Mixing
Show that $mathbb{P}(T>n)=p(1-q)^{n-2},$ for $ngeq 2.$
$begingroup$
Given the state space $S={1,2}$ and the transition matrix for
the two-state chain is given by
$$mathbf{P}=begin{pmatrix} 1-p & q \ p & 1-q end{pmatrix},$$
where $p$ and $q$ are not both $0$. Let $T$ be the first return time
to state $1$, for the chain started in $1$.
a) Show that $mathbb{P}(Tgeq n)=p(1-q)^{n-2},$ for $ngeq 2.$
b) Find $mathbb{E}[T]$ and verify that $mathbb{E}[T]=1/pi_1$ where $pi$ is the stationary distribution of the chain.
a) Since this is a discrete case, we have that
$$mathbb{P}(Tgeq n)=1-mathbb{P}(Tleq n)=1-(mathbb{P}(T=2)+ ... +mathbb{P}(T=n)),$$
but how do I compute the $mathbb{P}(T=k)$?
b) If I can obtain an expression for $mathbb{P}(T> n)$, and given that $T$ is a discrete random variable, i can compute this expectation as
$$mathbb{E}[T]=sum_{k=2}^{n}mathbb{P}(T>k),$$
is this correct? If it is, then I only need help to proceed with a).
markov-chains
asked Dec 30 '18 at 16:12
ParsevalParseval
2,7791718
$endgroup$
add a comment |
$begingroup$
Given the state space $S={1,2}$ and the transition matrix for
the two-state chain is given by
$$mathbf{P}=begin{pmatrix} 1-p & q \ p & 1-q end{pmatrix},$$
where $p$ and $q$ are not both $0$. Let $T$ be the first return time
to state $1$, for the chain started in $1$.
a) Show that $mathbb{P}(Tgeq n)=p(1-q)^{n-2},$ for $ngeq 2.$
b) Find $mathbb{E}[T]$ and verify that $mathbb{E}[T]=1/pi_1$ where $pi$ is the stationary distribution of the chain.
a) Since this is a discrete case, we have that
$$mathbb{P}(Tgeq n)=1-mathbb{P}(Tleq n)=1-(mathbb{P}(T=2)+ ... +mathbb{P}(T=n)),$$
but how do I compute the $mathbb{P}(T=k)$?
b) If I can obtain an expression for $mathbb{P}(T> n)$, and given that $T$ is a discrete random variable, i can compute this expectation as
$$mathbb{E}[T]=sum_{k=2}^{n}mathbb{P}(T>k),$$
is this correct? If it is, then I only need help to proceed with a).
markov-chains
asked Dec 30 '18 at 16:12
ParsevalParseval
2,7791718
$endgroup$
$begingroup$
Why would the chain go to state 2 immediately? It might as well stay in state 1 with probability $1-p.$ I don't understand where the geometric series comes in to play. what are you summing?
$endgroup$
– Parseval
Dec 30 '18 at 16:55
$begingroup$
I overlooked the possibility of staying in state 1 for a while. Still you have the probability that it stays in state 1 for $k$ steps before transitioning to state 2, then stays in state 2 for $n-1-k$ steps at least, then transitions back to state 1. For each $k$ the sum of the probabilities is a geometric series.
$endgroup$
– saulspatz
Dec 30 '18 at 17:04
add a comment |
$begingroup$
Given the state space $S={1,2}$ and the transition matrix for
the two-state chain is given by
$$mathbf{P}=begin{pmatrix} 1-p & q \ p & 1-q end{pmatrix},$$
where $p$ and $q$ are not both $0$. Let $T$ be the first return time
to state $1$, for the chain started in $1$.
a) Show that $mathbb{P}(Tgeq n)=p(1-q)^{n-2},$ for $ngeq 2.$
b) Find $mathbb{E}[T]$ and verify that $mathbb{E}[T]=1/pi_1$ where $pi$ is the stationary distribution of the chain.
a) Since this is a discrete case, we have that
$$mathbb{P}(Tgeq n)=1-mathbb{P}(Tleq n)=1-(mathbb{P}(T=2)+ ... +mathbb{P}(T=n)),$$
but how do I compute the $mathbb{P}(T=k)$?
b) If I can obtain an expression for $mathbb{P}(T> n)$, and given that $T$ is a discrete random variable, i can compute this expectation as
$$mathbb{E}[T]=sum_{k=2}^{n}mathbb{P}(T>k),$$
is this correct? If it is, then I only need help to proceed with a).
markov-chains
asked Dec 30 '18 at 16:12
ParsevalParseval
2,7791718
$endgroup$
Given the state space $S={1,2}$ and the transition matrix for
the two-state chain is given by
$$mathbf{P}=begin{pmatrix} 1-p & q \ p & 1-q end{pmatrix},$$
where $p$ and $q$ are not both $0$. Let $T$ be the first return time
to state $1$, for the chain started in $1$.
a) Show that $mathbb{P}(Tgeq n)=p(1-q)^{n-2},$ for $ngeq 2.$
b) Find $mathbb{E}[T]$ and verify that $mathbb{E}[T]=1/pi_1$ where $pi$ is the stationary distribution of the chain.
a) Since this is a discrete case, we have that
$$mathbb{P}(Tgeq n)=1-mathbb{P}(Tleq n)=1-(mathbb{P}(T=2)+ ... +mathbb{P}(T=n)),$$
but how do I compute the $mathbb{P}(T=k)$?
b) If I can obtain an expression for $mathbb{P}(T> n)$, and given that $T$ is a discrete random variable, i can compute this expectation as
$$mathbb{E}[T]=sum_{k=2}^{n}mathbb{P}(T>k),$$
is this correct? If it is, then I only need help to proceed with a).
markov-chains
markov-chains
asked Dec 30 '18 at 16:12
ParsevalParseval
2,7791718
asked Dec 30 '18 at 16:12
ParsevalParseval
2,7791718
edited Dec 30 '18 at 16:31
Parseval
asked Dec 30 '18 at 16:12
ParsevalParseval
2,7791718
asked Dec 30 '18 at 16:12
ParsevalParseval
2,7791718
asked Dec 30 '18 at 16:12
ParsevalParseval
2,7791718
2,7791718
$begingroup$
Why would the chain go to state 2 immediately? It might as well stay in state 1 with probability $1-p.$ I don't understand where the geometric series comes in to play. what are you summing?
$endgroup$
– Parseval
Dec 30 '18 at 16:55
$begingroup$
I overlooked the possibility of staying in state 1 for a while. Still you have the probability that it stays in state 1 for $k$ steps before transitioning to state 2, then stays in state 2 for $n-1-k$ steps at least, then transitions back to state 1. For each $k$ the sum of the probabilities is a geometric series.
$endgroup$
– saulspatz
Dec 30 '18 at 17:04
add a comment |
$begingroup$
Why would the chain go to state 2 immediately? It might as well stay in state 1 with probability $1-p.$ I don't understand where the geometric series comes in to play. what are you summing?
$endgroup$
– Parseval
Dec 30 '18 at 16:55
$begingroup$
I overlooked the possibility of staying in state 1 for a while. Still you have the probability that it stays in state 1 for $k$ steps before transitioning to state 2, then stays in state 2 for $n-1-k$ steps at least, then transitions back to state 1. For each $k$ the sum of the probabilities is a geometric series.
$endgroup$
– saulspatz
Dec 30 '18 at 17:04
$begingroup$
Why would the chain go to state 2 immediately? It might as well stay in state 1 with probability $1-p.$ I don't understand where the geometric series comes in to play. what are you summing?
$endgroup$
– Parseval
Dec 30 '18 at 16:55
$begingroup$
Why would the chain go to state 2 immediately? It might as well stay in state 1 with probability $1-p.$ I don't understand where the geometric series comes in to play. what are you summing?
$endgroup$
– Parseval
Dec 30 '18 at 16:55
$begingroup$
I overlooked the possibility of staying in state 1 for a while. Still you have the probability that it stays in state 1 for $k$ steps before transitioning to state 2, then stays in state 2 for $n-1-k$ steps at least, then transitions back to state 1. For each $k$ the sum of the probabilities is a geometric series.
$endgroup$
– saulspatz
Dec 30 '18 at 17:04
$begingroup$
I overlooked the possibility of staying in state 1 for a while. Still you have the probability that it stays in state 1 for $k$ steps before transitioning to state 2, then stays in state 2 for $n-1-k$ steps at least, then transitions back to state 1. For each $k$ the sum of the probabilities is a geometric series.
$endgroup$
– saulspatz
Dec 30 '18 at 17:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$T$ is a random variable taking values in the non-negative integers and infinity(the chain never returns to one).
In this case, we have the formula : $mathbb E[T] = sum_{k=1}^infty P(T geq k)$. Therefore, from part $a$ one can obtain the result of part $b$.
As for part $a$ itself, naturally $T geq n$ if and only if the chain is in state two from time $2$ to at least time $n-1$. This is by definition of what first return time means : the first return time to $1$ starting from $1$ is greater than or equal to $n$, if and only if the chain does not go to state $1$ from time $2$ to $n-1$, if and only if (in our case) the chain is in state two from time $2$ to time $n-1$.
Your comment does not apply : if the transition $1to 1$ occurs, then the chain will have returned to state $1$ at time $1$, so the return time is $1$. Remember, the first return time does not take into account repeats and then leaving and coming back : a repeat is a return time of $1$.
But, for the chain to be in state two from time $2$ to $n-1$ it has to transition like :
$$
1 xrightarrow{1} 2 xrightarrow{2} 2 xrightarrow{3} 2 xrightarrow{4} ... xrightarrow{n-2} 2 xrightarrow{n-1} mbox{anywhere}
$$
which means that the desired probability is the probability that the given transitions occur in that order, which is easily seen to be probability of $1 to 2$, which is $q$, times probability of $2 to 2$ exactly $n-2$ times, which is $(1-q)$. This results in $q times (1-q)^{n-2}$.
Therefore, by our expectation formula, the expected return time is $P(T geq 1) + sum_{k=2}^infty P(T geq k)$, which leads to $1 + qsum_{n=2}^infty(1-q)^{n-2}$ (The splitting of $k = 1$ and $kgeq 2$ is important here : $T geq 1$ happens with probability $1$). By the geometric series formula, this equals $1+frac qq = 2$.
Now, the stationary distribution for the given Markov chain can be found by definition of a stationary distribution: $$(pi_1,pi_2) = ((1-p)pi_1+ppi_2,qpi_1+(1-q)pi_2)$$
Noting that $pi_1+pi_2 = 1$ allows us to get $pi = (frac 12,frac 12)$. Verify part $b$ from here.
answered Jan 1 at 12:39
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
$endgroup$
$begingroup$
Thanks for the great explanation of the first return time. But if we start on $1$ and go to $2$, that is a probability of $q$, then we need to stay at state $2$ till time $n-1$. I don't see why you get a $p$ in the expression for the given transitions to occur? As I see it, at time $1$, we are at state $1$, so how can we stay in state $2$ form time $1$ as you have written?
$endgroup$
– Parseval
Jan 1 at 20:42
1
$begingroup$
I had made a few errors. Please check the edit. The question as stated has a wrong conclusion, but my idea was right.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 3:26
$begingroup$
No, your idea was right AND the conclusion as well. I made a typo in my transition matrix. Sorry about that, I'm editing now. Thanks for your help!
$endgroup$
– Parseval
Jan 2 at 14:02
$begingroup$
Welcome! I will edit the answer as well.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 15:42
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$T$ is a random variable taking values in the non-negative integers and infinity(the chain never returns to one).
In this case, we have the formula : $mathbb E[T] = sum_{k=1}^infty P(T geq k)$. Therefore, from part $a$ one can obtain the result of part $b$.
As for part $a$ itself, naturally $T geq n$ if and only if the chain is in state two from time $2$ to at least time $n-1$. This is by definition of what first return time means : the first return time to $1$ starting from $1$ is greater than or equal to $n$, if and only if the chain does not go to state $1$ from time $2$ to $n-1$, if and only if (in our case) the chain is in state two from time $2$ to time $n-1$.
Your comment does not apply : if the transition $1to 1$ occurs, then the chain will have returned to state $1$ at time $1$, so the return time is $1$. Remember, the first return time does not take into account repeats and then leaving and coming back : a repeat is a return time of $1$.
But, for the chain to be in state two from time $2$ to $n-1$ it has to transition like :
$$
1 xrightarrow{1} 2 xrightarrow{2} 2 xrightarrow{3} 2 xrightarrow{4} ... xrightarrow{n-2} 2 xrightarrow{n-1} mbox{anywhere}
$$
which means that the desired probability is the probability that the given transitions occur in that order, which is easily seen to be probability of $1 to 2$, which is $q$, times probability of $2 to 2$ exactly $n-2$ times, which is $(1-q)$. This results in $q times (1-q)^{n-2}$.
Therefore, by our expectation formula, the expected return time is $P(T geq 1) + sum_{k=2}^infty P(T geq k)$, which leads to $1 + qsum_{n=2}^infty(1-q)^{n-2}$ (The splitting of $k = 1$ and $kgeq 2$ is important here : $T geq 1$ happens with probability $1$). By the geometric series formula, this equals $1+frac qq = 2$.
Now, the stationary distribution for the given Markov chain can be found by definition of a stationary distribution: $$(pi_1,pi_2) = ((1-p)pi_1+ppi_2,qpi_1+(1-q)pi_2)$$
Noting that $pi_1+pi_2 = 1$ allows us to get $pi = (frac 12,frac 12)$. Verify part $b$ from here.
answered Jan 1 at 12:39
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
$endgroup$
$begingroup$
Thanks for the great explanation of the first return time. But if we start on $1$ and go to $2$, that is a probability of $q$, then we need to stay at state $2$ till time $n-1$. I don't see why you get a $p$ in the expression for the given transitions to occur? As I see it, at time $1$, we are at state $1$, so how can we stay in state $2$ form time $1$ as you have written?
$endgroup$
– Parseval
Jan 1 at 20:42
1
$begingroup$
I had made a few errors. Please check the edit. The question as stated has a wrong conclusion, but my idea was right.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 3:26
$begingroup$
No, your idea was right AND the conclusion as well. I made a typo in my transition matrix. Sorry about that, I'm editing now. Thanks for your help!
$endgroup$
– Parseval
Jan 2 at 14:02
$begingroup$
Welcome! I will edit the answer as well.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 15:42
add a comment |
$begingroup$
$T$ is a random variable taking values in the non-negative integers and infinity(the chain never returns to one).
In this case, we have the formula : $mathbb E[T] = sum_{k=1}^infty P(T geq k)$. Therefore, from part $a$ one can obtain the result of part $b$.
As for part $a$ itself, naturally $T geq n$ if and only if the chain is in state two from time $2$ to at least time $n-1$. This is by definition of what first return time means : the first return time to $1$ starting from $1$ is greater than or equal to $n$, if and only if the chain does not go to state $1$ from time $2$ to $n-1$, if and only if (in our case) the chain is in state two from time $2$ to time $n-1$.
Your comment does not apply : if the transition $1to 1$ occurs, then the chain will have returned to state $1$ at time $1$, so the return time is $1$. Remember, the first return time does not take into account repeats and then leaving and coming back : a repeat is a return time of $1$.
But, for the chain to be in state two from time $2$ to $n-1$ it has to transition like :
$$
1 xrightarrow{1} 2 xrightarrow{2} 2 xrightarrow{3} 2 xrightarrow{4} ... xrightarrow{n-2} 2 xrightarrow{n-1} mbox{anywhere}
$$
which means that the desired probability is the probability that the given transitions occur in that order, which is easily seen to be probability of $1 to 2$, which is $q$, times probability of $2 to 2$ exactly $n-2$ times, which is $(1-q)$. This results in $q times (1-q)^{n-2}$.
Therefore, by our expectation formula, the expected return time is $P(T geq 1) + sum_{k=2}^infty P(T geq k)$, which leads to $1 + qsum_{n=2}^infty(1-q)^{n-2}$ (The splitting of $k = 1$ and $kgeq 2$ is important here : $T geq 1$ happens with probability $1$). By the geometric series formula, this equals $1+frac qq = 2$.
Now, the stationary distribution for the given Markov chain can be found by definition of a stationary distribution: $$(pi_1,pi_2) = ((1-p)pi_1+ppi_2,qpi_1+(1-q)pi_2)$$
Noting that $pi_1+pi_2 = 1$ allows us to get $pi = (frac 12,frac 12)$. Verify part $b$ from here.
answered Jan 1 at 12:39
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
$endgroup$
$begingroup$
Thanks for the great explanation of the first return time. But if we start on $1$ and go to $2$, that is a probability of $q$, then we need to stay at state $2$ till time $n-1$. I don't see why you get a $p$ in the expression for the given transitions to occur? As I see it, at time $1$, we are at state $1$, so how can we stay in state $2$ form time $1$ as you have written?
$endgroup$
– Parseval
Jan 1 at 20:42
1
$begingroup$
I had made a few errors. Please check the edit. The question as stated has a wrong conclusion, but my idea was right.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 3:26
$begingroup$
No, your idea was right AND the conclusion as well. I made a typo in my transition matrix. Sorry about that, I'm editing now. Thanks for your help!
$endgroup$
– Parseval
Jan 2 at 14:02
$begingroup$
Welcome! I will edit the answer as well.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 15:42
add a comment |
$begingroup$
$T$ is a random variable taking values in the non-negative integers and infinity(the chain never returns to one).
In this case, we have the formula : $mathbb E[T] = sum_{k=1}^infty P(T geq k)$. Therefore, from part $a$ one can obtain the result of part $b$.
As for part $a$ itself, naturally $T geq n$ if and only if the chain is in state two from time $2$ to at least time $n-1$. This is by definition of what first return time means : the first return time to $1$ starting from $1$ is greater than or equal to $n$, if and only if the chain does not go to state $1$ from time $2$ to $n-1$, if and only if (in our case) the chain is in state two from time $2$ to time $n-1$.
Your comment does not apply : if the transition $1to 1$ occurs, then the chain will have returned to state $1$ at time $1$, so the return time is $1$. Remember, the first return time does not take into account repeats and then leaving and coming back : a repeat is a return time of $1$.
But, for the chain to be in state two from time $2$ to $n-1$ it has to transition like :
$$
1 xrightarrow{1} 2 xrightarrow{2} 2 xrightarrow{3} 2 xrightarrow{4} ... xrightarrow{n-2} 2 xrightarrow{n-1} mbox{anywhere}
$$
which means that the desired probability is the probability that the given transitions occur in that order, which is easily seen to be probability of $1 to 2$, which is $q$, times probability of $2 to 2$ exactly $n-2$ times, which is $(1-q)$. This results in $q times (1-q)^{n-2}$.
Therefore, by our expectation formula, the expected return time is $P(T geq 1) + sum_{k=2}^infty P(T geq k)$, which leads to $1 + qsum_{n=2}^infty(1-q)^{n-2}$ (The splitting of $k = 1$ and $kgeq 2$ is important here : $T geq 1$ happens with probability $1$). By the geometric series formula, this equals $1+frac qq = 2$.
Now, the stationary distribution for the given Markov chain can be found by definition of a stationary distribution: $$(pi_1,pi_2) = ((1-p)pi_1+ppi_2,qpi_1+(1-q)pi_2)$$
Noting that $pi_1+pi_2 = 1$ allows us to get $pi = (frac 12,frac 12)$. Verify part $b$ from here.
answered Jan 1 at 12:39
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
$endgroup$
$T$ is a random variable taking values in the non-negative integers and infinity(the chain never returns to one).
In this case, we have the formula : $mathbb E[T] = sum_{k=1}^infty P(T geq k)$. Therefore, from part $a$ one can obtain the result of part $b$.
As for part $a$ itself, naturally $T geq n$ if and only if the chain is in state two from time $2$ to at least time $n-1$. This is by definition of what first return time means : the first return time to $1$ starting from $1$ is greater than or equal to $n$, if and only if the chain does not go to state $1$ from time $2$ to $n-1$, if and only if (in our case) the chain is in state two from time $2$ to time $n-1$.
Your comment does not apply : if the transition $1to 1$ occurs, then the chain will have returned to state $1$ at time $1$, so the return time is $1$. Remember, the first return time does not take into account repeats and then leaving and coming back : a repeat is a return time of $1$.
But, for the chain to be in state two from time $2$ to $n-1$ it has to transition like :
$$
1 xrightarrow{1} 2 xrightarrow{2} 2 xrightarrow{3} 2 xrightarrow{4} ... xrightarrow{n-2} 2 xrightarrow{n-1} mbox{anywhere}
$$
which means that the desired probability is the probability that the given transitions occur in that order, which is easily seen to be probability of $1 to 2$, which is $q$, times probability of $2 to 2$ exactly $n-2$ times, which is $(1-q)$. This results in $q times (1-q)^{n-2}$.
Therefore, by our expectation formula, the expected return time is $P(T geq 1) + sum_{k=2}^infty P(T geq k)$, which leads to $1 + qsum_{n=2}^infty(1-q)^{n-2}$ (The splitting of $k = 1$ and $kgeq 2$ is important here : $T geq 1$ happens with probability $1$). By the geometric series formula, this equals $1+frac qq = 2$.
Now, the stationary distribution for the given Markov chain can be found by definition of a stationary distribution: $$(pi_1,pi_2) = ((1-p)pi_1+ppi_2,qpi_1+(1-q)pi_2)$$
Noting that $pi_1+pi_2 = 1$ allows us to get $pi = (frac 12,frac 12)$. Verify part $b$ from here.
answered Jan 1 at 12:39
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
edited Jan 2 at 3:26
answered Jan 1 at 12:39
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
answered Jan 1 at 12:39
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
answered Jan 1 at 12:39
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
37.4k33376
$begingroup$
Thanks for the great explanation of the first return time. But if we start on $1$ and go to $2$, that is a probability of $q$, then we need to stay at state $2$ till time $n-1$. I don't see why you get a $p$ in the expression for the given transitions to occur? As I see it, at time $1$, we are at state $1$, so how can we stay in state $2$ form time $1$ as you have written?
$endgroup$
– Parseval
Jan 1 at 20:42
1
$begingroup$
I had made a few errors. Please check the edit. The question as stated has a wrong conclusion, but my idea was right.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 3:26
$begingroup$
No, your idea was right AND the conclusion as well. I made a typo in my transition matrix. Sorry about that, I'm editing now. Thanks for your help!
$endgroup$
– Parseval
Jan 2 at 14:02
$begingroup$
Welcome! I will edit the answer as well.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 15:42
add a comment |
$begingroup$
Thanks for the great explanation of the first return time. But if we start on $1$ and go to $2$, that is a probability of $q$, then we need to stay at state $2$ till time $n-1$. I don't see why you get a $p$ in the expression for the given transitions to occur? As I see it, at time $1$, we are at state $1$, so how can we stay in state $2$ form time $1$ as you have written?
$endgroup$
– Parseval
Jan 1 at 20:42
1
$begingroup$
I had made a few errors. Please check the edit. The question as stated has a wrong conclusion, but my idea was right.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 3:26
$begingroup$
No, your idea was right AND the conclusion as well. I made a typo in my transition matrix. Sorry about that, I'm editing now. Thanks for your help!
$endgroup$
– Parseval
Jan 2 at 14:02
$begingroup$
Welcome! I will edit the answer as well.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 15:42
$begingroup$
Thanks for the great explanation of the first return time. But if we start on $1$ and go to $2$, that is a probability of $q$, then we need to stay at state $2$ till time $n-1$. I don't see why you get a $p$ in the expression for the given transitions to occur? As I see it, at time $1$, we are at state $1$, so how can we stay in state $2$ form time $1$ as you have written?
$endgroup$
– Parseval
Jan 1 at 20:42
$begingroup$
Thanks for the great explanation of the first return time. But if we start on $1$ and go to $2$, that is a probability of $q$, then we need to stay at state $2$ till time $n-1$. I don't see why you get a $p$ in the expression for the given transitions to occur? As I see it, at time $1$, we are at state $1$, so how can we stay in state $2$ form time $1$ as you have written?
$endgroup$
– Parseval
Jan 1 at 20:42
1
1
$begingroup$
I had made a few errors. Please check the edit. The question as stated has a wrong conclusion, but my idea was right.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 3:26
$begingroup$
I had made a few errors. Please check the edit. The question as stated has a wrong conclusion, but my idea was right.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 3:26
$begingroup$
No, your idea was right AND the conclusion as well. I made a typo in my transition matrix. Sorry about that, I'm editing now. Thanks for your help!
$endgroup$
– Parseval
Jan 2 at 14:02
$begingroup$
No, your idea was right AND the conclusion as well. I made a typo in my transition matrix. Sorry about that, I'm editing now. Thanks for your help!
$endgroup$
– Parseval
Jan 2 at 14:02
$begingroup$
Welcome! I will edit the answer as well.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 15:42
$begingroup$
Welcome! I will edit the answer as well.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 2 at 15:42
add a comment |
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$begingroup$
Why would the chain go to state 2 immediately? It might as well stay in state 1 with probability $1-p.$ I don't understand where the geometric series comes in to play. what are you summing?
$endgroup$
– Parseval
Dec 30 '18 at 16:55
$begingroup$
I overlooked the possibility of staying in state 1 for a while. Still you have the probability that it stays in state 1 for $k$ steps before transitioning to state 2, then stays in state 2 for $n-1-k$ steps at least, then transitions back to state 1. For each $k$ the sum of the probabilities is a geometric series.
$endgroup$
– saulspatz
Dec 30 '18 at 17:04