Does convergence in distribution imply asymptotic stationarity?
Let ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ be a (possibly non-stationary) stochastic sequence of $d$-dimensional random vectors that converges in distribution. Does it immediately follow that the stochastic sequence
$[{bf tilde{x}}_1, {bf tilde{x}}_2], [{bf tilde{x}}_3, {bf tilde{x}}_4], ldots$ converges in distribution as well? Why or why not? More generally, if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_{t+1}, {bf tilde{x}}_{t+2}, ldots, {bf tilde{x}}_{t+m}$ converges in distribution as $t rightarrow infty$ for fixed positive integer $m$ where $m > 2$? My interest in this problem is that if I have an algorithm that is generating a non-stationary stochastic sequence which converges in distribution to a random vector I want to assess convergence by examining the autocorrelation function which assumes weak stationarity. My interpretation the asymptotic behavior of the stochastic sequence would be stronger if I a priori knew that the tail end of the generated stochastic sequence was asymptotically stationary in the strong sense.
time-series stochastic-processes convergence asymptotics
asked Dec 29 '18 at 14:48
RMG
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Let ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ be a (possibly non-stationary) stochastic sequence of $d$-dimensional random vectors that converges in distribution. Does it immediately follow that the stochastic sequence
$[{bf tilde{x}}_1, {bf tilde{x}}_2], [{bf tilde{x}}_3, {bf tilde{x}}_4], ldots$ converges in distribution as well? Why or why not? More generally, if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_{t+1}, {bf tilde{x}}_{t+2}, ldots, {bf tilde{x}}_{t+m}$ converges in distribution as $t rightarrow infty$ for fixed positive integer $m$ where $m > 2$? My interest in this problem is that if I have an algorithm that is generating a non-stationary stochastic sequence which converges in distribution to a random vector I want to assess convergence by examining the autocorrelation function which assumes weak stationarity. My interpretation the asymptotic behavior of the stochastic sequence would be stronger if I a priori knew that the tail end of the generated stochastic sequence was asymptotically stationary in the strong sense.
time-series stochastic-processes convergence asymptotics
asked Dec 29 '18 at 14:48
RMG
184
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1
"...if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ are independent and identically distributed?" - that would seem to preclude the existence of CLTs for non-iid sequences, which evidently is not the case.
– Christoph Hanck
Dec 29 '18 at 15:33
I fixed the question...
– RMG
Dec 29 '18 at 16:21
add a comment |
Let ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ be a (possibly non-stationary) stochastic sequence of $d$-dimensional random vectors that converges in distribution. Does it immediately follow that the stochastic sequence
$[{bf tilde{x}}_1, {bf tilde{x}}_2], [{bf tilde{x}}_3, {bf tilde{x}}_4], ldots$ converges in distribution as well? Why or why not? More generally, if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_{t+1}, {bf tilde{x}}_{t+2}, ldots, {bf tilde{x}}_{t+m}$ converges in distribution as $t rightarrow infty$ for fixed positive integer $m$ where $m > 2$? My interest in this problem is that if I have an algorithm that is generating a non-stationary stochastic sequence which converges in distribution to a random vector I want to assess convergence by examining the autocorrelation function which assumes weak stationarity. My interpretation the asymptotic behavior of the stochastic sequence would be stronger if I a priori knew that the tail end of the generated stochastic sequence was asymptotically stationary in the strong sense.
time-series stochastic-processes convergence asymptotics
asked Dec 29 '18 at 14:48
RMG
184
New contributor
RMG is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ be a (possibly non-stationary) stochastic sequence of $d$-dimensional random vectors that converges in distribution. Does it immediately follow that the stochastic sequence
$[{bf tilde{x}}_1, {bf tilde{x}}_2], [{bf tilde{x}}_3, {bf tilde{x}}_4], ldots$ converges in distribution as well? Why or why not? More generally, if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_{t+1}, {bf tilde{x}}_{t+2}, ldots, {bf tilde{x}}_{t+m}$ converges in distribution as $t rightarrow infty$ for fixed positive integer $m$ where $m > 2$? My interest in this problem is that if I have an algorithm that is generating a non-stationary stochastic sequence which converges in distribution to a random vector I want to assess convergence by examining the autocorrelation function which assumes weak stationarity. My interpretation the asymptotic behavior of the stochastic sequence would be stronger if I a priori knew that the tail end of the generated stochastic sequence was asymptotically stationary in the strong sense.
time-series stochastic-processes convergence asymptotics
time-series stochastic-processes convergence asymptotics
asked Dec 29 '18 at 14:48
RMG
184
New contributor
RMG is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 29 '18 at 14:48
RMG
184
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RMG is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 29 '18 at 16:24
asked Dec 29 '18 at 14:48
RMG
184
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asked Dec 29 '18 at 14:48
RMG
184
asked Dec 29 '18 at 14:48
RMG
184
184
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RMG is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
RMG is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1
"...if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ are independent and identically distributed?" - that would seem to preclude the existence of CLTs for non-iid sequences, which evidently is not the case.
– Christoph Hanck
Dec 29 '18 at 15:33
I fixed the question...
– RMG
Dec 29 '18 at 16:21
add a comment |
1
"...if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ are independent and identically distributed?" - that would seem to preclude the existence of CLTs for non-iid sequences, which evidently is not the case.
– Christoph Hanck
Dec 29 '18 at 15:33
I fixed the question...
– RMG
Dec 29 '18 at 16:21
1
1
"...if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ are independent and identically distributed?" - that would seem to preclude the existence of CLTs for non-iid sequences, which evidently is not the case.
– Christoph Hanck
Dec 29 '18 at 15:33
"...if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ are independent and identically distributed?" - that would seem to preclude the existence of CLTs for non-iid sequences, which evidently is not the case.
– Christoph Hanck
Dec 29 '18 at 15:33
I fixed the question...
– RMG
Dec 29 '18 at 16:21
I fixed the question...
– RMG
Dec 29 '18 at 16:21
add a comment |
1 Answer
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No.
Here is a counterexample to help you develop some intuition. It concerns a sequence of identically distributed variables with a common Bernoulli$(1/2)$ distribution. It is defined by letting $(X_{4i+1}, X_{4i+2}, X_{4i+3}, X_{4i+4})$ equal $(0,0,0,1)$ with probability $1/2$ and otherwise letting it equal $(1,1,1,0)$ with probability $1/2$ for $i=1,2,3,ldots.$
The variables $(X_{4i+1}, X_{4i+2})$ are always equal whereas the variables $(X_{4i+3}, X_{4i+4})$ always differ.
The sequence $(X_{2j+1}, X_{2j+2})$ thus has two convergent subsequences, but they converge to distinctly different bivariate distributions, demonstrating this sequence of bivariate random variables does not converge in distribution.
You should have no difficulty generalizing this counterexample to blocks of length $m$ within the original sequence of random variables.
answered Dec 29 '18 at 18:27
whuber♦
201k33437805
add a comment |
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No.
Here is a counterexample to help you develop some intuition. It concerns a sequence of identically distributed variables with a common Bernoulli$(1/2)$ distribution. It is defined by letting $(X_{4i+1}, X_{4i+2}, X_{4i+3}, X_{4i+4})$ equal $(0,0,0,1)$ with probability $1/2$ and otherwise letting it equal $(1,1,1,0)$ with probability $1/2$ for $i=1,2,3,ldots.$
The variables $(X_{4i+1}, X_{4i+2})$ are always equal whereas the variables $(X_{4i+3}, X_{4i+4})$ always differ.
The sequence $(X_{2j+1}, X_{2j+2})$ thus has two convergent subsequences, but they converge to distinctly different bivariate distributions, demonstrating this sequence of bivariate random variables does not converge in distribution.
You should have no difficulty generalizing this counterexample to blocks of length $m$ within the original sequence of random variables.
answered Dec 29 '18 at 18:27
whuber♦
201k33437805
add a comment |
No.
Here is a counterexample to help you develop some intuition. It concerns a sequence of identically distributed variables with a common Bernoulli$(1/2)$ distribution. It is defined by letting $(X_{4i+1}, X_{4i+2}, X_{4i+3}, X_{4i+4})$ equal $(0,0,0,1)$ with probability $1/2$ and otherwise letting it equal $(1,1,1,0)$ with probability $1/2$ for $i=1,2,3,ldots.$
The variables $(X_{4i+1}, X_{4i+2})$ are always equal whereas the variables $(X_{4i+3}, X_{4i+4})$ always differ.
The sequence $(X_{2j+1}, X_{2j+2})$ thus has two convergent subsequences, but they converge to distinctly different bivariate distributions, demonstrating this sequence of bivariate random variables does not converge in distribution.
You should have no difficulty generalizing this counterexample to blocks of length $m$ within the original sequence of random variables.
answered Dec 29 '18 at 18:27
whuber♦
201k33437805
add a comment |
No.
Here is a counterexample to help you develop some intuition. It concerns a sequence of identically distributed variables with a common Bernoulli$(1/2)$ distribution. It is defined by letting $(X_{4i+1}, X_{4i+2}, X_{4i+3}, X_{4i+4})$ equal $(0,0,0,1)$ with probability $1/2$ and otherwise letting it equal $(1,1,1,0)$ with probability $1/2$ for $i=1,2,3,ldots.$
The variables $(X_{4i+1}, X_{4i+2})$ are always equal whereas the variables $(X_{4i+3}, X_{4i+4})$ always differ.
The sequence $(X_{2j+1}, X_{2j+2})$ thus has two convergent subsequences, but they converge to distinctly different bivariate distributions, demonstrating this sequence of bivariate random variables does not converge in distribution.
You should have no difficulty generalizing this counterexample to blocks of length $m$ within the original sequence of random variables.
answered Dec 29 '18 at 18:27
whuber♦
201k33437805
No.
Here is a counterexample to help you develop some intuition. It concerns a sequence of identically distributed variables with a common Bernoulli$(1/2)$ distribution. It is defined by letting $(X_{4i+1}, X_{4i+2}, X_{4i+3}, X_{4i+4})$ equal $(0,0,0,1)$ with probability $1/2$ and otherwise letting it equal $(1,1,1,0)$ with probability $1/2$ for $i=1,2,3,ldots.$
The variables $(X_{4i+1}, X_{4i+2})$ are always equal whereas the variables $(X_{4i+3}, X_{4i+4})$ always differ.
The sequence $(X_{2j+1}, X_{2j+2})$ thus has two convergent subsequences, but they converge to distinctly different bivariate distributions, demonstrating this sequence of bivariate random variables does not converge in distribution.
You should have no difficulty generalizing this counterexample to blocks of length $m$ within the original sequence of random variables.
answered Dec 29 '18 at 18:27
whuber♦
201k33437805
answered Dec 29 '18 at 18:27
whuber♦
201k33437805
answered Dec 29 '18 at 18:27
whuber♦
201k33437805
answered Dec 29 '18 at 18:27
whuber♦
201k33437805
201k33437805
add a comment |
add a comment |
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1
"...if ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ converges in distribution, then does it follow that ${bf tilde{x}}_1, {bf tilde{x}}_2, ldots$ are independent and identically distributed?" - that would seem to preclude the existence of CLTs for non-iid sequences, which evidently is not the case.
– Christoph Hanck
Dec 29 '18 at 15:33
I fixed the question...
– RMG
Dec 29 '18 at 16:21