Mixing
Convergence of $X_n(t_n)$ when $X_n(t)$ is weakly convergent to a continuous process
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Suppose a càdlàg univariate process $X_n(t)$, $tinmathbb{R}_+$ converges weakly to an a.s. continuous process $X(t)$ as $nuparrowinfty$, where $X(t)$ can be described by a SDE. Suppose that $X(t)$ has a stationary distribution.
Question: For arbitrary sequence ${t_n}_n$, consider ${X_n(t_n)}_n$,
1) If $t_nuparrowinfty$ as $nuparrowinfty$, does $X_n(t_n)$ converge to the stationary distribution of $X(t)$ as $nuparrowinfty$?
2) If $t_nto t_0$ for a fixed $t_0<infty$, does $X_n(t_n)$ weakly converge to $X(t_0)$?
Obviously, continuous mapping theorem implies $X(t_n)$ weakly converges to $X(t_0)$, but I am not sure how this would help here.
I would appreciate any comments on how to prove the two things above rigorously. Thank you.
probability probability-theory stochastic-processes stochastic-calculus
asked Jan 11 at 14:03
NickNick
527
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add a comment |
$begingroup$
Suppose a càdlàg univariate process $X_n(t)$, $tinmathbb{R}_+$ converges weakly to an a.s. continuous process $X(t)$ as $nuparrowinfty$, where $X(t)$ can be described by a SDE. Suppose that $X(t)$ has a stationary distribution.
Question: For arbitrary sequence ${t_n}_n$, consider ${X_n(t_n)}_n$,
1) If $t_nuparrowinfty$ as $nuparrowinfty$, does $X_n(t_n)$ converge to the stationary distribution of $X(t)$ as $nuparrowinfty$?
2) If $t_nto t_0$ for a fixed $t_0<infty$, does $X_n(t_n)$ weakly converge to $X(t_0)$?
Obviously, continuous mapping theorem implies $X(t_n)$ weakly converges to $X(t_0)$, but I am not sure how this would help here.
I would appreciate any comments on how to prove the two things above rigorously. Thank you.
probability probability-theory stochastic-processes stochastic-calculus
asked Jan 11 at 14:03
NickNick
527
$endgroup$
add a comment |
$begingroup$
Suppose a càdlàg univariate process $X_n(t)$, $tinmathbb{R}_+$ converges weakly to an a.s. continuous process $X(t)$ as $nuparrowinfty$, where $X(t)$ can be described by a SDE. Suppose that $X(t)$ has a stationary distribution.
Question: For arbitrary sequence ${t_n}_n$, consider ${X_n(t_n)}_n$,
1) If $t_nuparrowinfty$ as $nuparrowinfty$, does $X_n(t_n)$ converge to the stationary distribution of $X(t)$ as $nuparrowinfty$?
2) If $t_nto t_0$ for a fixed $t_0<infty$, does $X_n(t_n)$ weakly converge to $X(t_0)$?
Obviously, continuous mapping theorem implies $X(t_n)$ weakly converges to $X(t_0)$, but I am not sure how this would help here.
I would appreciate any comments on how to prove the two things above rigorously. Thank you.
probability probability-theory stochastic-processes stochastic-calculus
asked Jan 11 at 14:03
NickNick
527
$endgroup$
Suppose a càdlàg univariate process $X_n(t)$, $tinmathbb{R}_+$ converges weakly to an a.s. continuous process $X(t)$ as $nuparrowinfty$, where $X(t)$ can be described by a SDE. Suppose that $X(t)$ has a stationary distribution.
Question: For arbitrary sequence ${t_n}_n$, consider ${X_n(t_n)}_n$,
1) If $t_nuparrowinfty$ as $nuparrowinfty$, does $X_n(t_n)$ converge to the stationary distribution of $X(t)$ as $nuparrowinfty$?
2) If $t_nto t_0$ for a fixed $t_0<infty$, does $X_n(t_n)$ weakly converge to $X(t_0)$?
Obviously, continuous mapping theorem implies $X(t_n)$ weakly converges to $X(t_0)$, but I am not sure how this would help here.
I would appreciate any comments on how to prove the two things above rigorously. Thank you.
probability probability-theory stochastic-processes stochastic-calculus
probability probability-theory stochastic-processes stochastic-calculus
asked Jan 11 at 14:03
NickNick
527
asked Jan 11 at 14:03
NickNick
527
asked Jan 11 at 14:03
NickNick
527
asked Jan 11 at 14:03
NickNick
527
asked Jan 11 at 14:03
NickNick
527
527
add a comment |
add a comment |
1 Answer
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I am not sure if I am interpreting the convergence correctly, but if I am, the first one is false and the second one is true. These should hopefully be helpful tips:
1) For the first one you should construct a counterexample; I think a deterministic one would suffice; you should make sure $f_n$'s converge uniformly to 0 on arbitrary compact sets. This may sound complicated, but there is a simple family of functions that achieve this. Now, simply pick your $t_n$s to be so that $f_n(t_n)$ is not equal to 0 for any $n$.
2) For this one, we need to look at probabilities of sets $A = {X_m(t_n) > y}, B = {X(t_n) > y}$ and $C = {X(t_0) > y}$. Now, fix $n$, and now you can pick $m$ large enough so that sets A and B have similar probability. Now, pick $n$ large enough so that $t_0$ are $t_n$ are sufficiently close, and since $X$ is a.s. continuous, $B$ and $C$ should be comparable as well.
answered Jan 11 at 15:26
E-AE-A
2,1121414
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I am not sure if I am interpreting the convergence correctly, but if I am, the first one is false and the second one is true. These should hopefully be helpful tips:
1) For the first one you should construct a counterexample; I think a deterministic one would suffice; you should make sure $f_n$'s converge uniformly to 0 on arbitrary compact sets. This may sound complicated, but there is a simple family of functions that achieve this. Now, simply pick your $t_n$s to be so that $f_n(t_n)$ is not equal to 0 for any $n$.
2) For this one, we need to look at probabilities of sets $A = {X_m(t_n) > y}, B = {X(t_n) > y}$ and $C = {X(t_0) > y}$. Now, fix $n$, and now you can pick $m$ large enough so that sets A and B have similar probability. Now, pick $n$ large enough so that $t_0$ are $t_n$ are sufficiently close, and since $X$ is a.s. continuous, $B$ and $C$ should be comparable as well.
answered Jan 11 at 15:26
E-AE-A
2,1121414
$endgroup$
add a comment |
$begingroup$
I am not sure if I am interpreting the convergence correctly, but if I am, the first one is false and the second one is true. These should hopefully be helpful tips:
1) For the first one you should construct a counterexample; I think a deterministic one would suffice; you should make sure $f_n$'s converge uniformly to 0 on arbitrary compact sets. This may sound complicated, but there is a simple family of functions that achieve this. Now, simply pick your $t_n$s to be so that $f_n(t_n)$ is not equal to 0 for any $n$.
2) For this one, we need to look at probabilities of sets $A = {X_m(t_n) > y}, B = {X(t_n) > y}$ and $C = {X(t_0) > y}$. Now, fix $n$, and now you can pick $m$ large enough so that sets A and B have similar probability. Now, pick $n$ large enough so that $t_0$ are $t_n$ are sufficiently close, and since $X$ is a.s. continuous, $B$ and $C$ should be comparable as well.
answered Jan 11 at 15:26
E-AE-A
2,1121414
$endgroup$
add a comment |
$begingroup$
I am not sure if I am interpreting the convergence correctly, but if I am, the first one is false and the second one is true. These should hopefully be helpful tips:
1) For the first one you should construct a counterexample; I think a deterministic one would suffice; you should make sure $f_n$'s converge uniformly to 0 on arbitrary compact sets. This may sound complicated, but there is a simple family of functions that achieve this. Now, simply pick your $t_n$s to be so that $f_n(t_n)$ is not equal to 0 for any $n$.
2) For this one, we need to look at probabilities of sets $A = {X_m(t_n) > y}, B = {X(t_n) > y}$ and $C = {X(t_0) > y}$. Now, fix $n$, and now you can pick $m$ large enough so that sets A and B have similar probability. Now, pick $n$ large enough so that $t_0$ are $t_n$ are sufficiently close, and since $X$ is a.s. continuous, $B$ and $C$ should be comparable as well.
answered Jan 11 at 15:26
E-AE-A
2,1121414
$endgroup$
I am not sure if I am interpreting the convergence correctly, but if I am, the first one is false and the second one is true. These should hopefully be helpful tips:
1) For the first one you should construct a counterexample; I think a deterministic one would suffice; you should make sure $f_n$'s converge uniformly to 0 on arbitrary compact sets. This may sound complicated, but there is a simple family of functions that achieve this. Now, simply pick your $t_n$s to be so that $f_n(t_n)$ is not equal to 0 for any $n$.
2) For this one, we need to look at probabilities of sets $A = {X_m(t_n) > y}, B = {X(t_n) > y}$ and $C = {X(t_0) > y}$. Now, fix $n$, and now you can pick $m$ large enough so that sets A and B have similar probability. Now, pick $n$ large enough so that $t_0$ are $t_n$ are sufficiently close, and since $X$ is a.s. continuous, $B$ and $C$ should be comparable as well.
answered Jan 11 at 15:26
E-AE-A
2,1121414
answered Jan 11 at 15:26
E-AE-A
2,1121414
answered Jan 11 at 15:26
E-AE-A
2,1121414
answered Jan 11 at 15:26
E-AE-A
2,1121414
2,1121414
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