If $Y$ is a Markov chain and $h>0$, why is $(Y_{lfloor t/hrfloor})_{tge0}$ not a Markov process?
$begingroup$
Let $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a Markov chain for $ninmathbb N$, $(h_n)_{ninmathbb N}subseteq(0,infty)$ with $h_nxrightarrow{ntoinfty}infty$ and $$X^{(n)}_t:=Y^{(n)}_{lfloorfrac t{h_n}rfloor};;;text{for }tge0$$ for $ninmathbb N$.
Let $ninmathbb N$. In Remark 2 below Corollary 206 of these lecture notes, the author is noting that $X^{(n)}$ is not a Markov process.
I don't get that. If $mathcal F^Z$ denotes the filtration generated by a process $Z$, we should easily obtain $$mathcal F^{X^{(n)}}_t=mathcal F^{Y^{(n)}}_{lfloorfrac t{h_n}rfloor};;;text{for all }tge0tag1$$ and $$operatorname Pleft[X^{(n)}_tin Bmidmathcal F^{X^{(n)}}_sright]=operatorname Pleft[X^{(n)}_tin Bmid X^{(n)}_sright];;;text{almost surely}tag2.$$ So, $X^{(n)}$ should be a Markov process. What am I missing?
probability-theory stochastic-processes markov-chains markov-process stochastic-analysis
edited Dec 8 '18 at 18:44
Henning Makholm
240k17304541
asked Dec 8 '18 at 18:40
0xbadf00d0xbadf00d
1,89541531
$endgroup$
|
show 3 more comments
$begingroup$
Let $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a Markov chain for $ninmathbb N$, $(h_n)_{ninmathbb N}subseteq(0,infty)$ with $h_nxrightarrow{ntoinfty}infty$ and $$X^{(n)}_t:=Y^{(n)}_{lfloorfrac t{h_n}rfloor};;;text{for }tge0$$ for $ninmathbb N$.
Let $ninmathbb N$. In Remark 2 below Corollary 206 of these lecture notes, the author is noting that $X^{(n)}$ is not a Markov process.
I don't get that. If $mathcal F^Z$ denotes the filtration generated by a process $Z$, we should easily obtain $$mathcal F^{X^{(n)}}_t=mathcal F^{Y^{(n)}}_{lfloorfrac t{h_n}rfloor};;;text{for all }tge0tag1$$ and $$operatorname Pleft[X^{(n)}_tin Bmidmathcal F^{X^{(n)}}_sright]=operatorname Pleft[X^{(n)}_tin Bmid X^{(n)}_sright];;;text{almost surely}tag2.$$ So, $X^{(n)}$ should be a Markov process. What am I missing?
probability-theory stochastic-processes markov-chains markov-process stochastic-analysis
edited Dec 8 '18 at 18:44
Henning Makholm
240k17304541
asked Dec 8 '18 at 18:40
0xbadf00d0xbadf00d
1,89541531
$endgroup$
$begingroup$
What does the superscript $(n)$ in the notation $Y^{(n)}_k$ mean?
$endgroup$
– Henning Makholm
Dec 8 '18 at 18:47
$begingroup$
Shouldn't it be "not necessarily" Markov? An iid sequence is trivially Markovian.
$endgroup$
– BruceET
Dec 8 '18 at 18:48
$begingroup$
Also do your lecture notes assume that a "Markov process" must be homogenous in time?
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:10
1
$begingroup$
In particular, note that what Corollary 206 says is that $X_n$ is not a homogeneous Markov process.
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:16
1
$begingroup$
Yes the process is Markov and no, it is not homogeneous Markov. This is exactly what the authors (rightfully) mean.
$endgroup$
– Did
Dec 9 '18 at 9:48
|
show 3 more comments
$begingroup$
Let $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a Markov chain for $ninmathbb N$, $(h_n)_{ninmathbb N}subseteq(0,infty)$ with $h_nxrightarrow{ntoinfty}infty$ and $$X^{(n)}_t:=Y^{(n)}_{lfloorfrac t{h_n}rfloor};;;text{for }tge0$$ for $ninmathbb N$.
Let $ninmathbb N$. In Remark 2 below Corollary 206 of these lecture notes, the author is noting that $X^{(n)}$ is not a Markov process.
I don't get that. If $mathcal F^Z$ denotes the filtration generated by a process $Z$, we should easily obtain $$mathcal F^{X^{(n)}}_t=mathcal F^{Y^{(n)}}_{lfloorfrac t{h_n}rfloor};;;text{for all }tge0tag1$$ and $$operatorname Pleft[X^{(n)}_tin Bmidmathcal F^{X^{(n)}}_sright]=operatorname Pleft[X^{(n)}_tin Bmid X^{(n)}_sright];;;text{almost surely}tag2.$$ So, $X^{(n)}$ should be a Markov process. What am I missing?
probability-theory stochastic-processes markov-chains markov-process stochastic-analysis
edited Dec 8 '18 at 18:44
Henning Makholm
240k17304541
asked Dec 8 '18 at 18:40
0xbadf00d0xbadf00d
1,89541531
$endgroup$
Let $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a Markov chain for $ninmathbb N$, $(h_n)_{ninmathbb N}subseteq(0,infty)$ with $h_nxrightarrow{ntoinfty}infty$ and $$X^{(n)}_t:=Y^{(n)}_{lfloorfrac t{h_n}rfloor};;;text{for }tge0$$ for $ninmathbb N$.
Let $ninmathbb N$. In Remark 2 below Corollary 206 of these lecture notes, the author is noting that $X^{(n)}$ is not a Markov process.
I don't get that. If $mathcal F^Z$ denotes the filtration generated by a process $Z$, we should easily obtain $$mathcal F^{X^{(n)}}_t=mathcal F^{Y^{(n)}}_{lfloorfrac t{h_n}rfloor};;;text{for all }tge0tag1$$ and $$operatorname Pleft[X^{(n)}_tin Bmidmathcal F^{X^{(n)}}_sright]=operatorname Pleft[X^{(n)}_tin Bmid X^{(n)}_sright];;;text{almost surely}tag2.$$ So, $X^{(n)}$ should be a Markov process. What am I missing?
probability-theory stochastic-processes markov-chains markov-process stochastic-analysis
probability-theory stochastic-processes markov-chains markov-process stochastic-analysis
edited Dec 8 '18 at 18:44
Henning Makholm
240k17304541
asked Dec 8 '18 at 18:40
0xbadf00d0xbadf00d
1,89541531
edited Dec 8 '18 at 18:44
Henning Makholm
240k17304541
asked Dec 8 '18 at 18:40
0xbadf00d0xbadf00d
1,89541531
edited Dec 8 '18 at 18:44
Henning Makholm
240k17304541
edited Dec 8 '18 at 18:44
Henning Makholm
240k17304541
edited Dec 8 '18 at 18:44
Henning Makholm
240k17304541
240k17304541
asked Dec 8 '18 at 18:40
0xbadf00d0xbadf00d
1,89541531
asked Dec 8 '18 at 18:40
0xbadf00d0xbadf00d
1,89541531
asked Dec 8 '18 at 18:40
0xbadf00d0xbadf00d
1,89541531
1,89541531
$begingroup$
What does the superscript $(n)$ in the notation $Y^{(n)}_k$ mean?
$endgroup$
– Henning Makholm
Dec 8 '18 at 18:47
$begingroup$
Shouldn't it be "not necessarily" Markov? An iid sequence is trivially Markovian.
$endgroup$
– BruceET
Dec 8 '18 at 18:48
$begingroup$
Also do your lecture notes assume that a "Markov process" must be homogenous in time?
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:10
1
$begingroup$
In particular, note that what Corollary 206 says is that $X_n$ is not a homogeneous Markov process.
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:16
1
$begingroup$
Yes the process is Markov and no, it is not homogeneous Markov. This is exactly what the authors (rightfully) mean.
$endgroup$
– Did
Dec 9 '18 at 9:48
|
show 3 more comments
$begingroup$
What does the superscript $(n)$ in the notation $Y^{(n)}_k$ mean?
$endgroup$
– Henning Makholm
Dec 8 '18 at 18:47
$begingroup$
Shouldn't it be "not necessarily" Markov? An iid sequence is trivially Markovian.
$endgroup$
– BruceET
Dec 8 '18 at 18:48
$begingroup$
Also do your lecture notes assume that a "Markov process" must be homogenous in time?
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:10
1
$begingroup$
In particular, note that what Corollary 206 says is that $X_n$ is not a homogeneous Markov process.
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:16
1
$begingroup$
Yes the process is Markov and no, it is not homogeneous Markov. This is exactly what the authors (rightfully) mean.
$endgroup$
– Did
Dec 9 '18 at 9:48
$begingroup$
What does the superscript $(n)$ in the notation $Y^{(n)}_k$ mean?
$endgroup$
– Henning Makholm
Dec 8 '18 at 18:47
$begingroup$
What does the superscript $(n)$ in the notation $Y^{(n)}_k$ mean?
$endgroup$
– Henning Makholm
Dec 8 '18 at 18:47
$begingroup$
Shouldn't it be "not necessarily" Markov? An iid sequence is trivially Markovian.
$endgroup$
– BruceET
Dec 8 '18 at 18:48
$begingroup$
Shouldn't it be "not necessarily" Markov? An iid sequence is trivially Markovian.
$endgroup$
– BruceET
Dec 8 '18 at 18:48
$begingroup$
Also do your lecture notes assume that a "Markov process" must be homogenous in time?
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:10
$begingroup$
Also do your lecture notes assume that a "Markov process" must be homogenous in time?
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:10
1
1
$begingroup$
In particular, note that what Corollary 206 says is that $X_n$ is not a homogeneous Markov process.
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:16
$begingroup$
In particular, note that what Corollary 206 says is that $X_n$ is not a homogeneous Markov process.
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:16
1
1
$begingroup$
Yes the process is Markov and no, it is not homogeneous Markov. This is exactly what the authors (rightfully) mean.
$endgroup$
– Did
Dec 9 '18 at 9:48
$begingroup$
Yes the process is Markov and no, it is not homogeneous Markov. This is exactly what the authors (rightfully) mean.
$endgroup$
– Did
Dec 9 '18 at 9:48
|
show 3 more comments
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$begingroup$
What does the superscript $(n)$ in the notation $Y^{(n)}_k$ mean?
$endgroup$
– Henning Makholm
Dec 8 '18 at 18:47
$begingroup$
Shouldn't it be "not necessarily" Markov? An iid sequence is trivially Markovian.
$endgroup$
– BruceET
Dec 8 '18 at 18:48
$begingroup$
Also do your lecture notes assume that a "Markov process" must be homogenous in time?
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:10
1
$begingroup$
In particular, note that what Corollary 206 says is that $X_n$ is not a homogeneous Markov process.
$endgroup$
– Henning Makholm
Dec 8 '18 at 19:16
1
$begingroup$
Yes the process is Markov and no, it is not homogeneous Markov. This is exactly what the authors (rightfully) mean.
$endgroup$
– Did
Dec 9 '18 at 9:48