose value is $sum_{j=1}^{infty} jp_j$?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:12










  • $begingroup$
    $1+(1-p_1)+(1-p_1-p_2)+(1-p_1-p_2-p_3)+dots=sum_{j=1}^{infty} jp_j$
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:17










  • $begingroup$
    How? I got $$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j.$$ You may check. How do you get your expression? Would you please share @saulspatz?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:29










  • $begingroup$
    Yeah I have understood. Anyway same expression. Actually I love multiple sums.😀
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:30


















$begingroup$
I get $sum_{j=1}^infty{jp_j}$ The value of this will depend on the $p_j$ of course. I think you've solved the problem already.
$endgroup$
– saulspatz
Dec 5 '18 at 15:02




$begingroup$
I get $sum_{j=1}^infty{jp_j}$ The value of this will depend on the $p_j$ of course. I think you've solved the problem already.
$endgroup$
– saulspatz
Dec 5 '18 at 15:02












$begingroup$
Whose value is $sum_{j=1}^{infty} jp_j$?
$endgroup$
– Dbchatto67
Dec 5 '18 at 15:12




$begingroup$
Whose value is $sum_{j=1}^{infty} jp_j$?
$endgroup$
– Dbchatto67
Dec 5 '18 at 15:12












$begingroup$
$1+(1-p_1)+(1-p_1-p_2)+(1-p_1-p_2-p_3)+dots=sum_{j=1}^{infty} jp_j$
$endgroup$
– saulspatz
Dec 5 '18 at 15:17




$begingroup$
$1+(1-p_1)+(1-p_1-p_2)+(1-p_1-p_2-p_3)+dots=sum_{j=1}^{infty} jp_j$
$endgroup$
– saulspatz
Dec 5 '18 at 15:17












$begingroup$
How? I got $$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j.$$ You may check. How do you get your expression? Would you please share @saulspatz?
$endgroup$
– Dbchatto67
Dec 5 '18 at 15:29




$begingroup$
How? I got $$sumlimits_{k=1}^{infty} sumlimits_{j=k}^{infty} p_j.$$ You may check. How do you get your expression? Would you please share @saulspatz?
$endgroup$
– Dbchatto67
Dec 5 '18 at 15:29












$begingroup$
Yeah I have understood. Anyway same expression. Actually I love multiple sums.😀
$endgroup$
– Dbchatto67
Dec 5 '18 at 15:30






$begingroup$
Yeah I have understood. Anyway same expression. Actually I love multiple sums.😀
$endgroup$
– Dbchatto67
Dec 5 '18 at 15:30












1 Answer
1






active

oldest

votes


















1












$begingroup$

$$begin{align}
sum_{j=1}^infty{jp_j}&=sum_{j=1}^infty{p_j}+sum_{j=2}^infty{(j-1)p_j}\
&=1+sum_{j=2}^infty{p_j}+sum_{j=3}^infty{(j-2)p_j}\
&=1+(1-p_1)+sum_{j=3}^infty{p_j}+sum_{j=4}^infty{(j-3)p_j}\
&vdots
end{align}$$



EDIT



This is true if the given sum converges. As discussed in the accepted answer to this question, it need not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have understood it already. Thanks for your answer.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:37










  • $begingroup$
    @Dbchatto67 Right, I'd just finished typing it when I saw your comment, so I went ahead and posted it. An interesting point is that this problem shows if $p_jge0$ and $sum{p_j}=1,$ then $sum{jp_j}$ converges. Is there a more direct proof?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:44












  • $begingroup$
    Hmm. Good observation. May be there are direct approaches but I am not familiar with those. Sorry for that. Do you know some of them? Then please don't forget to share your thoughts. That will also be helpful for me.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:54












  • $begingroup$
    @Dbchatto67 There must be an error here. Look at math.stackexchange.com/questions/3027282/…
    $endgroup$
    – saulspatz
    Dec 5 '18 at 16:30










  • $begingroup$
    Then what is the error lying in the question?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 16:36











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

$$begin{align}
sum_{j=1}^infty{jp_j}&=sum_{j=1}^infty{p_j}+sum_{j=2}^infty{(j-1)p_j}\
&=1+sum_{j=2}^infty{p_j}+sum_{j=3}^infty{(j-2)p_j}\
&=1+(1-p_1)+sum_{j=3}^infty{p_j}+sum_{j=4}^infty{(j-3)p_j}\
&vdots
end{align}$$



EDIT



This is true if the given sum converges. As discussed in the accepted answer to this question, it need not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have understood it already. Thanks for your answer.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:37










  • $begingroup$
    @Dbchatto67 Right, I'd just finished typing it when I saw your comment, so I went ahead and posted it. An interesting point is that this problem shows if $p_jge0$ and $sum{p_j}=1,$ then $sum{jp_j}$ converges. Is there a more direct proof?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:44












  • $begingroup$
    Hmm. Good observation. May be there are direct approaches but I am not familiar with those. Sorry for that. Do you know some of them? Then please don't forget to share your thoughts. That will also be helpful for me.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:54












  • $begingroup$
    @Dbchatto67 There must be an error here. Look at math.stackexchange.com/questions/3027282/…
    $endgroup$
    – saulspatz
    Dec 5 '18 at 16:30










  • $begingroup$
    Then what is the error lying in the question?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 16:36
















1












$begingroup$

$$begin{align}
sum_{j=1}^infty{jp_j}&=sum_{j=1}^infty{p_j}+sum_{j=2}^infty{(j-1)p_j}\
&=1+sum_{j=2}^infty{p_j}+sum_{j=3}^infty{(j-2)p_j}\
&=1+(1-p_1)+sum_{j=3}^infty{p_j}+sum_{j=4}^infty{(j-3)p_j}\
&vdots
end{align}$$



EDIT



This is true if the given sum converges. As discussed in the accepted answer to this question, it need not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have understood it already. Thanks for your answer.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:37










  • $begingroup$
    @Dbchatto67 Right, I'd just finished typing it when I saw your comment, so I went ahead and posted it. An interesting point is that this problem shows if $p_jge0$ and $sum{p_j}=1,$ then $sum{jp_j}$ converges. Is there a more direct proof?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:44












  • $begingroup$
    Hmm. Good observation. May be there are direct approaches but I am not familiar with those. Sorry for that. Do you know some of them? Then please don't forget to share your thoughts. That will also be helpful for me.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:54












  • $begingroup$
    @Dbchatto67 There must be an error here. Look at math.stackexchange.com/questions/3027282/…
    $endgroup$
    – saulspatz
    Dec 5 '18 at 16:30










  • $begingroup$
    Then what is the error lying in the question?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 16:36














1












1








1





$begingroup$

$$begin{align}
sum_{j=1}^infty{jp_j}&=sum_{j=1}^infty{p_j}+sum_{j=2}^infty{(j-1)p_j}\
&=1+sum_{j=2}^infty{p_j}+sum_{j=3}^infty{(j-2)p_j}\
&=1+(1-p_1)+sum_{j=3}^infty{p_j}+sum_{j=4}^infty{(j-3)p_j}\
&vdots
end{align}$$



EDIT



This is true if the given sum converges. As discussed in the accepted answer to this question, it need not.






share|cite|improve this answer











$endgroup$



$$begin{align}
sum_{j=1}^infty{jp_j}&=sum_{j=1}^infty{p_j}+sum_{j=2}^infty{(j-1)p_j}\
&=1+sum_{j=2}^infty{p_j}+sum_{j=3}^infty{(j-2)p_j}\
&=1+(1-p_1)+sum_{j=3}^infty{p_j}+sum_{j=4}^infty{(j-3)p_j}\
&vdots
end{align}$$



EDIT



This is true if the given sum converges. As discussed in the accepted answer to this question, it need not.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 18:00

























answered Dec 5 '18 at 15:36









saulspatzsaulspatz

14.4k21329




14.4k21329












  • $begingroup$
    I have understood it already. Thanks for your answer.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:37










  • $begingroup$
    @Dbchatto67 Right, I'd just finished typing it when I saw your comment, so I went ahead and posted it. An interesting point is that this problem shows if $p_jge0$ and $sum{p_j}=1,$ then $sum{jp_j}$ converges. Is there a more direct proof?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:44












  • $begingroup$
    Hmm. Good observation. May be there are direct approaches but I am not familiar with those. Sorry for that. Do you know some of them? Then please don't forget to share your thoughts. That will also be helpful for me.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:54












  • $begingroup$
    @Dbchatto67 There must be an error here. Look at math.stackexchange.com/questions/3027282/…
    $endgroup$
    – saulspatz
    Dec 5 '18 at 16:30










  • $begingroup$
    Then what is the error lying in the question?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 16:36


















  • $begingroup$
    I have understood it already. Thanks for your answer.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:37










  • $begingroup$
    @Dbchatto67 Right, I'd just finished typing it when I saw your comment, so I went ahead and posted it. An interesting point is that this problem shows if $p_jge0$ and $sum{p_j}=1,$ then $sum{jp_j}$ converges. Is there a more direct proof?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 15:44












  • $begingroup$
    Hmm. Good observation. May be there are direct approaches but I am not familiar with those. Sorry for that. Do you know some of them? Then please don't forget to share your thoughts. That will also be helpful for me.
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 15:54












  • $begingroup$
    @Dbchatto67 There must be an error here. Look at math.stackexchange.com/questions/3027282/…
    $endgroup$
    – saulspatz
    Dec 5 '18 at 16:30










  • $begingroup$
    Then what is the error lying in the question?
    $endgroup$
    – Dbchatto67
    Dec 5 '18 at 16:36
















$begingroup$
I have understood it already. Thanks for your answer.
$endgroup$
– Dbchatto67
Dec 5 '18 at 15:37




$begingroup$
I have understood it already. Thanks for your answer.
$endgroup$
– Dbchatto67
Dec 5 '18 at 15:37












$begingroup$
@Dbchatto67 Right, I'd just finished typing it when I saw your comment, so I went ahead and posted it. An interesting point is that this problem shows if $p_jge0$ and $sum{p_j}=1,$ then $sum{jp_j}$ converges. Is there a more direct proof?
$endgroup$
– saulspatz
Dec 5 '18 at 15:44






$begingroup$
@Dbchatto67 Right, I'd just finished typing it when I saw your comment, so I went ahead and posted it. An interesting point is that this problem shows if $p_jge0$ and $sum{p_j}=1,$ then $sum{jp_j}$ converges. Is there a more direct proof?
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– saulspatz
Dec 5 '18 at 15:44














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Hmm. Good observation. May be there are direct approaches but I am not familiar with those. Sorry for that. Do you know some of them? Then please don't forget to share your thoughts. That will also be helpful for me.
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– Dbchatto67
Dec 5 '18 at 15:54






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Hmm. Good observation. May be there are direct approaches but I am not familiar with those. Sorry for that. Do you know some of them? Then please don't forget to share your thoughts. That will also be helpful for me.
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– Dbchatto67
Dec 5 '18 at 15:54














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@Dbchatto67 There must be an error here. Look at math.stackexchange.com/questions/3027282/…
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– saulspatz
Dec 5 '18 at 16:30




$begingroup$
@Dbchatto67 There must be an error here. Look at math.stackexchange.com/questions/3027282/…
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– saulspatz
Dec 5 '18 at 16:30












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Then what is the error lying in the question?
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– Dbchatto67
Dec 5 '18 at 16:36




$begingroup$
Then what is the error lying in the question?
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– Dbchatto67
Dec 5 '18 at 16:36


















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